Straight Lines Class 11 Mathematics NCERT Solutions

Q1EXERCISE 9.1

Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4, 5), (0, 7), (5, -5)$ and $(-4, -2)$ . Also, find its area.

Solution

Given: The vertices of the quadrilateral are A

(-4, 5)

, B

(0, 7)

, C

(5, -5)

, and D

(-4, -2)

.

To Find: The area of the quadrilateral ABCD.

Solution: We can find the area of the quadrilateral by dividing it into two triangles, say

\triangle

ABC and

\triangle

ACD.

The formula for the area of a triangle with vertices

(x_1, y_1)

,

(x_2, y_2)

, and

(x_3, y_3)

is:

\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

Area of $\triangle$ ABC: Vertices are A

(-4, 5)

, B

(0, 7)

, C

(5, -5)

.

\text{Area}(\triangle ABC) = \frac{1}{2} |(-4)(7 - (-5)) + 0(-5 - 5) + 5(5 - 7)|

= \frac{1}{2} |(-4)(12) + 0(-10) + 5(-2)|

= \frac{1}{2} |-48 + 0 - 10|

= \frac{1}{2} |-58| = 29 \text{ square units}

Area of $\triangle$ ACD: Vertices are A

(-4, 5)

, C

(5, -5)

, D

(-4, -2)

.

\text{Area}(\triangle ACD) = \frac{1}{2} |(-4)(-5 - (-2)) + 5(-2 - 5) + (-4)(5 - (-5))|

= \frac{1}{2} |(-4)(-3) + 5(-7) + (-4)(10)|

= \frac{1}{2} |12 - 35 - 40|

= \frac{1}{2} |-63| = 31.5 \text{ square units}

Area of Quadrilateral ABCD:

\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD)

= 29 + 31.5 = 60.5 \text{ square units}

Final Answer: The area of the quadrilateral is 60.5 square units.

Q2EXERCISE 9.1

The base of an equilateral triangle with side $2a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Solution

Given: An equilateral triangle with side length

2a

. The base lies along the y-axis. The mid-point of the base is at the origin (0, 0).

To Find: The vertices of the triangle.

Solution: Let the equilateral triangle be ABC, with base BC on the y-axis. Since the midpoint of the base BC is the origin, the coordinates of B and C are

(0, a)

and

(0, -a)

respectively. The length of the base BC is

\sqrt{(0-0)^2 + (a - (-a))^2} = \sqrt{(2a)^2} = 2a

.

Let the third vertex be A

(x, 0)

. Since the triangle is equilateral, the side length AB must be

2a

. Using the distance formula for AB:

AB^2 = (x - 0)^2 + (0 - a)^2 = (2a)^2

x^2 + a^2 = 4a^2

x^2 = 3a^2

x = \pm \sqrt{3}a

So, the coordinates of the third vertex A can be

(\sqrt{3}a, 0)

or

(-\sqrt{3}a, 0)

.

Thus, there are two possible sets of vertices for the triangle:

$(0, a), (0, -a), (\sqrt{3}a, 0)$
$(0, a), (0, -a), (-\sqrt{3}a, 0)$

Final Answer: The vertices of the triangle are

(0, a), (0, -a), (\sqrt{3}a, 0)

or

(0, a), (0, -a), (-\sqrt{3}a, 0)

.

Q3EXERCISE 9.1

Find the distance between $\mathrm{P}(x_{1}, y_{1})$ and $\mathrm{Q}(x_{2}, y_{2})$ when : (i) PQ is parallel to the $y$ -axis, (ii) PQ is parallel to the $x$ -axis.

Solution

Given: Two points P

(x_1, y_1)

and Q

(x_2, y_2)

. The distance formula is

PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

.

To Find: The distance PQ under two conditions.

Solution: (i) PQ is parallel to the y-axis: If a line segment is parallel to the y-axis, the x-coordinates of its endpoints are equal. Therefore,

x_1 = x_2

. Substituting this into the distance formula:

PQ = \sqrt{(x_1 - x_1)^2 + (y_2 - y_1)^2}

PQ = \sqrt{0^2 + (y_2 - y_1)^2}

PQ = \sqrt{(y_2 - y_1)^2}

PQ = |y_2 - y_1|

(ii) PQ is parallel to the x-axis: If a line segment is parallel to the x-axis, the y-coordinates of its endpoints are equal. Therefore,

y_1 = y_2

. Substituting this into the distance formula:

PQ = \sqrt{(x_2 - x_1)^2 + (y_1 - y_1)^2}

PQ = \sqrt{(x_2 - x_1)^2 + 0^2}

PQ = \sqrt{(x_2 - x_1)^2}

PQ = |x_2 - x_1|

Final Answer:

(i)

When PQ is parallel to the y-axis, the distance is

|y_2 - y_1|

.

(ii)

When PQ is parallel to the x-axis, the distance is

|x_2 - x_1|

.

Q4EXERCISE 9.1

Find a point on the $x$ -axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$ .

Solution

Given: Two points A

(7, 6)

and B

(3, 4)

.

To Find: A point on the x-axis that is equidistant from A and B.

Solution: Let the point on the x-axis be P

(x, 0)

. The condition is that P is equidistant from A and B, which means PA = PB. This implies

PA^2 = PB^2

.

Using the distance formula,

d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

.

For

PA^2

:

PA^2 = (x - 7)^2 + (0 - 6)^2

PA^2 = (x^2 - 14x + 49) + 36

PA^2 = x^2 - 14x + 85

For

PB^2

:

PB^2 = (x - 3)^2 + (0 - 4)^2

PB^2 = (x^2 - 6x + 9) + 16

PB^2 = x^2 - 6x + 25

Now, setting

PA^2 = PB^2

:

x^2 - 14x + 85 = x^2 - 6x + 25

-14x + 85 = -6x + 25

85 - 25 = -6x + 14x

60 = 8x

x = \frac{60}{8} = \frac{15}{2}

The required point on the x-axis is

(\frac{15}{2}, 0)

.

Final Answer: The point on the x-axis is

(\frac{15}{2}, 0)

.

Q5EXERCISE 9.1

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $\mathrm{P}(0,-4)$ and $\mathrm{B}(8,0)$ .

Solution

Given: Points P

(0, -4)

and B

(8, 0)

. The line passes through the origin O

(0, 0)

and the midpoint of the segment PB.

To Find: The slope of the line.

Solution: First, we find the coordinates of the midpoint M of the line segment joining P and B. Midpoint formula:

M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

M = (\frac{0 + 8}{2}, \frac{-4 + 0}{2})

M = (\frac{8}{2}, \frac{-4}{2})

M = (4, -2)

Now, we need to find the slope of the line passing through the origin O

(0, 0)

and the midpoint M

(4, -2)

. Slope formula:

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{-2 - 0}{4 - 0}

m = \frac{-2}{4} = -\frac{1}{2}

Final Answer: The slope of the line is

-\frac{1}{2}

.

Q6EXERCISE 9.1

Without using the Pythagoras theorem, show that the points $(4, 4), (3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

Solution

Given: The points A

(4, 4)

, B

(3, 5)

, and C

(-1, -1)

.

To Show: The points are the vertices of a right-angled triangle, without using the Pythagoras theorem.

Method: We will use the concept of slopes. If two lines are perpendicular, the product of their slopes is -1. A triangle with two perpendicular sides is a right-angled triangle.

Solution: Let's find the slopes of the sides AB, BC, and AC. Slope formula:

m = \frac{y_2 - y_1}{x_2 - x_1}

Slope of AB (

m_{AB}

):

m_{AB} = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1

Slope of BC (

m_{BC}

):

m_{BC} = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}

Slope of AC (

m_{AC}

):

m_{AC} = \frac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1

Now, let's check the product of the slopes:

m_{AB} \times m_{BC} = (-1) \times (\frac{3}{2}) = -\frac{3}{2} \neq -1

m_{BC} \times m_{AC} = (\frac{3}{2}) \times (1) = \frac{3}{2} \neq -1

m_{AB} \times m_{AC} = (-1) \times (1) = -1

Since the product of the slopes of sides AB and AC is -1, the side AB is perpendicular to the side AC. Therefore, the triangle formed by the points A, B, and C is a right-angled triangle with the right angle at vertex A.

Hence Shown.

Q7EXERCISE 9.1

Find the slope of the line, which makes an angle of $30^{\circ}$ with the positive direction of $y$ -axis measured anticlockwise.

Solution

Given: A line makes an angle of

30^{\circ}

with the positive direction of the y-axis, measured anticlockwise.

To Find: The slope of the line.

Solution: The slope of a line is given by

m = \tan \theta

, where

\theta

is the inclination of the line, which is the angle it makes with the positive direction of the x-axis, measured anticlockwise.

The positive direction of the y-axis makes an angle of

90^{\circ}

with the positive direction of the x-axis. The given line makes a further angle of

30^{\circ}

with the positive y-axis in the anticlockwise direction.

Therefore, the total angle

\theta

that the line makes with the positive x-axis is:

\theta = 90^{\circ} + 30^{\circ} = 120^{\circ}

Now, we can find the slope

m

:

m = \tan(120^{\circ})

m = \tan(180^{\circ} - 60^{\circ})

m = -\tan(60^{\circ})

m = -\sqrt{3}

Final Answer: The slope of the line is

-\sqrt{3}

.

Q8EXERCISE 9.1

Without using distance formula, show that points $(-2, -1), (4, 0), (3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

Solution

Given: The points A

(-2, -1)

, B

(4, 0)

, C

(3, 3)

, and D

(-3, 2)

.

To Show: The points are the vertices of a parallelogram, without using the distance formula.

Method: A quadrilateral is a parallelogram if its opposite sides are parallel. Two lines are parallel if their slopes are equal.

Solution: We will calculate the slopes of all four sides. Slope formula:

m = \frac{y_2 - y_1}{x_2 - x_1}

Slope of AB (

m_{AB}

):

m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}

Slope of BC (

m_{BC}

):

m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3

Slope of CD (

m_{CD}

):

m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}

Slope of DA (

m_{DA}

):

m_{DA} = \frac{-1 - 2}{-2 - (-3)} = \frac{-3}{1} = -3

Comparing the slopes:

Slope of AB ( $m_{AB}$ ) = Slope of CD ( $m_{CD}$ ) = $\frac{1}{6}$ . This means side AB is parallel to side CD.
Slope of BC ( $m_{BC}$ ) = Slope of DA ( $m_{DA}$ ) = $-3$ . This means side BC is parallel to side DA.

Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

Hence Shown.

Q9EXERCISE 9.1

Find the angle between the $x$ -axis and the line joining the points $(3, -1)$ and $(4, -2)$ .

Solution

Given: Two points on a line: P

(3, -1)

and Q

(4, -2)

.

To Find: The angle between the x-axis and the line PQ.

Solution: The angle between the x-axis and a line is the inclination

\theta

of the line. The slope of the line,

m

, is equal to

\tan \theta

.

First, let's find the slope of the line joining P and Q. Slope formula:

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{-2 - (-1)}{4 - 3} = \frac{-1}{1} = -1

Now, we have

\tan \theta = m = -1

. Since the slope is negative, the angle of inclination

\theta

is obtuse (

90^{\circ} < \theta < 180^{\circ}

). We know that

\tan(180^{\circ} - 45^{\circ}) = -\tan(45^{\circ}) = -1

. Therefore,

\theta = 180^{\circ} - 45^{\circ} = 135^{\circ}

.

The angle can also be expressed in radians:

135^{\circ} = \frac{3\pi}{4}

radians.

Final Answer: The angle between the x-axis and the line is

135^{\circ}

.

Q10EXERCISE 9.1

The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$ , find the slopes of the lines.

Solution

Given: Let the slopes of the two lines be

m_1

and

m_2

.

m_1 = 2m_2

. The tangent of the angle

\theta

between the lines is

\tan \theta = \frac{1}{3}

.

To Find: The slopes of the lines,

m_1

and

m_2

.

Formula: The tangent of the angle

\theta

between two lines with slopes

m_1

and

m_2

is given by:

\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

Solution: Substitute the given values into the formula:

\frac{1}{3} = \left| \frac{2m_2 - m_2}{1 + (2m_2)m_2} \right|

\frac{1}{3} = \left| \frac{m_2}{1 + 2m_2^2} \right|

This gives two possible cases:

Case 1:

\frac{m_2}{1 + 2m_2^2} = \frac{1}{3}

3m_2 = 1 + 2m_2^2

2m_2^2 - 3m_2 + 1 = 0

Factorizing the quadratic equation:

2m_2^2 - 2m_2 - m_2 + 1 = 0

2m_2(m_2 - 1) - 1(m_2 - 1) = 0

(2m_2 - 1)(m_2 - 1) = 0

So,

m_2 = \frac{1}{2}

or

m_2 = 1

. If

m_2 = \frac{1}{2}

, then

m_1 = 2m_2 = 1

. The slopes are

\frac{1}{2}

and

1

. If

m_2 = 1

, then

m_1 = 2m_2 = 2

. The slopes are

1

and

2

.

Case 2:

\frac{m_2}{1 + 2m_2^2} = -\frac{1}{3}

3m_2 = -(1 + 2m_2^2)

3m_2 = -1 - 2m_2^2

2m_2^2 + 3m_2 + 1 = 0

Factorizing the quadratic equation:

2m_2^2 + 2m_2 + m_2 + 1 = 0

2m_2(m_2 + 1) + 1(m_2 + 1) = 0

(2m_2 + 1)(m_2 + 1) = 0

So,

m_2 = -\frac{1}{2}

or

m_2 = -1

. If

m_2 = -\frac{1}{2}

, then

m_1 = 2m_2 = -1

. The slopes are

-\frac{1}{2}

and

-1

. If

m_2 = -1

, then

m_1 = 2m_2 = -2

. The slopes are

-1

and

-2

.

Final Answer: The possible pairs of slopes are

(\frac{1}{2}, 1)

,

(1, 2)

,

(-\frac{1}{2}, -1)

, and

(-1, -2)

.

Q11EXERCISE 9.1

A line passes through $(x_1, y_1)$ and $(h, k)$ . If slope of the line is $m$ , show that $k-y_1=m(h-x_1)$ .

Solution

Given: A line passes through the points P

(x_1, y_1)

and Q

(h, k)

. The slope of the line is

m

.

To Show:

k - y_1 = m(h - x_1)

.

Proof: By the definition of the slope of a line passing through two points, the slope

m

is given by the formula:

m = \frac{\text{change in y}}{\text{change in x}}

m = \frac{k - y_1}{h - x_1}

To derive the required relationship, we can multiply both sides of the equation by

(h - x_1)

:

m(h - x_1) = \frac{k - y_1}{h - x_1} \times (h - x_1)

m(h - x_1) = k - y_1

Rearranging the terms, we get:

k - y_1 = m(h - x_1)

This is the point-slope form of the equation of a line.

Hence Shown.

Q1EXERCISE 9.2

Write the equations for the $x$ -and $y$ -axes.

Solution

To Find: The equations of the x-axis and the y-axis.

Solution:

Equation of the x-axis: For any point on the x-axis, its y-coordinate is always 0. For example, points like (1, 0), (-5, 0), (0, 0) are on the x-axis. The condition that defines the x-axis is that the ordinate (y-value) is zero. Therefore, the equation of the x-axis is

y = 0

.

Equation of the y-axis: For any point on the y-axis, its x-coordinate is always 0. For example, points like (0, 2), (0, -3), (0, 0) are on the y-axis. The condition that defines the y-axis is that the abscissa (x-value) is zero. Therefore, the equation of the y-axis is

x = 0

.

Final Answer: The equation of the x-axis is

y = 0

. The equation of the y-axis is

x = 0

.

Q2EXERCISE 9.2

Passing through the point $(-4, 3)$ with slope $\frac{1}{2}$ .

Solution

Given: A point on the line

(x_0, y_0) = (-4, 3)

. The slope of the line

m = \frac{1}{2}

.

To Find: The equation of the line.

Formula: The point-slope form of the equation of a line is:

y - y_0 = m(x - x_0)

Solution: Substitute the given values into the formula:

y - 3 = \frac{1}{2}(x - (-4))

y - 3 = \frac{1}{2}(x + 4)

To simplify, multiply both sides by 2:

2(y - 3) = x + 4

2y - 6 = x + 4

Rearranging the terms to the general form

Ax + By + C = 0

:

x - 2y + 4 + 6 = 0

x - 2y + 10 = 0

Final Answer: The equation of the line is

x - 2y + 10 = 0

.

Q3EXERCISE 9.2

Passing through $(0, 0)$ with slope $m$ .

Solution

Given: A point on the line

(x_0, y_0) = (0, 0)

. The slope of the line is

m

.

To Find: The equation of the line.

Formula: The point-slope form of the equation of a line is:

y - y_0 = m(x - x_0)

Solution: Substitute the given values into the formula:

y - 0 = m(x - 0)

y = mx

This is the standard equation for a line passing through the origin.

Final Answer: The equation of the line is

y = mx

.

Q4EXERCISE 9.2

Passing through $(2, 2\sqrt{3})$ and inclined with the $x$ -axis at an angle of $75^{\circ}$ .

Solution

Given: A point on the line

(x_0, y_0) = (2, 2\sqrt{3})

. The angle of inclination with the x-axis is

\theta = 75^{\circ}

.

To Find: The equation of the line.

Solution: First, we need to find the slope

m

of the line.

m = \tan \theta = \tan(75^{\circ})

We can write

75^{\circ}

as

45^{\circ} + 30^{\circ}

. Using the tangent addition formula,

\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

:

m = \tan(45^{\circ} + 30^{\circ}) = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}}

m = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

To rationalize the denominator, multiply the numerator and denominator by

(\sqrt{3} + 1)

:

m = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3})^2 + 2\sqrt{3} + 1^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}

Now, use the point-slope form

y - y_0 = m(x - x_0)

:

y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)

y - 2\sqrt{3} = (2 + \sqrt{3})x - 2(2 + \sqrt{3})

y - 2\sqrt{3} = (2 + \sqrt{3})x - 4 - 2\sqrt{3}

y = (2 + \sqrt{3})x - 4

Rearranging to the general form:

(2 + \sqrt{3})x - y - 4 = 0

Final Answer: The equation of the line is

(2 + \sqrt{3})x - y - 4 = 0

.

Q5EXERCISE 9.2

Intersecting the $x$ -axis at a distance of 3 units to the left of origin with slope -2.

Solution

Given: The slope of the line

m = -2

. The line intersects the x-axis at a distance of 3 units to the left of the origin.

To Find: The equation of the line.

Solution: The point where the line intersects the x-axis is its x-intercept. A distance of 3 units to the left of the origin means the x-coordinate is -3. So, the line passes through the point

(x_0, y_0) = (-3, 0)

.

Using the point-slope form of the equation of a line:

y - y_0 = m(x - x_0)

y - 0 = -2(x - (-3))

y = -2(x + 3)

y = -2x - 6

Rearranging to the general form:

2x + y + 6 = 0

Final Answer: The equation of the line is

2x + y + 6 = 0

.

Q6EXERCISE 9.2

Intersecting the $y$ -axis at a distance of 2 units above the origin and making an angle of $30^{\circ}$ with positive direction of the $x$ -axis.

Solution

Given: The line intersects the y-axis at a distance of 2 units above the origin. This means the y-intercept is

c = 2

. The angle of inclination with the positive x-axis is

\theta = 30^{\circ}

.

To Find: The equation of the line.

Solution: First, find the slope

m

of the line:

m = \tan \theta = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}

Now, use the slope-intercept form of the equation of a line:

y = mx + c

y = \frac{1}{\sqrt{3}}x + 2

To eliminate the fraction, multiply the entire equation by

\sqrt{3}

:

\sqrt{3}y = x + 2\sqrt{3}

Rearranging to the general form:

x - \sqrt{3}y + 2\sqrt{3} = 0

Final Answer: The equation of the line is

x - \sqrt{3}y + 2\sqrt{3} = 0

.

Q7EXERCISE 9.2

Passing through the points $(-1, 1)$ and $(2, -4)$ .

Solution

Given: Two points on the line:

(x_1, y_1) = (-1, 1)

and

(x_2, y_2) = (2, -4)

.

To Find: The equation of the line.

Formula: The two-point form of the equation of a line is:

y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

Solution: First, let's calculate the slope:

m = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}

Now, substitute the slope and one of the points (e.g.,

(-1, 1)

) into the formula:

y - 1 = -\frac{5}{3}(x - (-1))

y - 1 = -\frac{5}{3}(x + 1)

Multiply both sides by 3 to clear the fraction:

3(y - 1) = -5(x + 1)

3y - 3 = -5x - 5

Rearranging to the general form:

5x + 3y - 3 + 5 = 0

5x + 3y + 2 = 0

Final Answer: The equation of the line is

5x + 3y + 2 = 0

.

Q8EXERCISE 9.2

The vertices of $\triangle \mathrm{PQR}$ are $\mathrm{P}(2, 1), \mathrm{Q}(-2, 3)$ and $\mathrm{R}(4, 5)$ . Find equation of the median through the vertex R.

Solution

Given: The vertices of

\triangle

PQR are P

(2, 1)

, Q

(-2, 3)

, and R

(4, 5)

.

To Find: The equation of the median through the vertex R.

Solution: The median through vertex R is the line segment that connects R to the midpoint of the opposite side, PQ. Let S be the midpoint of PQ.

First, find the coordinates of the midpoint S of PQ. Midpoint formula:

S = (\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2})

S = (\frac{2 + (-2)}{2}, \frac{1 + 3}{2})

S = (\frac{0}{2}, \frac{4}{2})

S = (0, 2)

Now, we need to find the equation of the line passing through the points R

(4, 5)

and S

(0, 2)

. We use the two-point form of the equation of a line:

y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

Let

(x_1, y_1) = (0, 2)

and

(x_2, y_2) = (4, 5)

. The slope is

m = \frac{5 - 2}{4 - 0} = \frac{3}{4}

.

Using the point-slope form with point S

(0, 2)

(which is also the y-intercept):

y - 2 = \frac{3}{4}(x - 0)

y - 2 = \frac{3}{4}x

Multiply by 4 to clear the fraction:

4(y - 2) = 3x

4y - 8 = 3x

Rearranging to the general form:

3x - 4y + 8 = 0

Final Answer: The equation of the median through the vertex R is

3x - 4y + 8 = 0

.

Q9EXERCISE 9.2

Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points $(2, 5)$ and $(-3, 6)$ .

Solution

Given: The required line passes through the point P

(-3, 5)

. It is perpendicular to the line passing through A

(2, 5)

and B

(-3, 6)

.

To Find: The equation of the required line.

Solution: First, find the slope of the line passing through points A and B. Let's call this slope

m_1

.

m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}

Let the slope of the required line be

m_2

. Since the required line is perpendicular to the line AB, the product of their slopes must be -1.

m_1 \times m_2 = -1

-\frac{1}{5} \times m_2 = -1

m_2 = 5

Now we have the slope of the required line (

m_2 = 5

) and a point it passes through P

(-3, 5)

. We can use the point-slope form of the equation of a line:

y - y_0 = m_2(x - x_0)

y - 5 = 5(x - (-3))

y - 5 = 5(x + 3)

y - 5 = 5x + 15

Rearranging to the general form:

5x - y + 15 + 5 = 0

5x - y + 20 = 0

Final Answer: The equation of the line is

5x - y + 20 = 0

.

Q10EXERCISE 9.2

A line perpendicular to the line segment joining the points $(1, 0)$ and $(2, 3)$ divides it in the ratio $1:n$ . Find the equation of the line.

Solution

Given: Points A

(1, 0)

and B

(2, 3)

. A line is perpendicular to the segment AB. This line divides the segment AB in the ratio

1:n

.

To Find: The equation of the perpendicular line.

Solution: Step 1: Find the slope of the perpendicular line. First, find the slope of the segment AB (

m_{AB}

).

m_{AB} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3

Let the slope of the required perpendicular line be

m_{perp}

.

m_{AB} \times m_{perp} = -1

3 \times m_{perp} = -1

m_{perp} = -\frac{1}{3}

Step 2: Find the point of intersection. Let the point P that divides the segment AB in the ratio

1:n

be

(x, y)

. Using the section formula:

x = \frac{1(2) + n(1)}{1 + n} = \frac{2 + n}{1 + n}

y = \frac{1(3) + n(0)}{1 + n} = \frac{3}{1 + n}

So, the point of intersection is P

(\frac{2 + n}{1 + n}, \frac{3}{1 + n})

.

Step 3: Find the equation of the line. Now we have the slope

m_{perp} = -\frac{1}{3}

and a point P

(\frac{2 + n}{1 + n}, \frac{3}{1 + n})

that the line passes through. Using the point-slope form

y - y_0 = m(x - x_0)

:

y - \frac{3}{1 + n} = -\frac{1}{3} \left( x - \frac{2 + n}{1 + n} \right)

Multiply both sides by

3(1+n)

to clear the denominators:

3(1 + n) \left( y - \frac{3}{1 + n} \right) = - (1 + n) \left( x - \frac{2 + n}{1 + n} \right)

3(1 + n)y - 3(3) = - (1 + n)x + (2 + n)

3(1 + n)y - 9 = - (1 + n)x + 2 + n

Rearranging to the general form:

(1 + n)x + 3(1 + n)y - 9 - 2 - n = 0

(1 + n)x + 3(1 + n)y - (11 + n) = 0

Final Answer: The equation of the line is

(1 + n)x + 3(1 + n)y - (11 + n) = 0

.

Q11EXERCISE 9.2

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2, 3)$ .

Solution

Given: The line cuts off equal intercepts on the coordinate axes. The line passes through the point

(2, 3)

.

To Find: The equation of the line.

Solution: Let the x-intercept be 'a' and the y-intercept be 'b'. The condition of equal intercepts means

a = b

(where

a, b \neq 0

).

The intercept form of the equation of a line is:

\frac{x}{a} + \frac{y}{b} = 1

Since

a=b

, the equation becomes:

\frac{x}{a} + \frac{y}{a} = 1

x + y = a

We are given that this line passes through the point

(2, 3)

. Substituting these values for x and y into the equation:

2 + 3 = a

a = 5

So, the equation of the line is:

x + y = 5

Note: There is another possibility that the intercepts are equal in magnitude but opposite in sign, i.e.,

b = -a

. In this case, the equation would be

\frac{x}{a} + \frac{y}{-a} = 1

, which simplifies to

x - y = a

. Since it passes through

(2, 3)

, we would have

2 - 3 = a

, so

a = -1

. The equation would be

x - y = -1

. However, the phrase "equal intercepts" usually implies they are identical (

a=b

).

Let's consider the primary interpretation.

Final Answer: The equation of the line is

x + y = 5

.

Q12EXERCISE 9.2

Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is 9.

Solution

Given: The line passes through the point

(2, 2)

. The sum of the intercepts on the axes is 9.

To Find: The equation of the line.

Solution: Let the x-intercept be 'a' and the y-intercept be 'b'. We are given that

a + b = 9

, which means

b = 9 - a

.

The intercept form of the equation of a line is:

\frac{x}{a} + \frac{y}{b} = 1

Substitute

b = 9 - a

into the equation:

\frac{x}{a} + \frac{y}{9 - a} = 1

We are given that the line passes through the point

(2, 2)

. Substituting

x = 2

and

y = 2

into this equation:

\frac{2}{a} + \frac{2}{9 - a} = 1

To solve for 'a', multiply the entire equation by

a(9 - a)

:

2(9 - a) + 2a = a(9 - a)

18 - 2a + 2a = 9a - a^2

18 = 9a - a^2

a^2 - 9a + 18 = 0

Factorize the quadratic equation:

(a - 3)(a - 6) = 0

This gives two possible values for 'a':

a = 3

or

a = 6

.

Case 1:

a = 3

If

a = 3

, then

b = 9 - a = 9 - 3 = 6

. The equation is

\frac{x}{3} + \frac{y}{6} = 1

. Multiplying by 6 gives

2x + y = 6

, or

2x + y - 6 = 0

.

Case 2:

a = 6

If

a = 6

, then

b = 9 - a = 9 - 6 = 3

. The equation is

\frac{x}{6} + \frac{y}{3} = 1

. Multiplying by 6 gives

x + 2y = 6

, or

x + 2y - 6 = 0

.

There are two such lines.

Final Answer: The equations of the lines are

2x + y - 6 = 0

and

x + 2y - 6 = 0

.

Q13EXERCISE 9.2

Find equation of the line through the point $(0, 2)$ making an angle $\frac{2 \pi}{3}$ with the positive $x$ -axis. Also, find the equation of line parallel to it and crossing the $y$ -axis at a distance of 2 units below the origin.

Solution

Part 1: Find the equation of the first line.

Given: A point on the line

(x_0, y_0) = (0, 2)

. The angle of inclination with the positive x-axis is

\theta = \frac{2\pi}{3} = 120^{\circ}

.

Solution: First, find the slope

m

of the line:

m = \tan(\frac{2\pi}{3}) = \tan(120^{\circ}) = -\sqrt{3}

The point

(0, 2)

is the y-intercept, so

c = 2

. Using the slope-intercept form

y = mx + c

:

y = -\sqrt{3}x + 2

Rearranging to the general form:

\sqrt{3}x + y - 2 = 0

Part 2: Find the equation of the parallel line.

Given: The second line is parallel to the first line. It crosses the y-axis at a distance of 2 units below the origin.

Solution: Since the second line is parallel to the first, it has the same slope:

m = -\sqrt{3}

Crossing the y-axis 2 units below the origin means the y-intercept is

c = -2

. Using the slope-intercept form

y = mx + c

:

y = -\sqrt{3}x - 2

Rearranging to the general form:

\sqrt{3}x + y + 2 = 0

Final Answer: The equation of the first line is

\sqrt{3}x + y - 2 = 0

. The equation of the parallel line is

\sqrt{3}x + y + 2 = 0

.

Q14EXERCISE 9.2

The perpendicular from the origin to a line meets it at the point $(-2, 9)$ , find the equation of the line.

Solution

Given: The perpendicular from the origin O

(0, 0)

to a line meets the line at the point P

(-2, 9)

.

To Find: The equation of the line.

Solution: The required line passes through the point P

(-2, 9)

. Also, the required line is perpendicular to the line segment OP, which connects the origin to the point P.

First, let's find the slope of the line segment OP (

m_{OP}

).

m_{OP} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}

Let the slope of the required line be

m_{line}

. Since the line is perpendicular to OP, the product of their slopes is -1.

m_{OP} \times m_{line} = -1

-\frac{9}{2} \times m_{line} = -1

m_{line} = \frac{2}{9}

Now we have the slope of the required line (

m_{line} = \frac{2}{9}

) and a point it passes through P

(-2, 9)

. We can use the point-slope form of the equation of a line:

y - y_0 = m(x - x_0)

y - 9 = \frac{2}{9}(x - (-2))

y - 9 = \frac{2}{9}(x + 2)

Multiply by 9 to clear the fraction:

9(y - 9) = 2(x + 2)

9y - 81 = 2x + 4

Rearranging to the general form:

2x - 9y + 4 + 81 = 0

2x - 9y + 85 = 0

Final Answer: The equation of the line is

2x - 9y + 85 = 0

.

Q15EXERCISE 9.2

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if $\mathrm{L}=124.942$ when $\mathrm{C}=20$ and $\mathrm{L}=125.134$ when $\mathrm{C}=110$ , express L in terms of C.

Solution

Given: Length L is a linear function of Temperature C. We have two data points: Point 1:

(C_1, L_1) = (20, 124.942)

Point 2:

(C_2, L_2) = (110, 125.134)

To Find: An equation that expresses L in terms of C.

Solution: Since the relationship is linear, we can model it with the equation of a straight line. We can use the two-point form, treating C as the x-variable and L as the y-variable.

Two-point form:

L - L_1 = \frac{L_2 - L_1}{C_2 - C_1}(C - C_1)

First, calculate the slope (the rate of change of length with respect to temperature):

m = \frac{L_2 - L_1}{C_2 - C_1} = \frac{125.134 - 124.942}{110 - 20} = \frac{0.192}{90}

Now, substitute the slope and Point 1 into the point-slope form:

L - 124.942 = \frac{0.192}{90}(C - 20)

This is a valid expression for L in terms of C. We can simplify it further.

L = \frac{0.192}{90}(C - 20) + 124.942

Let's expand and combine the constants:

L = \frac{0.192}{90}C - \frac{0.192 \times 20}{90} + 124.942

L = \frac{0.192}{90}C - \frac{3.84}{90} + 124.942

L = 0.002133...C - 0.04266... + 124.942

L = \frac{0.192}{90}C + 124.89933...

Let's keep the equation in the initial simplified form for precision.

Final Answer: The relationship between L and C is given by the equation

L = \frac{0.192}{90}(C - 20) + 124.942

.

Q16EXERCISE 9.2

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Solution

Given: A linear relationship between selling price (P) and demand (D). Point 1:

(P_1, D_1) = (14, 980)

(Price in Rs/litre, Demand in litres) Point 2:

(P_2, D_2) = (16, 1220)

To Find: The demand (D) when the selling price (P) is Rs 17/litre.

Solution: Step 1: Find the linear equation relating D and P. We can use the two-point form of a line, with P as the x-variable and D as the y-variable.

D - D_1 = \frac{D_2 - D_1}{P_2 - P_1}(P - P_1)

First, calculate the slope (change in demand per unit change in price):

m = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120 \text{ litres/Rs}

Now, use the point-slope form with Point 1:

D - 980 = 120(P - 14)

D = 120P - 120 \times 14 + 980

D = 120P - 1680 + 980

D = 120P - 700

This is the linear relationship between demand and price.

Step 2: Calculate the demand for a price of Rs 17/litre. Substitute

P = 17

into the equation:

D = 120(17) - 700

D = 2040 - 700

D = 1340

Final Answer: The owner could sell 1340 litres of milk weekly at Rs 17/litre.

Q17EXERCISE 9.2

$\mathrm{P}(a, b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a}+\frac{y}{b}=2$ .

Solution

Given: A line segment lies between the coordinate axes. The midpoint of this segment is P

(a, b)

.

To Show: The equation of the line is

\frac{x}{a} + \frac{y}{b} = 2

.

Proof: Let the line intersect the x-axis at point A and the y-axis at point B. The coordinates of A will be

(x_0, 0)

and the coordinates of B will be

(0, y_0)

. Here,

x_0

is the x-intercept and

y_0

is the y-intercept.

The line segment is AB. We are given that P

(a, b)

is the midpoint of AB. Using the midpoint formula:

P(a, b) = (\frac{x_0 + 0}{2}, \frac{0 + y_0}{2})

P(a, b) = (\frac{x_0}{2}, \frac{y_0}{2})

Equating the coordinates:

a = \frac{x_0}{2} \implies x_0 = 2a

b = \frac{y_0}{2} \implies y_0 = 2b

So, the x-intercept of the line is

2a

and the y-intercept is

2b

.

The intercept form of the equation of a line is:

\frac{x}{\text{x-intercept}} + \frac{y}{\text{y-intercept}} = 1

Substituting the intercepts we found:

\frac{x}{2a} + \frac{y}{2b} = 1

To get the desired form, multiply the entire equation by 2:

2(\frac{x}{2a} + \frac{y}{2b}) = 2(1)

\frac{x}{a} + \frac{y}{b} = 2

Hence Shown.

Q18EXERCISE 9.2

Point $\mathrm{R}(h, k)$ divides a line segment between the axes in the ratio $1:2$ . Find equation of the line.

Solution

Given: A line segment lies between the coordinate axes. A point R

(h, k)

divides this segment in the ratio

1:2

.

To Find: The equation of the line.

Solution: Let the line intersect the x-axis at point A and the y-axis at point B. The coordinates of A will be

(a, 0)

and the coordinates of B will be

(0, b)

. Here, 'a' is the x-intercept and 'b' is the y-intercept.

The point R

(h, k)

divides the line segment AB in the ratio

1:2

.

Using the section formula for a point

(x, y)

that divides the segment from

(x_1, y_1)

to

(x_2, y_2)

in the ratio

m:n

:

x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}

Here,

(x_1, y_1) = A(a, 0)

,

(x_2, y_2) = B(0, b)

,

(x, y) = R(h, k)

, and the ratio is

m:n = 1:2

.

For the x-coordinate:

h = \frac{1(0) + 2(a)}{1 + 2} = \frac{2a}{3} \implies a = \frac{3h}{2}

For the y-coordinate:

k = \frac{1(b) + 2(0)}{1 + 2} = \frac{b}{3} \implies b = 3k

Now we have the x-intercept (

a = \frac{3h}{2}

) and the y-intercept (

b = 3k

).

The intercept form of the equation of a line is:

\frac{x}{a} + \frac{y}{b} = 1

Substitute the expressions for a and b:

\frac{x}{3h/2} + \frac{y}{3k} = 1

\frac{2x}{3h} + \frac{y}{3k} = 1

To clear the denominators, we can multiply the equation by

3hk

:

2xk + yh = 3hk

Final Answer: The equation of the line is

\frac{2x}{3h} + \frac{y}{3k} = 1

, or

2kx + hy = 3hk

.

Q19EXERCISE 9.2

By using the concept of equation of a line, prove that the three points $(3, 0), (-2, -2)$ and $(8, 2)$ are collinear.

Solution

Given: Three points A

(3, 0)

, B

(-2, -2)

, and C

(8, 2)

.

To Prove: The points A, B, and C are collinear using the concept of the equation of a line.

Method: We will find the equation of the line passing through two of the points (say, A and B). Then, we will show that the third point (C) satisfies this equation. If it does, it must lie on the same line, proving collinearity.

Proof: Step 1: Find the equation of the line passing through A(3, 0) and B(-2, -2). Using the two-point form:

y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

Let

(x_1, y_1) = (3, 0)

and

(x_2, y_2) = (-2, -2)

.

First, find the slope:

m = \frac{-2 - 0}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5}

Now, use the point-slope form with point A(3, 0):

y - 0 = \frac{2}{5}(x - 3)

y = \frac{2}{5}(x - 3)

5y = 2(x - 3)

5y = 2x - 6

2x - 5y - 6 = 0

This is the equation of the line passing through points A and B.

Step 2: Check if point C(8, 2) satisfies this equation. Substitute

x = 8

and

y = 2

into the equation

2x - 5y - 6 = 0

:

2(8) - 5(2) - 6 = 16 - 10 - 6 = 6 - 6 = 0

Since the coordinates of point C satisfy the equation of the line passing through A and B, point C must lie on the same line.

Therefore, the three points A, B, and C are collinear.

Hence Proved.

Q1EXERCISE 9.3

Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

(i)

$x + 7y = 0$ ,

(ii)

$6x + 3y - 5 = 0$ ,

(iii)

$y = 0$ .

Solution

To Do: Convert the given equations to the slope-intercept form,

y = mx + c

, and identify the slope (

m

) and y-intercept (

c

).

(i) $x + 7y = 0$

Solution: To convert to slope-intercept form, we isolate

y

.

7y = -x

y = -\frac{1}{7}x + 0

Comparing this with

y = mx + c

, we get: Slope,

m = -\frac{1}{7}

y-intercept,

c = 0

(ii) $6x + 3y - 5 = 0$

Solution: Isolate

y

.

3y = -6x + 5

y = \frac{-6}{3}x + \frac{5}{3}

y = -2x + \frac{5}{3}

Comparing this with

y = mx + c

, we get: Slope,

m = -2

y-intercept,

c = \frac{5}{3}

(iii) $y = 0$

Solution: This equation is already in a form that can be compared to

y = mx + c

. We can write it as:

y = 0 \cdot x + 0

Comparing this with

y = mx + c

, we get: Slope,

m = 0

y-intercept,

c = 0

Final Answer:

(i)

Slope-intercept form:

y = -\frac{1}{7}x + 0

; Slope

m = -\frac{1}{7}

; y-intercept

c = 0

.

(ii)

Slope-intercept form:

y = -2x + \frac{5}{3}

; Slope

m = -2

; y-intercept

c = \frac{5}{3}

.

(iii)

Slope-intercept form:

y = 0x + 0

; Slope

m = 0

; y-intercept

c = 0

.

Q2EXERCISE 9.3

Reduce the following equations into intercept form and find their intercepts on the axes.

(i)

$3x + 2y - 12 = 0$ ,

(ii)

$4x - 3y = 6$ ,

(iii)

$3y + 2 = 0$ .

Solution

To Do: Convert the given equations to the intercept form,

\frac{x}{a} + \frac{y}{b} = 1

, and identify the x-intercept (

a

) and y-intercept (

b

).

(i) $3x + 2y - 12 = 0$

Solution: First, move the constant term to the right side.

3x + 2y = 12

Now, divide the entire equation by 12 to make the right side equal to 1.

\frac{3x}{12} + \frac{2y}{12} = \frac{12}{12}

\frac{x}{4} + \frac{y}{6} = 1

Comparing this with

\frac{x}{a} + \frac{y}{b} = 1

, we get: x-intercept,

a = 4

y-intercept,

b = 6

(ii) $4x - 3y = 6$

Solution: The constant term is already on the right side. Divide the entire equation by 6.

\frac{4x}{6} - \frac{3y}{6} = \frac{6}{6}

\frac{2x}{3} - \frac{y}{2} = 1

To get it into the standard intercept form, rewrite the terms:

\frac{x}{3/2} + \frac{y}{-2} = 1

Comparing this with

\frac{x}{a} + \frac{y}{b} = 1

, we get: x-intercept,

a = \frac{3}{2}

y-intercept,

b = -2

(iii) $3y + 2 = 0$

Solution: First, isolate the term with

y

.

3y = -2

y = -\frac{2}{3}

This is the equation of a horizontal line. It is parallel to the x-axis and therefore does not intersect it. Hence, there is no x-intercept. The y-intercept is clearly

-\frac{2}{3}

. Because there is no x-intercept, this equation cannot be written in the standard intercept form

\frac{x}{a} + \frac{y}{b} = 1

.

Final Answer:

(i)

Intercept form:

\frac{x}{4} + \frac{y}{6} = 1

; x-intercept

a=4

; y-intercept

b=6

.

(ii)

Intercept form:

\frac{x}{3/2} + \frac{y}{-2} = 1

; x-intercept

a=\frac{3}{2}

; y-intercept

b=-2

.

(iii)

The equation cannot be reduced to intercept form. The y-intercept is

-\frac{2}{3}

, and there is no x-intercept.

Q3EXERCISE 9.3

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$ .

Solution

Given: A point P

(x_1, y_1) = (-1, 1)

. A line with equation

12(x + 6) = 5(y - 2)

.

To Find: The perpendicular distance from the point to the line.

Formula: The distance

d

of a point

(x_1, y_1)

from a line

Ax + By + C = 0

is given by:

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Solution: Step 1: Convert the equation of the line to the general form $Ax + By + C = 0$ .

12(x + 6) = 5(y - 2)

12x + 72 = 5y - 10

12x - 5y + 72 + 10 = 0

12x - 5y + 82 = 0

From this equation, we have

A = 12

,

B = -5

, and

C = 82

.

Step 2: Apply the distance formula. The point is

(x_1, y_1) = (-1, 1)

.

d = \frac{|12(-1) + (-5)(1) + 82|}{\sqrt{(12)^2 + (-5)^2}}

d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}}

d = \frac{|-17 + 82|}{\sqrt{169}}

d = \frac{|65|}{13}

d = 5

Final Answer: The distance of the point from the line is 5 units.

Q4EXERCISE 9.3

Find the points on the $x$ -axis, whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are 4 units.

Solution

Given: A line with equation

\frac{x}{3} + \frac{y}{4} = 1

. The distance from the required points to this line is 4 units. The points lie on the x-axis.

To Find: The coordinates of the points.

Solution: Step 1: Define the points and the line equation. Any point on the x-axis can be represented as P

(x_0, 0)

. Convert the line equation to the general form

Ax + By + C = 0

.

\frac{x}{3} + \frac{y}{4} = 1

Multiply by 12 (the LCM of 3 and 4) to clear the fractions:

4x + 3y = 12

4x + 3y - 12 = 0

Here,

A = 4

,

B = 3

, and

C = -12

.

Step 2: Use the distance formula. The distance

d

from a point

(x_0, 0)

to this line is given as 4.

d = \frac{|Ax_0 + B(0) + C|}{\sqrt{A^2 + B^2}} = 4

\frac{|4x_0 + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = 4

\frac{|4x_0 - 12|}{\sqrt{16 + 9}} = 4

\frac{|4x_0 - 12|}{\sqrt{25}} = 4

\frac{|4x_0 - 12|}{5} = 4

|4x_0 - 12| = 20

Step 3: Solve for $x_0$ . The absolute value equation gives two possible cases:

Case 1:

4x_0 - 12 = 20

4x_0 = 32

x_0 = 8

The point is

(8, 0)

.

Case 2:

4x_0 - 12 = -20

4x_0 = -8

x_0 = -2

The point is

(-2, 0)

.

Final Answer: The points on the x-axis are

(8, 0)

and

(-2, 0)

.

Q5EXERCISE 9.3

Find the distance between parallel lines

(i)

$15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$

(ii)

$l(x + y) + p = 0$ and $l(x + y) - r = 0$ .

Solution

Formula: The distance

d

between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by:

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

(i) $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$

Solution: The lines are parallel because the coefficients of x and y are the same. Here,

A = 15

,

B = 8

,

C_1 = -34

, and

C_2 = 31

.

Using the formula:

d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}}

d = \frac{|-65|}{\sqrt{225 + 64}}

d = \frac{65}{\sqrt{289}}

d = \frac{65}{17}

(ii) $l(x + y) + p = 0$ and $l(x + y) - r = 0$

Solution: First, expand the equations to the standard form. Line 1:

lx + ly + p = 0

Line 2:

lx + ly - r = 0

The lines are parallel. Here,

A = l

,

B = l

,

C_1 = p

, and

C_2 = -r

.

Using the formula:

d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}}

d = \frac{|p + r|}{\sqrt{2l^2}}

d = \frac{|p + r|}{|l|\sqrt{2}}

Final Answer:

(i)

The distance between the lines is

\frac{65}{17}

units.

(ii)

The distance between the lines is

\frac{|p + r|}{|l|\sqrt{2}}

units.

Q6EXERCISE 9.3

Find equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(-2, 3)$ .

Solution

Given: A line

3x - 4y + 2 = 0

. The required line is parallel to this line. The required line passes through the point

(-2, 3)

.

To Find: The equation of the required line.

Solution: Any line parallel to

3x - 4y + 2 = 0

will have the same coefficients for x and y, but a different constant term. So, the equation of the required line will be of the form:

3x - 4y + k = 0

where

k

is a constant.

We are given that this line passes through the point

(-2, 3)

. To find the value of

k

, we substitute

x = -2

and

y = 3

into the equation:

3(-2) - 4(3) + k = 0

-6 - 12 + k = 0

-18 + k = 0

k = 18

Now, substitute the value of

k

back into the equation of the line:

3x - 4y + 18 = 0

Final Answer: The equation of the line is

3x - 4y + 18 = 0

.

Q7EXERCISE 9.3

Find equation of the line perpendicular to the line $x - 7y + 5 = 0$ and having $x$ intercept 3.

Solution

Given: A line

x - 7y + 5 = 0

. The required line is perpendicular to this line. The required line has an x-intercept of 3.

To Find: The equation of the required line.

Solution: For a line with equation

Ax + By + C = 0

, any line perpendicular to it has the form

Bx - Ay + k = 0

.

The given line is

1x - 7y + 5 = 0

. So,

A = 1

and

B = -7

. The equation of a perpendicular line will be of the form:

-7x - 1y + k = 0

7x + y - k = 0

Let's use a new constant

C = -k

. The equation is:

7x + y + C = 0

We are given that the line has an x-intercept of 3. This means the line passes through the point

(3, 0)

. To find the value of

C

, we substitute

x = 3

and

y = 0

into the equation:

7(3) + 0 + C = 0

21 + C = 0

C = -21

Now, substitute the value of

C

back into the equation of the line:

7x + y - 21 = 0

Final Answer: The equation of the line is

7x + y - 21 = 0

.

Q8EXERCISE 9.3

Find angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$ .

Solution

Given: Two lines: Line 1:

\sqrt{3}x + y = 1

Line 2:

x + \sqrt{3}y = 1

To Find: The angles between these two lines.

Formula: If

\theta

is the acute angle between two lines with slopes

m_1

and

m_2

, then:

\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Solution: Step 1: Find the slopes of the lines. Convert each equation to the slope-intercept form

y = mx + c

.

For Line 1:

\sqrt{3}x + y = 1 \implies y = -\sqrt{3}x + 1

. So, the slope is

m_1 = -\sqrt{3}

.

For Line 2:

x + \sqrt{3}y = 1 \implies \sqrt{3}y = -x + 1 \implies y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}

. So, the slope is

m_2 = -\frac{1}{\sqrt{3}}

.

Step 2: Apply the angle formula.

\tan \theta = \left| \frac{-\frac{1}{\sqrt{3}} - (-\sqrt{3})}{1 + (-\sqrt{3})(-\frac{1}{\sqrt{3}})} \right|

\tan \theta = \left| \frac{-\frac{1}{\sqrt{3}} + \sqrt{3}}{1 + 1} \right|

\tan \theta = \left| \frac{\frac{-1 + 3}{\sqrt{3}}}{2} \right|

\tan \theta = \left| \frac{\frac{2}{\sqrt{3}}}{2} \right| = \left| \frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}}

Since

\tan \theta = \frac{1}{\sqrt{3}}

, the acute angle

\theta

is

30^{\circ}

(or

\frac{\pi}{6}

radians).

When two lines intersect, they form two pairs of vertically opposite angles. One angle is acute (

"theta

) and the other is obtuse (

"phi

). The obtuse angle is given by

\phi = 180^{\circ} - \theta

.

\phi = 180^{\circ} - 30^{\circ} = 150^{\circ}

Final Answer: The angles between the lines are

30^{\circ}

and

150^{\circ}

.

Q9EXERCISE 9.3

The line through the points $(h, 3)$ and $(4, 1)$ intersects the line $7x - 9y - 19 = 0$ . at right angle. Find the value of $h$ .

Solution

Given: Line 1 passes through points A

(h, 3)

and B

(4, 1)

. Line 2 has the equation

7x - 9y - 19 = 0

. Line 1 and Line 2 intersect at a right angle (they are perpendicular).

To Find: The value of

h

.

Condition for Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1. That is,

m_1 \times m_2 = -1

.

Solution: Step 1: Find the slope of Line 1 ( $m_1$ ). Using the slope formula for points A

(h, 3)

and B

(4, 1)

:

m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h}

Step 2: Find the slope of Line 2 ( $m_2$ ). Convert the equation

7x - 9y - 19 = 0

to slope-intercept form (

y = mx + c

).

-9y = -7x + 19

9y = 7x - 19

y = \frac{7}{9}x - \frac{19}{9}

The slope of Line 2 is

m_2 = \frac{7}{9}

.

Step 3: Apply the perpendicularity condition.

m_1 \times m_2 = -1

\left( \frac{-2}{4 - h} \right) \times \left( \frac{7}{9} \right) = -1

\frac{-14}{9(4 - h)} = -1

-14 = -9(4 - h)

14 = 9(4 - h)

14 = 36 - 9h

9h = 36 - 14

9h = 22

h = \frac{22}{9}

Final Answer: The value of

h

is

\frac{22}{9}

.

Q10EXERCISE 9.3

Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax + By + C = 0$ is $A(x - x_1) + B(y - y_1) = 0$ .

Solution

Given: A point

(x_1, y_1)

. A line

L_1

with equation

Ax + By + C = 0

. A second line

L_2

which passes through

(x_1, y_1)

and is parallel to

L_1

.

To Prove: The equation of line

L_2

is

A(x - x_1) + B(y - y_1) = 0

.

Proof: Step 1: Find the slope of the given line $L_1$ . We can find the slope by rearranging the equation

Ax + By + C = 0

to the slope-intercept form (

y=mx+c

).

By = -Ax - C

y = -\frac{A}{B}x - \frac{C}{B}

The slope of line

L_1

is

m_1 = -\frac{A}{B}

.

Step 2: Determine the slope of the required line $L_2$ . Since line

L_2

is parallel to line

L_1

, they must have the same slope. So, the slope of line

L_2

is

m_2 = m_1 = -\frac{A}{B}

.

Step 3: Find the equation of line $L_2$ . We know that line

L_2

passes through the point

(x_1, y_1)

and has a slope of

m_2 = -\frac{A}{B}

. Using the point-slope form of a line,

y - y_0 = m(x - x_0)

:

y - y_1 = -\frac{A}{B}(x - x_1)

Step 4: Rearrange the equation to the required form. Multiply both sides by B:

B(y - y_1) = -A(x - x_1)

Move all terms to one side:

A(x - x_1) + B(y - y_1) = 0

This is the required equation.

Hence Proved.

Q11EXERCISE 9.3

Two lines passing through the point $(2, 3)$ intersects each other at an angle of $60^{\circ}$ . If slope of one line is 2, find equation of the other line.

Solution

Given: Two lines intersect at the point

(2, 3)

. The angle between the lines is

\theta = 60^{\circ}

. The slope of one line is

m_1 = 2

.

To Find: The equation of the other line.

Formula: The tangent of the angle

\theta

between two lines with slopes

m_1

and

m_2

is:

\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Solution: Step 1: Find the slope of the other line ( $m_2$ ). Let the slope of the other line be

m

. We have

\theta = 60^{\circ}

, so

\tan 60^{\circ} = \sqrt{3}

.

\sqrt{3} = \left| \frac{m - 2}{1 + 2m} \right|

This gives two possible cases:

Case 1:

\frac{m - 2}{1 + 2m} = \sqrt{3}

m - 2 = \sqrt{3}(1 + 2m)

m - 2 = \sqrt{3} + 2\sqrt{3}m

m - 2\sqrt{3}m = 2 + \sqrt{3}

m(1 - 2\sqrt{3}) = 2 + \sqrt{3}

m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}

Case 2:

\frac{m - 2}{1 + 2m} = -\sqrt{3}

m - 2 = -\sqrt{3}(1 + 2m)

m - 2 = -\sqrt{3} - 2\sqrt{3}m

m + 2\sqrt{3}m = 2 - \sqrt{3}

m(1 + 2\sqrt{3}) = 2 - \sqrt{3}

m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}

Step 2: Find the equations of the lines. Both lines pass through the point

(2, 3)

. We use the point-slope form

y - y_0 = m(x - x_0)

.

For Case 1:

m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}

The equation is

y - 3 = \left( \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \right) (x - 2)

.

For Case 2:

m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}

The equation is

y - 3 = \left( \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \right) (x - 2)

.

Final Answer: There are two possible equations for the other line:

$y - 3 = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} (x - 2)$
$y - 3 = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} (x - 2)$

Q12EXERCISE 9.3

Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1, 2)$ .

Solution

Given: Two points on a line segment: A

(3, 4)

and B

(-1, 2)

.

To Find: The equation of the right bisector of the segment AB.

Method: The right bisector of a segment has two properties:

It is perpendicular to the segment.
It passes through the midpoint of the segment.

Solution: Step 1: Find the midpoint of the segment AB. Let the midpoint be M.

M = (\frac{3 + (-1)}{2}, \frac{4 + 2}{2}) = (\frac{2}{2}, \frac{6}{2}) = (1, 3)

Step 2: Find the slope of the segment AB ( $m_{AB}$ ).

m_{AB} = \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}

Step 3: Find the slope of the right bisector ( $m_{perp}$ ). Since the right bisector is perpendicular to AB, the product of their slopes is -1.

m_{AB} \times m_{perp} = -1

\frac{1}{2} \times m_{perp} = -1

m_{perp} = -2

Step 4: Find the equation of the right bisector. We have the slope (

m_{perp} = -2

) and a point it passes through (the midpoint M

(1, 3)

). Using the point-slope form

y - y_0 = m(x - x_0)

:

y - 3 = -2(x - 1)

y - 3 = -2x + 2

Rearranging to the general form:

2x + y - 3 - 2 = 0

2x + y - 5 = 0

Final Answer: The equation of the right bisector is

2x + y - 5 = 0

.

Q13EXERCISE 9.3

Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$ .

Solution

Given: A point P

(-1, 3)

. A line L:

3x - 4y - 16 = 0

.

To Find: The coordinates of the foot of the perpendicular from P to L.

Method: Let the foot of the perpendicular be M

(h, k)

.

M lies on the line L, so its coordinates satisfy the equation of L.
The line segment PM is perpendicular to the line L.

Solution: Step 1: Use the property that M(h, k) lies on the line L. Substitute

(h, k)

into the equation of L:

3h - 4k - 16 = 0 \quad ...(1)

Step 2: Use the perpendicularity condition. First, find the slope of the line L (

m_L

).

3x - 16 = 4y \implies y = \frac{3}{4}x - 4

So,

m_L = \frac{3}{4}

.

Next, find the slope of the line segment PM (

m_{PM}

).

m_{PM} = \frac{k - 3}{h - (-1)} = \frac{k - 3}{h + 1}

Since PM is perpendicular to L, the product of their slopes is -1.

m_L \times m_{PM} = -1

\frac{3}{4} \times \frac{k - 3}{h + 1} = -1

3(k - 3) = -4(h + 1)

3k - 9 = -4h - 4

4h + 3k = 5 \quad ...(2)

Step 3: Solve the system of linear equations (1) and (2). From equation (2), we can express h:

h = \frac{5 - 3k}{4}

. Substitute this into equation (1):

3(\frac{5 - 3k}{4}) - 4k - 16 = 0

Multiply by 4 to clear the fraction:

3(5 - 3k) - 16k - 64 = 0

15 - 9k - 16k - 64 = 0

-25k - 49 = 0

-25k = 49 \implies k = -\frac{49}{25}

Now substitute the value of k back into the expression for h:

h = \frac{5 - 3(-\frac{49}{25})}{4} = \frac{5 + \frac{147}{25}}{4} = \frac{\frac{125 + 147}{25}}{4} = \frac{\frac{272}{25}}{4} = \frac{272}{100} = \frac{68}{25}

The coordinates of the foot of the perpendicular are

(\frac{68}{25}, -\frac{49}{25})

.

Final Answer: The coordinates of the foot of the perpendicular are

(\frac{68}{25}, -\frac{49}{25})

.

Q14EXERCISE 9.3

The perpendicular from the origin to the line $y = mx + c$ meets it at the point $(-1, 2)$ . Find the values of $m$ and $c$ .

Solution

Given: A line L:

y = mx + c

. A perpendicular from the origin O

(0, 0)

meets L at the point P

(-1, 2)

.

To Find: The values of

m

and

c

.

Solution: Step 1: Use the fact that point P(-1, 2) lies on the line L. Substitute the coordinates of P into the equation of the line:

2 = m(-1) + c

2 = -m + c \quad ...(1)

Step 2: Use the perpendicularity condition. The line segment OP is perpendicular to the line L. First, find the slope of the line segment OP (

m_{OP}

).

m_{OP} = \frac{2 - 0}{-1 - 0} = -2

The slope of the line L is

m

. Since OP is perpendicular to L, the product of their slopes is -1.

m_{OP} \times m = -1

-2 \times m = -1

m = \frac{1}{2}

Step 3: Find the value of c. Substitute the value of

m = \frac{1}{2}

into equation (1):

2 = -(\frac{1}{2}) + c

c = 2 + \frac{1}{2}

c = \frac{4+1}{2} = \frac{5}{2}

Final Answer: The values are

m = \frac{1}{2}

and

c = \frac{5}{2}

.

Q15EXERCISE 9.3

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x \cos \theta - y \sin \theta = k \cos 2\theta$ and $x \sec \theta + y \operatorname{cosec} \theta = k$ , respectively, prove that $p^2 + 4q^2 = k^2$ .

Solution

Given: Line 1:

x \cos \theta - y \sin \theta = k \cos 2\theta

Line 2:

x \sec \theta + y \operatorname{cosec} \theta = k

p

is the length of the perpendicular from the origin

(0,0)

to Line 1.

q

is the length of the perpendicular from the origin

(0,0)

to Line 2.

To Prove:

p^2 + 4q^2 = k^2

.

Formula: The perpendicular distance

d

from the origin

(0,0)

to the line

Ax + By + C = 0

is

d = \frac{|C|}{\sqrt{A^2 + B^2}}

.

Proof: Step 1: Calculate $p$ for Line 1. Rewrite Line 1 in the general form:

x \cos \theta - y \sin \theta - k \cos 2\theta = 0

. Here,

A = \cos \theta

,

B = -\sin \theta

,

C = -k \cos 2\theta

.

p = \frac{|-k \cos 2\theta|}{\sqrt{(\cos \theta)^2 + (-\sin \theta)^2}} = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{|-k \cos 2\theta|}{\sqrt{1}} = |k \cos 2\theta|

Squaring both sides:

p^2 = k^2 \cos^2 2\theta \quad ...(1)

Step 2: Calculate $q$ for Line 2. Rewrite Line 2 in the general form:

x \sec \theta + y \operatorname{cosec} \theta - k = 0

\frac{x}{\cos \theta} + \frac{y}{\sin \theta} - k = 0

Multiply by

\sin \theta \cos \theta

to clear fractions:

x \sin \theta + y \cos \theta - k \sin \theta \cos \theta = 0

Here,

A = \sin \theta

,

B = \cos \theta

,

C = -k \sin \theta \cos \theta

.

q = \frac{|-k \sin \theta \cos \theta|}{\sqrt{(\sin \theta)^2 + (\cos \theta)^2}} = \frac{|-k \sin \theta \cos \theta|}{\sqrt{1}} = |k \sin \theta \cos \theta|

Using the double angle identity

\sin 2\theta = 2 \sin \theta \cos \theta

, we have

\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta

.

q = |k \cdot \frac{1}{2} \sin 2\theta| = \frac{|k|}{2} |\sin 2\theta|

Squaring both sides:

q^2 = \frac{k^2}{4} \sin^2 2\theta

4q^2 = k^2 \sin^2 2\theta \quad ...(2)

Step 3: Combine the results. Add equation (1) and equation (2):

p^2 + 4q^2 = k^2 \cos^2 2\theta + k^2 \sin^2 2\theta

Factor out

k^2

:

p^2 + 4q^2 = k^2 (\cos^2 2\theta + \sin^2 2\theta)

Using the identity

\cos^2 A + \sin^2 A = 1

:

p^2 + 4q^2 = k^2(1)

p^2 + 4q^2 = k^2

Hence Proved.

Q16EXERCISE 9.3

In the triangle ABC with vertices $\mathrm{A}(2, 3), \mathrm{B}(4, -1)$ and $\mathrm{C}(1, 2)$ , find the equation and length of altitude from the vertex A.

Solution

Given: Vertices of

\triangle

ABC: A

(2, 3)

, B

(4, -1)

, C

(1, 2)

.

To Find:

The equation of the altitude from vertex A.
The length of the altitude from vertex A.

Method: The altitude from A is a line segment from A that is perpendicular to the opposite side BC.

Part 1: Equation of the altitude from A.

Step 1.1: Find the slope of the side BC ( $m_{BC}$ ).

m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1

Step 1.2: Find the slope of the altitude from A ( $m_{alt}$ ). The altitude is perpendicular to BC, so

m_{alt} \times m_{BC} = -1

.

m_{alt} \times (-1) = -1 \implies m_{alt} = 1

Step 1.3: Find the equation of the altitude. The altitude passes through point A

(2, 3)

and has a slope of 1. Using the point-slope form

y - y_0 = m(x - x_0)

:

y - 3 = 1(x - 2)

y - 3 = x - 2

x - y + 1 = 0

Part 2: Length of the altitude from A.

The length of the altitude from A is the perpendicular distance from point A to the line containing side BC.

Step 2.1: Find the equation of the line BC. We have the slope

m_{BC} = -1

and it passes through C

(1, 2)

. Using the point-slope form:

y - 2 = -1(x - 1)

y - 2 = -x + 1

x + y - 3 = 0

Step 2.2: Find the perpendicular distance from A(2, 3) to the line BC. Using the distance formula

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

with point

(x_1, y_1) = (2, 3)

and line

x + y - 3 = 0

. Here

A=1, B=1, C=-3

.

\text{Length} = \frac{|1(2) + 1(3) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 3|}{\sqrt{2}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

Final Answer: The equation of the altitude from vertex A is

x - y + 1 = 0

. The length of the altitude from vertex A is

\sqrt{2}

units.

Q17EXERCISE 9.3

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$ , then show that $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$ .

Solution

Given: A line has x-intercept 'a' and y-intercept 'b'.

p

is the length of the perpendicular from the origin

(0, 0)

to this line.

To Show:

\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}

.

Proof: Step 1: Write the equation of the line. Using the intercept form, the equation of the line is:

\frac{x}{a} + \frac{y}{b} = 1

Step 2: Convert the equation to the general form $Ax + By + C = 0$ . To clear the fractions, multiply by

ab

:

bx + ay = ab

bx + ay - ab = 0

Step 3: Calculate the perpendicular distance $p$ from the origin (0, 0). Using the formula

d = \frac{|C|}{\sqrt{A^2 + B^2}}

, with

A=b, B=a, C=-ab

:

p = \frac{|-ab|}{\sqrt{b^2 + a^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}

Step 4: Manipulate the equation to get the desired result. Square both sides of the equation for

p

:

p^2 = \frac{(ab)^2}{a^2 + b^2} = \frac{a^2 b^2}{a^2 + b^2}

Now, take the reciprocal of both sides:

\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}

Split the fraction on the right side:

\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}

Simplify the terms:

\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}

Rearranging gives the required form:

\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}

Hence Shown.

Q1Miscellaneous Exercise on Chapter 9

Find the values of $k$ for which the line $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is

(a)

Parallel to the

x

-axis,

(b)

Parallel to the

y

-axis,

(c)

Passing through the origin.

Solution

Given: The equation of a line:

(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0

. This is in the form

Ax + By + C = 0

, where:

A = k-3

B = -(4-k^2) = k^2 - 4

C = k^2 - 7k + 6

The slope of this line is

m = -\frac{A}{B} = -\frac{k-3}{k^2-4} = \frac{k-3}{4-k^2}

.

(a) Parallel to the x-axis

Condition: A line is parallel to the x-axis if its slope is 0. This occurs when the coefficient of x is zero (

A=0

) and the coefficient of y is non-zero (

B \neq 0

).

Solution: Set the coefficient of x to zero:

k - 3 = 0 \implies k = 3

. We must also check that for

k=3

, the coefficient of y is not zero.

B = k^2 - 4 = 3^2 - 4 = 9 - 4 = 5 \neq 0

. So, the condition is satisfied.

Answer (a):

k = 3

.

(b) Parallel to the y-axis

Condition: A line is parallel to the y-axis if its slope is undefined. This occurs when the coefficient of y is zero (

B=0

) and the coefficient of x is non-zero (

A \neq 0

).

Solution: Set the coefficient of y to zero:

k^2 - 4 = 0 \implies (k-2)(k+2) = 0 \implies k = 2

or

k = -2

. We must check that for these values, the coefficient of x is not zero. If

k=2

,

A = 2-3 = -1 \neq 0

. This is a valid solution. If

k=-2

,

A = -2-3 = -5 \neq 0

. This is also a valid solution.

Answer (b):

k = 2

or

k = -2

.

(c) Passing through the origin

Condition: A line passes through the origin

(0, 0)

if its constant term is zero (

C=0

).

Solution: Set the constant term to zero:

k^2 - 7k + 6 = 0

. Factorize the quadratic equation:

(k-1)(k-6) = 0

. So,

k=1

or

k=6

.

Answer (c):

k = 1

or

k = 6

.

Q2Miscellaneous Exercise on Chapter 9

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively.

Solution

Given: Let the x-intercept be 'a' and the y-intercept be 'b'. Sum of intercepts:

a + b = 1

Product of intercepts:

ab = -6

To Find: The equations of the lines.

Solution: Step 1: Find the values of the intercepts a and b. We have a system of two equations:

$a + b = 1$
$ab = -6$

From equation (1),

b = 1 - a

. Substitute this into equation (2):

a(1 - a) = -6

a - a^2 = -6

a^2 - a - 6 = 0

Factorize the quadratic equation:

(a - 3)(a + 2) = 0

This gives two possible values for 'a':

a = 3

or

a = -2

.

Case 1: If

a = 3

Then

b = 1 - a = 1 - 3 = -2

. The intercepts are

(a, b) = (3, -2)

.

Case 2: If

a = -2

Then

b = 1 - a = 1 - (-2) = 3

. The intercepts are

(a, b) = (-2, 3)

.

Step 2: Find the equations of the lines using the intercept form. The intercept form is

\frac{x}{a} + \frac{y}{b} = 1

.

For Case 1 (a=3, b=-2):

\frac{x}{3} + \frac{y}{-2} = 1

Multiply by 6 to clear the fractions:

2x - 3y = 6 \implies 2x - 3y - 6 = 0

For Case 2 (a=-2, b=3):

\frac{x}{-2} + \frac{y}{3} = 1

Multiply by -6 to clear the fractions:

3x - 2y = -6 \implies 3x - 2y + 6 = 0

Final Answer: The equations of the lines are

2x - 3y - 6 = 0

and

3x - 2y + 6 = 0

.

Q3Miscellaneous Exercise on Chapter 9

What are the points on the $y$ -axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

Solution

Given: A line with equation

\frac{x}{3} + \frac{y}{4} = 1

. The distance from the required points to this line is 4 units. The points lie on the y-axis.

To Find: The coordinates of the points.

Solution: Step 1: Define the points and the line equation. Any point on the y-axis can be represented as P

(0, y_0)

. Convert the line equation to the general form

Ax + By + C = 0

.

\frac{x}{3} + \frac{y}{4} = 1

Multiply by 12 (the LCM of 3 and 4):

4x + 3y = 12

4x + 3y - 12 = 0

Here,

A = 4

,

B = 3

, and

C = -12

.

Step 2: Use the distance formula. The distance

d

from a point

(0, y_0)

to this line is given as 4.

d = \frac{|A(0) + By_0 + C|}{\sqrt{A^2 + B^2}} = 4

\frac{|4(0) + 3y_0 - 12|}{\sqrt{4^2 + 3^2}} = 4

\frac{|3y_0 - 12|}{\sqrt{16 + 9}} = 4

\frac{|3y_0 - 12|}{\sqrt{25}} = 4

\frac{|3y_0 - 12|}{5} = 4

|3y_0 - 12| = 20

Step 3: Solve for $y_0$ . The absolute value equation gives two possible cases:

Case 1:

3y_0 - 12 = 20

3y_0 = 32

y_0 = \frac{32}{3}

The point is

(0, \frac{32}{3})

.

Case 2:

3y_0 - 12 = -20

3y_0 = -8

y_0 = -\frac{8}{3}

The point is

(0, -\frac{8}{3})

.

Final Answer: The points on the y-axis are

(0, \frac{32}{3})

and

(0, -\frac{8}{3})

.

Q4Miscellaneous Exercise on Chapter 9

Find perpendicular distance from the origin to the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ .

Solution

Given: Two points on a line: P

(\cos \theta, \sin \theta)

and Q

(\cos \phi, \sin \phi)

.

To Find: The perpendicular distance from the origin

(0,0)

to the line PQ.

Solution: Step 1: Find the equation of the line passing through P and Q. Using the two-point form

y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

:

y - \sin \theta = \frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta}(x - \cos \theta)

Multiply both sides by

(\cos \phi - \cos \theta)

:

(y - \sin \theta)(\cos \phi - \cos \theta) = (x - \cos \theta)(\sin \phi - \sin \theta)

y(\cos \phi - \cos \theta) - \sin \theta \cos \phi + \sin \theta \cos \theta = x(\sin \phi - \sin \theta) - \cos \theta \sin \phi + \cos \theta \sin \theta

y(\cos \phi - \cos \theta) - \sin \theta \cos \phi = x(\sin \phi - \sin \theta) - \cos \theta \sin \phi

Rearrange to the general form

Ax+By+C=0

:

x(\sin \phi - \sin \theta) - y(\cos \phi - \cos \theta) + \sin \theta \cos \phi - \cos \theta \sin \phi = 0

x(\sin \phi - \sin \theta) + y(\cos \theta - \cos \phi) + \sin(\theta - \phi) = 0

Step 2: Apply the distance formula from the origin. The distance

d

from

(0,0)

is

d = \frac{|C|}{\sqrt{A^2 + B^2}}

. Here,

C = \sin(\theta - \phi)

,

A = \sin \phi - \sin \theta

,

B = \cos \theta - \cos \phi

.

d = \frac{|\sin(\theta - \phi)|}{\sqrt{(\sin \phi - \sin \theta)^2 + (\cos \theta - \cos \phi)^2}}

Let's expand the denominator:

A^2 = \sin^2 \phi - 2\sin \phi \sin \theta + \sin^2 \theta

B^2 = \cos^2 \theta - 2\cos \theta \cos \phi + \cos^2 \phi

A^2 + B^2 = (\sin^2 \phi + \cos^2 \phi) + (\sin^2 \theta + \cos^2 \theta) - 2(\cos \theta \cos \phi + \sin \theta \sin \phi)

A^2 + B^2 = 1 + 1 - 2\cos(\theta - \phi) = 2 - 2\cos(\theta - \phi)

Using the identity

1 - \cos(2x) = 2\sin^2(x)

, we have

1 - \cos(\theta - \phi) = 2\sin^2(\frac{\theta - \phi}{2})

. So,

A^2 + B^2 = 2(1 - \cos(\theta - \phi)) = 2(2\sin^2(\frac{\theta - \phi}{2})) = 4\sin^2(\frac{\theta - \phi}{2})

. Therefore,

\sqrt{A^2 + B^2} = \sqrt{4\sin^2(\frac{\theta - \phi}{2})} = 2|\sin(\frac{\theta - \phi}{2})|

.

Now substitute this back into the distance formula:

d = \frac{|\sin(\theta - \phi)|}{2|\sin(\frac{\theta - \phi}{2})|}

Using the identity

\sin(2x) = 2\sin(x)\cos(x)

, we have

\sin(\theta - \phi) = 2\sin(\frac{\theta - \phi}{2})\cos(\frac{\theta - \phi}{2})

.

d = \frac{|2\sin(\frac{\theta - \phi}{2})\cos(\frac{\theta - \phi}{2})|}{2|\sin(\frac{\theta - \phi}{2})|} = |\cos(\frac{\theta - \phi}{2})|

Final Answer: The perpendicular distance is

|\cos(\frac{\theta - \phi}{2})|

.

Q5Miscellaneous Exercise on Chapter 9

Find the equation of the line parallel to $y$ -axis and drawn through the point of intersection of the lines $x - 7y + 5 = 0$ and $3x + y = 0$ .

Solution

Given: Line 1:

x - 7y + 5 = 0

Line 2:

3x + y = 0

The required line is parallel to the y-axis.

To Find: The equation of the line passing through the intersection of Line 1 and Line 2.

Solution: Step 1: Find the point of intersection of the two given lines. We need to solve the system of equations:

$x - 7y = -5$
$3x + y = 0$

From equation (2), we can express

y

in terms of

x

:

y = -3x

. Substitute this into equation (1):

x - 7(-3x) = -5

x + 21x = -5

22x = -5

x = -\frac{5}{22}

Now find the corresponding y-value:

y = -3x = -3(-\frac{5}{22}) = \frac{15}{22}

.

The point of intersection is

(-\frac{5}{22}, \frac{15}{22})

.

Step 2: Find the equation of the required line. The required line is parallel to the y-axis. A line parallel to the y-axis is a vertical line, and its equation is of the form

x = k

, where

k

is a constant. Since the line must pass through the point of intersection

(-\frac{5}{22}, \frac{15}{22})

, its x-coordinate must be constant and equal to

-\frac{5}{22}

.

So, the equation of the line is

x = -\frac{5}{22}

. This can be rewritten as

22x = -5

, or

22x + 5 = 0

.

Final Answer: The equation of the line is

22x + 5 = 0

.

Q6Miscellaneous Exercise on Chapter 9

Find the equation of a line drawn perpendicular to the line $\frac{x}{4} + \frac{y}{6} = 1$ through the point, where it meets the $y$ -axis.

Solution

Given: A line L1:

\frac{x}{4} + \frac{y}{6} = 1

. The required line L2 is perpendicular to L1. L2 passes through the point where L1 meets the y-axis.

To Find: The equation of line L2.

Solution: Step 1: Find the point of intersection of L1 with the y-axis. A line meets the y-axis where the x-coordinate is 0. Substitute

x=0

into the equation of L1:

\frac{0}{4} + \frac{y}{6} = 1

\frac{y}{6} = 1 \implies y = 6

. So, the point of intersection is

(0, 6)

. This is the point through which L2 passes.

Step 2: Find the slope of L1 ( $m_1$ ). Rewrite the equation of L1 in the slope-intercept form (

y = mx + c

).

\frac{x}{4} + \frac{y}{6} = 1

Multiply by 12 to clear fractions:

3x + 2y = 12

2y = -3x + 12

y = -\frac{3}{2}x + 6

The slope of L1 is

m_1 = -\frac{3}{2}

.

Step 3: Find the slope of the perpendicular line L2 ( $m_2$ ). Since L2 is perpendicular to L1, the product of their slopes is -1.

m_1 \times m_2 = -1

-\frac{3}{2} \times m_2 = -1

m_2 = \frac{2}{3}

.

Step 4: Find the equation of line L2. We know that L2 passes through the point

(0, 6)

and has a slope of

m_2 = \frac{2}{3}

. The point

(0, 6)

is the y-intercept, so

c = 6

. Using the slope-intercept form

y = mx + c

:

y = \frac{2}{3}x + 6

To write in general form, multiply by 3:

3y = 2x + 18

2x - 3y + 18 = 0

.

Final Answer: The equation of the line is

2x - 3y + 18 = 0

.

Q7Miscellaneous Exercise on Chapter 9

Find the area of the triangle formed by the lines $y - x = 0, x + y = 0$ and $x - k = 0$ .

Solution

Given: The equations of the three lines forming a triangle:

$y - x = 0 \implies y = x$
$x + y = 0 \implies y = -x$
$x - k = 0 \implies x = k$

To Find: The area of the triangle.

Solution: Step 1: Find the vertices of the triangle by finding the points of intersection of the lines.

Vertex A (Intersection of line 1 and 2):

y = x

and

y = -x

x = -x \implies 2x = 0 \implies x = 0

. Then

y = 0

. So, Vertex A is

(0, 0)

.

Vertex B (Intersection of line 1 and 3):

y = x

and

x = k

So,

y = k

. Vertex B is

(k, k)

.

Vertex C (Intersection of line 2 and 3):

y = -x

and

x = k

So,

y = -k

. Vertex C is

(k, -k)

.

Step 2: Calculate the area of the triangle with vertices A(0, 0), B(k, k), C(k, -k). Using the area formula for vertices

(x_1, y_1), (x_2, y_2), (x_3, y_3)

:

\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

\text{Area} = \frac{1}{2} |0(k - (-k)) + k(-k - 0) + k(0 - k)|

\text{Area} = \frac{1}{2} |0 + k(-k) + k(-k)|

\text{Area} = \frac{1}{2} |-k^2 - k^2|

\text{Area} = \frac{1}{2} |-2k^2|

Since area must be positive, and

k^2

is always non-negative:

\text{Area} = \frac{1}{2} (2k^2) = k^2

Final Answer: The area of the triangle is

k^2

square units.

Q8Miscellaneous Exercise on Chapter 9

Find the value of $p$ so that the three lines $3x + y - 2 = 0, px + 2y - 3 = 0$ and $2x - y - 3 = 0$ may intersect at one point.

Solution

Given: Three lines:

$3x + y - 2 = 0$
$px + 2y - 3 = 0$
$2x - y - 3 = 0$ The three lines intersect at one point (they are concurrent).

To Find: The value of

p

.

Method: If the three lines are concurrent, they all pass through the same point. We can find the point of intersection of two of the lines (the ones without

p

) and then substitute that point into the third line's equation to find

p

.

Solution: Step 1: Find the point of intersection of lines 1 and 3.

3x + y - 2 = 0 \quad ...(i)

2x - y - 3 = 0 \quad ...(ii)

Add equation (i) and (ii) to eliminate

y

:

(3x + y) + (2x - y) = 2 + 3

5x = 5

x = 1

Substitute

x=1

into equation (i) to find

y

:

3(1) + y - 2 = 0

3 + y - 2 = 0

y + 1 = 0

y = -1

The point of intersection is

(1, -1)

.

Step 2: Substitute the intersection point into the second line's equation. Since the three lines are concurrent, the point

(1, -1)

must lie on the second line,

px + 2y - 3 = 0

. Substitute

x=1

and

y=-1

into this equation:

p(1) + 2(-1) - 3 = 0

p - 2 - 3 = 0

p - 5 = 0

p = 5

Final Answer: The value of

p

is 5.

Q9Miscellaneous Exercise on Chapter 9

If three lines whose equations are $y = m_1 x + c_1, y = m_2 x + c_2$ and $y = m_3 x + c_3$ are concurrent, then show that $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$ .

Solution

Given: Three concurrent lines:

$y = m_1 x + c_1$
$y = m_2 x + c_2$
$y = m_3 x + c_3$

To Show:

m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0

.

Proof: Step 1: Find the point of intersection of lines 1 and 2. Since the lines are concurrent, they intersect at a single point

(x_0, y_0)

. At this point:

y_0 = m_1 x_0 + c_1

and

y_0 = m_2 x_0 + c_2

. Setting the expressions for

y_0

equal:

m_1 x_0 + c_1 = m_2 x_0 + c_2

m_1 x_0 - m_2 x_0 = c_2 - c_1

x_0(m_1 - m_2) = c_2 - c_1

x_0 = \frac{c_2 - c_1}{m_1 - m_2}

Now find

y_0

by substituting

x_0

back into the equation for line 1:

y_0 = m_1 (\frac{c_2 - c_1}{m_1 - m_2}) + c_1

y_0 = \frac{m_1(c_2 - c_1) + c_1(m_1 - m_2)}{m_1 - m_2}

y_0 = \frac{m_1 c_2 - m_1 c_1 + c_1 m_1 - c_1 m_2}{m_1 - m_2}

y_0 = \frac{m_1 c_2 - c_1 m_2}{m_1 - m_2}

So the point of intersection is

(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1 c_2 - m_2 c_1}{m_1 - m_2})

.

Step 2: Substitute this point into the equation for line 3. Since line 3 also passes through this point, its coordinates must satisfy

y = m_3 x + c_3

.

\frac{m_1 c_2 - m_2 c_1}{m_1 - m_2} = m_3 (\frac{c_2 - c_1}{m_1 - m_2}) + c_3

Multiply the entire equation by

(m_1 - m_2)

:

m_1 c_2 - m_2 c_1 = m_3 (c_2 - c_1) + c_3 (m_1 - m_2)

m_1 c_2 - m_2 c_1 = m_3 c_2 - m_3 c_1 + c_3 m_1 - c_3 m_2

Step 3: Rearrange the terms to get the desired expression. Move all terms to one side:

m_1 c_2 - m_2 c_1 - m_3 c_2 + m_3 c_1 - c_3 m_1 + c_3 m_2 = 0

Group the terms by

m_1, m_2, m_3

:

m_1(c_2 - c_3) - m_2(c_1 - c_3) + m_3(c_1 - c_2) = 0

m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0

Hence Shown.

Q10Miscellaneous Exercise on Chapter 9

Find the equation of the lines through the point $(3, 2)$ which make an angle of $45^{\circ}$ with the line $x - 2y = 3$ .

Solution

Given: Required lines pass through the point

(3, 2)

. The angle with the line

x - 2y = 3

is

\theta = 45^{\circ}

.

To Find: The equations of the lines.

Solution: Step 1: Find the slope of the given line. Line:

x - 2y = 3 \implies 2y = x - 3 \implies y = \frac{1}{2}x - \frac{3}{2}

. The slope of this line is

m_1 = \frac{1}{2}

.

Step 2: Find the slopes of the required lines. Let the slope of a required line be

m

. The angle between the lines is

45^{\circ}

. Using the angle formula:

\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|

. We know

\tan 45^{\circ} = 1

.

1 = \left| \frac{m - \frac{1}{2}}{1 + m(\frac{1}{2})} \right| = \left| \frac{\frac{2m - 1}{2}}{\frac{2 + m}{2}} \right| = \left| \frac{2m - 1}{m + 2} \right|

This gives two possible cases:

Case 1:

\frac{2m - 1}{m + 2} = 1

2m - 1 = m + 2

m = 3

Case 2:

\frac{2m - 1}{m + 2} = -1

2m - 1 = -(m + 2)

2m - 1 = -m - 2

3m = -1

m = -\frac{1}{3}

So, there are two possible slopes for the required lines:

3

and

-\frac{1}{3}

.

Step 3: Find the equations of the lines. Both lines pass through the point

(3, 2)

. Use the point-slope form

y - y_0 = m(x - x_0)

.

Line 1 (with slope m = 3):

y - 2 = 3(x - 3)

y - 2 = 3x - 9

3x - y - 7 = 0

Line 2 (with slope m = -1/3):

y - 2 = -\frac{1}{3}(x - 3)

3(y - 2) = -(x - 3)

3y - 6 = -x + 3

x + 3y - 9 = 0

Final Answer: The equations of the two lines are

3x - y - 7 = 0

and

x + 3y - 9 = 0

.

Q11Miscellaneous Exercise on Chapter 9

Find the equation of the line passing through the point of intersection of the lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ that has equal intercepts on the axes.

Solution

Given: Line 1:

L_1 \equiv 4x + 7y - 3 = 0

Line 2:

L_2 \equiv 2x - 3y + 1 = 0

The required line has equal intercepts on the axes.

To Find: The equation of the required line.

Method 1: Find the intersection point first.

Step 1.1: Solve the system of equations.

$4x + 7y = 3$
$2x - 3y = -1$ Multiply equation (2) by 2: $4x - 6y = -2$ . Subtract this new equation from equation (1): $(4x + 7y) - (4x - 6y) = 3 - (-2)$ $13y = 5 \implies y = \frac{5}{13}$ . Substitute $y$ back into equation (2): $2x - 3(\frac{5}{13}) = -1 \implies 2x - \frac{15}{13} = -1 \implies 2x = \frac{15}{13} - 1 = \frac{2}{13} \implies x = \frac{1}{13}$ . The intersection point is $(\frac{1}{13}, \frac{5}{13})$ .

Step 1.2: Use the equal intercepts condition. A line with equal intercepts 'a' has the equation

\frac{x}{a} + \frac{y}{a} = 1

, which is

x + y = a

. Since this line passes through

(\frac{1}{13}, \frac{5}{13})

:

\frac{1}{13} + \frac{5}{13} = a \implies a = \frac{6}{13}

. The equation is

x + y = \frac{6}{13}

, or

13x + 13y = 6

.

Another case for equal intercepts is

b = -a

, leading to

x - y = a

.

(\frac{1}{13}) - (\frac{5}{13}) = a \implies a = -\frac{4}{13}

. The equation is

x - y = -\frac{4}{13}

, or

13x - 13y = -4

.

Method 2: Using the family of lines.

Step 2.1: Write the equation of the family of lines. Any line passing through the intersection of

L_1

and

L_2

can be written as

L_1 + k L_2 = 0

.

(4x + 7y - 3) + k(2x - 3y + 1) = 0

(4 + 2k)x + (7 - 3k)y + (k - 3) = 0

.

Step 2.2: Apply the equal intercepts condition. For a line

Ax + By + C = 0

, the x-intercept is

-C/A

and y-intercept is

-C/B

. If intercepts are equal:

-\frac{C}{A} = -\frac{C}{B} \implies A = B

(assuming

C \neq 0

).

4 + 2k = 7 - 3k \implies 5k = 3 \implies k = \frac{3}{5}

. Substitute

k=3/5

back into the family equation:

(4 + 2(\frac{3}{5}))x + (7 - 3(\frac{3}{5}))y + (\frac{3}{5} - 3) = 0

(\frac{20+6}{5})x + (\frac{35-9}{5})y + (\frac{3-15}{5}) = 0

\frac{26}{5}x + \frac{26}{5}y - \frac{12}{5} = 0

26x + 26y - 12 = 0 \implies 13x + 13y - 6 = 0

.

If intercepts are equal in magnitude but opposite in sign:

A = -B

.

4 + 2k = -(7 - 3k) \implies 4 + 2k = -7 + 3k \implies k = 11

. Substitute

k=11

:

(4 + 22)x + (7 - 33)y + (11 - 3) = 0

26x - 26y + 8 = 0 \implies 13x - 13y + 4 = 0

.

Final Answer: The equations of the lines are

13x + 13y - 6 = 0

and

13x - 13y + 4 = 0

.

Q12Miscellaneous Exercise on Chapter 9

Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is $\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}$ .

Solution

Given: A line

L_1

:

y = mx + c

. Its slope is

m_1 = m

. A required line

L_2

passes through the origin

(0, 0)

. The angle between

L_1

and

L_2

is

\theta

.

To Show: The equation of

L_2

is

\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}

.

Proof: Step 1: Define the required line $L_2$ . Any line passing through the origin has the form

y = m'x

, where

m'

is its slope. We can write this as

\frac{y}{x} = m'

. So, we need to find the possible values for

m'

.

Step 2: Use the angle formula. The tangent of the angle between two lines with slopes

m

and

m'

is given by:

\tan \theta = \left| \frac{m' - m}{1 + m'm} \right|

This absolute value equation leads to two possibilities:

Case 1:

\tan \theta = \frac{m' - m}{1 + m'm}

an \theta (1 + m'm) = m' - m

an \theta + m'm \tan \theta = m' - m

m + \tan \theta = m' - m'm \tan \theta

m + \tan \theta = m'(1 - m \tan \theta)

m' = \frac{m + \tan \theta}{1 - m \tan \theta}

Case 2:

\tan \theta = -\left( \frac{m' - m}{1 + m'm} \right) = \frac{m - m'}{1 + m'm}

an \theta (1 + m'm) = m - m'

an \theta + m'm \tan \theta = m - m'

m' + m'm \tan \theta = m - \tan \theta

m'(1 + m \tan \theta) = m - \tan \theta

m' = \frac{m - \tan \theta}{1 + m \tan \theta}

Step 3: Combine the results. We have two possible values for the slope

m'

of the line passing through the origin:

m' = \frac{m + \tan \theta}{1 - m \tan \theta} \quad \text{and} \quad m' = \frac{m - \tan \theta}{1 + m \tan \theta}

These can be written compactly as:

m' = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}

Since the equation of the line is

y = m'x

, we have

m' = \frac{y}{x}

. Therefore,

\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}

Hence Shown.

Q13Miscellaneous Exercise on Chapter 9

In what ratio, the line joining $(-1, 1)$ and $(5, 7)$ is divided by the line $x + y = 4$ ?

Solution

Given: A line segment with endpoints A

(-1, 1)

and B

(5, 7)

. A dividing line L:

x + y = 4

.

To Find: The ratio in which the line L divides the segment AB.

Solution: Let the line

x + y = 4

divide the line segment joining A and B at a point P in the ratio

k:1

.

Using the section formula, the coordinates of the point P are:

P = \left( \frac{k(5) + 1(-1)}{k+1}, \frac{k(7) + 1(1)}{k+1} \right)

P = \left( \frac{5k - 1}{k+1}, \frac{7k + 1}{k+1} \right)

Since the point P lies on the line

x + y = 4

, its coordinates must satisfy this equation. Substitute the x and y coordinates of P into the line equation:

\left( \frac{5k - 1}{k+1} \right) + \left( \frac{7k + 1}{k+1} \right) = 4

Since the denominators are the same, we can add the numerators:

\frac{(5k - 1) + (7k + 1)}{k+1} = 4

\frac{12k}{k+1} = 4

Now, solve for

k

(assuming

k+1 \neq 0

):

12k = 4(k+1)

12k = 4k + 4

8k = 4

k = \frac{4}{8} = \frac{1}{2}

The ratio is

k:1

, which is

\frac{1}{2}:1

. To express this with integers, we can multiply both sides by 2, giving the ratio

1:2

.

Since

k

is positive, the division is internal.

Final Answer: The line divides the segment in the ratio

1:2

.

Q14Miscellaneous Exercise on Chapter 9

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$ .

Solution

Given: A point P

(1, 2)

. A target line

L_1

:

4x + 7y + 5 = 0

. A path line

L_2

:

2x - y = 0

.

To Find: The distance from P to

L_1

measured along

L_2

. This is the distance between point P and the intersection of lines

L_1

and

L_2

.

Solution: Step 1: Find the point of intersection of lines $L_1$ and $L_2$ . Let the intersection point be Q.

$4x + 7y = -5$
$2x - y = 0 \implies y = 2x$

Substitute the expression for

y

from equation (2) into equation (1):

4x + 7(2x) = -5

4x + 14x = -5

18x = -5

x = -\frac{5}{18}

Now find the corresponding y-value:

y = 2x = 2(-\frac{5}{18}) = -\frac{10}{18} = -\frac{5}{9}

.

The point of intersection is Q

(-\frac{5}{18}, -\frac{5}{9})

.

Step 2: Find the distance between point P(1, 2) and point Q(-5/18, -5/9). Using the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

:

d = \sqrt{\left(-\frac{5}{18} - 1\right)^2 + \left(-\frac{5}{9} - 2\right)^2}

d = \sqrt{\left(-\frac{5}{18} - \frac{18}{18}\right)^2 + \left(-\frac{5}{9} - \frac{18}{9}\right)^2}

d = \sqrt{\left(-\frac{23}{18}\right)^2 + \left(-\frac{23}{9}\right)^2}

d = \sqrt{\frac{23^2}{18^2} + \frac{23^2}{9^2}}

Factor out

23^2

:

d = \sqrt{23^2 \left(\frac{1}{18^2} + \frac{1}{9^2}\right)} = 23 \sqrt{\frac{1}{324} + \frac{1}{81}}

Find a common denominator:

d = 23 \sqrt{\frac{1}{324} + \frac{4}{324}} = 23 \sqrt{\frac{5}{324}}

d = 23 \frac{\sqrt{5}}{\sqrt{324}} = 23 \frac{\sqrt{5}}{18} = \frac{23\sqrt{5}}{18}

Final Answer: The distance is

\frac{23\sqrt{5}}{18}

units.

Q15Miscellaneous Exercise on Chapter 9

Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of 3 units from this point.

Solution

Given: Starting point P

(-1, 2)

. Target line L:

x + y = 4

. The distance from P to the intersection point Q on L is 3 units.

To Find: The direction of the line PQ, which can be represented by its slope.

Solution: Let the required line be drawn at an angle

\theta

with the x-axis. Its slope is

m = \tan \theta

. The parametric form of a line passing through

(x_1, y_1)

at an angle

\theta

is:

\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta} = r

, where

r

is the distance from

(x_1, y_1)

.

The coordinates of any point Q on this line at a distance

r

from P

(-1, 2)

are:

x = x_1 + r \cos \theta = -1 + r \cos \theta

y = y_1 + r \sin \theta = 2 + r \sin \theta

We are given that the distance

r=3

. So the intersection point Q has coordinates:

Q(-1 + 3\cos \theta, 2 + 3\sin \theta)

.

This point Q must lie on the line

x + y = 4

. Substitute the coordinates of Q into this equation:

(-1 + 3\cos \theta) + (2 + 3\sin \theta) = 4

1 + 3\cos \theta + 3\sin \theta = 4

3(\cos \theta + \sin \theta) = 3

\cos \theta + \sin \theta = 1

To solve for

\theta

, we can square both sides, but it is easier to use the R-formula. Divide by

\sqrt{1^2 + 1^2} = \sqrt{2}

:

\frac{1}{\sqrt{2}}\cos \theta + \frac{1}{\sqrt{2}}\sin \theta = \frac{1}{\sqrt{2}}

\cos 45^{\circ} \cos \theta + \sin 45^{\circ} \sin \theta = \frac{1}{\sqrt{2}}

\cos(\theta - 45^{\circ}) = \cos 45^{\circ}

This gives two possibilities:

$\theta - 45^{\circ} = 45^{\circ} \implies \theta = 90^{\circ}$
$\theta - 45^{\circ} = -45^{\circ} \implies \theta = 0^{\circ}$

The direction is given by the angle

\theta

. An angle of

0^{\circ}

means the line is horizontal (parallel to x-axis). An angle of

90^{\circ}

means the line is vertical (parallel to y-axis).

The slope

m = \tan \theta

. If

\theta = 0^{\circ}

, the slope is

m = \tan 0^{\circ} = 0

. If

\theta = 90^{\circ}

, the slope is undefined.

Final Answer: The line must be drawn either horizontally (parallel to the x-axis, slope 0) or vertically (parallel to the y-axis, slope undefined).

Q16Miscellaneous Exercise on Chapter 9

The hypotenuse of a right angled triangle has its ends at the points $(1, 3)$ and $(-4, 1)$ . Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Solution

Given: A right-angled triangle. The endpoints of the hypotenuse are A

(1, 3)

and B

(-4, 1)

. The legs of the triangle are parallel to the coordinate axes.

To Find: The equations of the legs.

Solution: Let the vertices of the triangle be A, B, and C, where C is the vertex with the right angle. Since the legs are parallel to the axes, one leg must be a horizontal line and the other must be a vertical line.

The horizontal leg will have the equation

y = \text{constant}

. The vertical leg will have the equation

x = \text{constant}

.

The vertex C where the legs meet must share its x-coordinate with one of the hypotenuse endpoints and its y-coordinate with the other.

There are two possibilities for the coordinates of vertex C:

Case 1: C has the x-coordinate of A and the y-coordinate of B. Coordinates of C are

(1, 1)

. The legs are the segments AC and BC.

Leg AC: This is a vertical line segment connecting A $(1, 3)$ and C $(1, 1)$ . The equation of the line containing this leg is $x = 1$ .
Leg BC: This is a horizontal line segment connecting B $(-4, 1)$ and C $(1, 1)$ . The equation of the line containing this leg is $y = 1$ .

Case 2: C has the x-coordinate of B and the y-coordinate of A. Coordinates of C are

(-4, 3)

. The legs are the segments AC and BC.

Leg AC: This is a horizontal line segment connecting A $(1, 3)$ and C $(-4, 3)$ . The equation of the line containing this leg is $y = 3$ .
Leg BC: This is a vertical line segment connecting B $(-4, 1)$ and C $(-4, 3)$ . The equation of the line containing this leg is $x = -4$ .

So, there are two possible sets of legs for the triangle.

Final Answer: The equations of the legs are either (

x=1

and

y=1

) or (

x=-4

and

y=3

).

Q17Miscellaneous Exercise on Chapter 9

Find the image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ assuming the line to be a plane mirror.

Solution

Given: The point P

(3, 8)

. The line (mirror) L:

x + 3y - 7 = 0

.

To Find: The image of point P, let's call it Q

(h, k)

.

Method: The line L is the perpendicular bisector of the segment PQ.

The midpoint of PQ lies on the line L.
The line segment PQ is perpendicular to the line L.

Solution: Step 1: Use the midpoint condition. The midpoint M of PQ is

M = (\frac{3+h}{2}, \frac{8+k}{2})

. Since M lies on the line

x + 3y - 7 = 0

, its coordinates satisfy the equation:

(\frac{3+h}{2}) + 3(\frac{8+k}{2}) - 7 = 0

Multiply by 2 to clear the denominators:

(3+h) + 3(8+k) - 14 = 0

3 + h + 24 + 3k - 14 = 0

h + 3k + 13 = 0 \quad ...(1)

Step 2: Use the perpendicularity condition. First, find the slope of the line L (

m_L

).

3y = -x + 7 \implies y = -\frac{1}{3}x + \frac{7}{3}

. So,

m_L = -\frac{1}{3}

.

Next, find the slope of the segment PQ (

m_{PQ}

).

m_{PQ} = \frac{k - 8}{h - 3}

Since PQ is perpendicular to L, the product of their slopes is -1.

m_{PQ} \times m_L = -1

\frac{k - 8}{h - 3} \times (-\frac{1}{3}) = -1

\frac{k - 8}{h - 3} = 3

k - 8 = 3(h - 3)

k - 8 = 3h - 9

3h - k - 1 = 0 \quad ...(2)

Step 3: Solve the system of linear equations (1) and (2). From equation (2),

k = 3h - 1

. Substitute this into equation (1):

h + 3(3h - 1) + 13 = 0

h + 9h - 3 + 13 = 0

10h + 10 = 0

10h = -10 \implies h = -1

.

Now find

k

using

k = 3h - 1

:

k = 3(-1) - 1 = -3 - 1 = -4

.

The coordinates of the image point Q are

(-1, -4)

.

Final Answer: The image of the point

(3, 8)

is

(-1, -4)

.

Q18Miscellaneous Exercise on Chapter 9

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$ , find the value of $m$ .

Solution

Given: Line 1 (

L_1

):

y = 3x + 1

, with slope

m_1 = 3

. Line 2 (

L_2

):

2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2}

, with slope

m_2 = \frac{1}{2}

. Line 3 (

L_3

):

y = mx + 4

, with slope

m_3 = m

.

L_1

and

L_2

are equally inclined to

L_3

.

To Find: The value of

m

.

Condition: If two lines are equally inclined to a third line, then the angle between

L_1

and

L_3

is equal to the angle between

L_2

and

L_3

. Let

\theta_1

be the angle between

L_1

and

L_3

, and

\theta_2

be the angle between

L_2

and

L_3

. Then

\tan \theta_1 = \tan \theta_2

.

\left| \frac{m - m_1}{1 + m m_1} \right| = \left| \frac{m - m_2}{1 + m m_2} \right|

Solution: Substitute the slope values:

\left| \frac{m - 3}{1 + 3m} \right| = \left| \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} \right| = \left| \frac{2m - 1}{2 + m} \right|

This gives two possible cases:

Case 1:

\frac{m - 3}{1 + 3m} = \frac{2m - 1}{m + 2}

(m - 3)(m + 2) = (2m - 1)(1 + 3m)

m^2 - m - 6 = 2m + 6m^2 - 1 - 3m

m^2 - m - 6 = 6m^2 - m - 1

5m^2 + 5 = 0

m^2 = -1

This has no real solutions for

m

.

Case 2:

\frac{m - 3}{1 + 3m} = -\left( \frac{2m - 1}{m + 2} \right) = \frac{1 - 2m}{m + 2}

(m - 3)(m + 2) = (1 - 2m)(1 + 3m)

m^2 - m - 6 = 1 + 3m - 2m - 6m^2

m^2 - m - 6 = -6m^2 + m + 1

7m^2 - 2m - 7 = 0

This is a quadratic equation for

m

. We can use the quadratic formula

m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

:

a = 7, b = -2, c = -7

m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)}

m = \frac{2 \pm \sqrt{4 + 196}}{14}

m = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}

Final Answer: The value of

m

is

\frac{1 \pm 5\sqrt{2}}{7}

.

Q19Miscellaneous Exercise on Chapter 9

If sum of the perpendicular distances of a variable point $\mathrm{P}(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always 10. Show that P must move on a line.

Solution

Given: A variable point P

(x, y)

. Line 1 (

L_1

):

x + y - 5 = 0

. Line 2 (

L_2

):

3x - 2y + 7 = 0

. The sum of the perpendicular distances from P to

L_1

and

L_2

is always 10.

To Show: P must move on a line.

Proof: Let

d_1

be the distance from P to

L_1

and

d_2

be the distance from P to

L_2

.

d_1 = \frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}

d_2 = \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{13}}

We are given that

d_1 + d_2 = 10

.

\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10

The expressions inside the absolute value signs,

(x+y-5)

and

(3x-2y+7)

, can be positive or negative. This leads to four possible cases depending on the signs.

Case 1: Assume

(x + y - 5) \geq 0

and

(3x - 2y + 7) \geq 0

.

\frac{x + y - 5}{\sqrt{2}} + \frac{3x - 2y + 7}{\sqrt{13}} = 10

Multiply by

\sqrt{2}\sqrt{13}

to clear denominators:

\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}

(\sqrt{13} + 3\sqrt{2})x + (\sqrt{13} - 2\sqrt{2})y + (7\sqrt{2} - 5\sqrt{13} - 10\sqrt{26}) = 0

This is an equation of the form

Ax + By + C = 0

, which represents a straight line.

Case 2: Assume

(x + y - 5) < 0

and

(3x - 2y + 7) \geq 0

.

\frac{-(x + y - 5)}{\sqrt{2}} + \frac{3x - 2y + 7}{\sqrt{13}} = 10

-\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}

(3\sqrt{2} - \sqrt{13})x - (\sqrt{13} + 2\sqrt{2})y + (5\sqrt{13} + 7\sqrt{2} - 10\sqrt{26}) = 0

This is also an equation of a straight line.

Case 3: Assume

(x + y - 5) \geq 0

and

(3x - 2y + 7) < 0

.

\frac{x + y - 5}{\sqrt{2}} - \frac{3x - 2y + 7}{\sqrt{13}} = 10

This will also result in a linear equation.

Case 4: Assume

(x + y - 5) < 0

and

(3x - 2y + 7) < 0

.

-\frac{x + y - 5}{\sqrt{2}} - \frac{3x - 2y + 7}{\sqrt{13}} = 10

This will also result in a linear equation.

In all possible cases, the locus of the point P

(x,y)

is described by a linear equation of the form

Ax + By + C = 0

. Therefore, the point P must move on a straight line (or a path composed of segments of these lines).

Hence Shown.

Q20Miscellaneous Exercise on Chapter 9

Find equation of the line which is equidistant from parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$ .

Solution

Given: Two parallel lines: Line 1 (

L_1

):

9x + 6y - 7 = 0

Line 2 (

L_2

):

3x + 2y + 6 = 0

To Find: The equation of the line that is equidistant from

L_1

and

L_2

.

Method: The line equidistant from two parallel lines is itself parallel to them and lies exactly in the middle. Its constant term will be the average of the constant terms of the given lines (when their x and y coefficients are made identical).

Solution: Step 1: Make the coefficients of x and y identical for both lines. The equation for

L_1

is

9x + 6y - 7 = 0

. The equation for

L_2

is

3x + 2y + 6 = 0

. We can multiply this by 3 to match the coefficients of

L_1

. Modified

L_2

:

3(3x + 2y + 6) = 3(0) \implies 9x + 6y + 18 = 0

.

Now we have two parallel lines in the form

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

:

L_1: 9x + 6y - 7 = 0

(so

C_1 = -7

)

L_2: 9x + 6y + 18 = 0

(so

C_2 = 18

)

Step 2: Find the equation of the equidistant line. The required line will have the form

9x + 6y + C = 0

, where

C

is the arithmetic mean of

C_1

and

C_2

.

C = \frac{C_1 + C_2}{2} = \frac{-7 + 18}{2} = \frac{11}{2}

So the equation of the equidistant line is:

9x + 6y + \frac{11}{2} = 0

To remove the fraction, multiply the entire equation by 2:

18x + 12y + 11 = 0

Final Answer: The equation of the line is

18x + 12y + 11 = 0

.

Q21Miscellaneous Exercise on Chapter 9

A ray of light passing through the point $(1, 2)$ reflects on the $x$ -axis at point A and the reflected ray passes through the point $(5, 3)$ . Find the coordinates of A.

Solution

Given: Initial point of the light ray: P

(1, 2)

. The ray reflects on the x-axis at point A. Point on the reflected ray: Q

(5, 3)

.

To Find: The coordinates of point A.

Method: The principle of reflection states that the angle of incidence equals the angle of reflection. In coordinate geometry, this implies that the reflected ray, when extended backward, appears to come from the image of the initial point reflected across the mirror line.

Solution: Step 1: Find the image of the initial point P with respect to the mirror (x-axis). The mirror is the x-axis, whose equation is

y=0

. The image of a point

(x, y)

reflected across the x-axis is

(x, -y)

. So, the image of P

(1, 2)

is P'

(1, -2)

.

Step 2: The reflected ray lies on the line connecting the image point P' and the point Q. The points P', A, and Q are collinear. Point A lies on the x-axis, so its coordinates are

(x, 0)

. We can find the equation of the line passing through P'

(1, -2)

and Q

(5, 3)

.

First, find the slope of the line P'Q:

m = \frac{3 - (-2)}{5 - 1} = \frac{5}{4}

Now, use the point-slope form with point P'

(1, -2)

:

y - (-2) = \frac{5}{4}(x - 1)

y + 2 = \frac{5}{4}(x - 1)

Step 3: Find the coordinates of point A. Point A is the intersection of this line with the x-axis. At the x-axis,

y=0

. Substitute

y=0

into the equation of the line:

0 + 2 = \frac{5}{4}(x - 1)

2 = \frac{5}{4}(x - 1)

8 = 5(x - 1)

8 = 5x - 5

13 = 5x

x = \frac{13}{5}

So, the coordinates of point A are

(\frac{13}{5}, 0)

.

Final Answer: The coordinates of A are

(\frac{13}{5}, 0)

.

Q22Miscellaneous Exercise on Chapter 9

Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^2 - b^2}, 0)$ and $(-\sqrt{a^2 - b^2}, 0)$ to the line $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$ is $b^2$ .

Solution

Given: Point 1:

P_1(\sqrt{a^2 - b^2}, 0)

Point 2:

P_2(-\sqrt{a^2 - b^2}, 0)

Line L:

\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1

To Prove: The product of the lengths of the perpendiculars from

P_1

and

P_2

to L is

b^2

.

Proof: Step 1: Write the equation of the line in the general form $Ax+By+C=0$ .

\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta - 1 = 0

Here,

A = \frac{\cos \theta}{a}

,

B = \frac{\sin \theta}{b}

, and

C = -1

.

Step 2: Calculate the length of the perpendicular from $P_1$ ( $d_1$ ). Using the distance formula

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

:

d_1 = \frac{|\frac{\cos \theta}{a}(\sqrt{a^2 - b^2}) + \frac{\sin \theta}{b}(0) - 1|}{\sqrt{(\frac{\cos \theta}{a})^2 + (\frac{\sin \theta}{b})^2}} = \frac{|\frac{\sqrt{a^2 - b^2}\cos \theta}{a} - 1|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}}

Step 3: Calculate the length of the perpendicular from $P_2$ ( $d_2$ ).

d_2 = \frac{|\frac{\cos \theta}{a}(-\sqrt{a^2 - b^2}) + \frac{\sin \theta}{b}(0) - 1|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} = \frac{|-\frac{\sqrt{a^2 - b^2}\cos \theta}{a} - 1|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}}

Step 4: Calculate the product $d_1 \times d_2$ .

d_1 d_2 = \frac{\left|\left(\frac{\sqrt{a^2 - b^2}\cos \theta}{a} - 1\right) \left(-\frac{\sqrt{a^2 - b^2}\cos \theta}{a} - 1\right)\right|}{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}

The numerator is of the form

|(X-1)(-X-1)| = |-(X-1)(X+1)| = |-(X^2-1)| = |1-X^2|

. Let

X = \frac{\sqrt{a^2 - b^2}\cos \theta}{a}

. Numerator =

\left| 1 - \left(\frac{\sqrt{a^2 - b^2}\cos \theta}{a}\right)^2 \right| = \left| 1 - \frac{(a^2 - b^2)\cos^2 \theta}{a^2} \right|

Numerator =

\left| \frac{a^2 - (a^2 - b^2)\cos^2 \theta}{a^2} \right| = \left| \frac{a^2 - a^2\cos^2 \theta + b^2\cos^2 \theta}{a^2} \right|

Numerator =

\left| \frac{a^2(1 - \cos^2 \theta) + b^2\cos^2 \theta}{a^2} \right| = \left| \frac{a^2\sin^2 \theta + b^2\cos^2 \theta}{a^2} \right|

Since the terms are squared, the expression is positive, so we can remove the absolute value. Numerator =

\frac{a^2\sin^2 \theta + b^2\cos^2 \theta}{a^2}

.

Denominator =

\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} = \frac{b^2\cos^2 \theta + a^2\sin^2 \theta}{a^2 b^2}

.

Now, the product is:

d_1 d_2 = \frac{\frac{a^2\sin^2 \theta + b^2\cos^2 \theta}{a^2}}{\frac{b^2\cos^2 \theta + a^2\sin^2 \theta}{a^2 b^2}}

d_1 d_2 = \frac{a^2\sin^2 \theta + b^2\cos^2 \theta}{a^2} \times \frac{a^2 b^2}{b^2\cos^2 \theta + a^2\sin^2 \theta}

The term

(a^2\sin^2 \theta + b^2\cos^2 \theta)

cancels out.

d_1 d_2 = \frac{a^2 b^2}{a^2} = b^2

Hence Proved.

Q23Miscellaneous Exercise on Chapter 9

A person standing at the junction (crossing) of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find equation of the path that he should follow.

Solution

Given: Path 1:

2x - 3y + 4 = 0

Path 2:

3x + 4y - 5 = 0

Destination Path:

6x - 7y + 8 = 0

To Find: The equation of the path the person should follow for the least time.

Method: The path of least time is the shortest path, which is the perpendicular path from the person's starting point to the destination path.

Solution: Step 1: Find the starting point of the person (the junction of Path 1 and Path 2). We need to solve the system of equations:

$2x - 3y = -4$
$3x + 4y = 5$

Multiply equation (1) by 4 and equation (2) by 3:

4 \times (2x - 3y = -4) \implies 8x - 12y = -16

3 \times (3x + 4y = 5) \implies 9x + 12y = 15

Add the two new equations to eliminate y:

(8x - 12y) + (9x + 12y) = -16 + 15

17x = -1

x = -\frac{1}{17}

Substitute

x = -1/17

into equation (1):

2(-\frac{1}{17}) - 3y = -4

-\frac{2}{17} - 3y = -4

-3y = -4 + \frac{2}{17} = \frac{-68 + 2}{17} = -\frac{66}{17}

y = \frac{-66}{17 \times (-3)} = \frac{22}{17}

The starting point is P

(-\frac{1}{17}, \frac{22}{17})

.

Step 2: Find the equation of the path to follow. This path must pass through P and be perpendicular to the destination path

6x - 7y + 8 = 0

.

First, find the slope of the destination path (

m_{dest}

).

6x + 8 = 7y \implies y = \frac{6}{7}x + \frac{8}{7}

. So,

m_{dest} = \frac{6}{7}

.

Let the slope of the required path be

m_{path}

. Since it's perpendicular:

m_{path} \times m_{dest} = -1

m_{path} \times \frac{6}{7} = -1 \implies m_{path} = -\frac{7}{6}

.

Now, find the equation of the path using the point-slope form with point P

(-\frac{1}{17}, \frac{22}{17})

and slope

m_{path} = -\frac{7}{6}

.

y - \frac{22}{17} = -\frac{7}{6} \left( x - (-\frac{1}{17}) \right)

y - \frac{22}{17} = -\frac{7}{6} \left( x + \frac{1}{17} \right)

\frac{17y - 22}{17} = -\frac{7}{6} \left( \frac{17x + 1}{17} \right)

Multiply both sides by 17:

17y - 22 = -\frac{7}{6} (17x + 1)

Multiply both sides by 6:

6(17y - 22) = -7(17x + 1)

102y - 132 = -119x - 7

119x + 102y - 132 + 7 = 0

119x + 102y - 125 = 0

Final Answer: The equation of the path he should follow is

119x + 102y - 125 = 0

.

Straight LinesClass 11 Mathematics NCERT Solutions

Draw a quadrilateral in the Cartesian plane, whose vertices are (−4,5),(0,7),(5,−5)(-4, 5), (0, 7), (5, -5)(−4,5),(0,7),(5,−5) and (−4,−2)(-4, -2)(−4,−2). Also, find its area.

Solution

The base of an equilateral triangle with side 2a2a2a lies along the yyy-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Solution

Find the distance between P(x1,y1)\mathrm{P}(x_{1}, y_{1})P(x1​,y1​) and Q(x2,y2)\mathrm{Q}(x_{2}, y_{2})Q(x2​,y2​) when : (i) PQ is parallel to the yyy-axis, (ii) PQ is parallel to the xxx-axis.

Solution

Find a point on the xxx-axis, which is equidistant from the points (7,6)(7, 6)(7,6) and (3,4)(3, 4)(3,4).

Solution

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,−4)\mathrm{P}(0,-4)P(0,−4) and B(8,0)\mathrm{B}(8,0)B(8,0).

Solution

Without using the Pythagoras theorem, show that the points (4,4),(3,5)(4, 4), (3, 5)(4,4),(3,5) and (−1,−1)(-1, -1)(−1,−1) are the vertices of a right angled triangle.

Solution

Find the slope of the line, which makes an angle of 30∘30^{\circ}30∘ with the positive direction of yyy-axis measured anticlockwise.

Solution

Without using distance formula, show that points (−2,−1),(4,0),(3,3)(-2, -1), (4, 0), (3, 3)(−2,−1),(4,0),(3,3) and (−3,2)(-3, 2)(−3,2) are the vertices of a parallelogram.

Solution

Find the angle between the xxx-axis and the line joining the points (3,−1)(3, -1)(3,−1) and (4,−2)(4, -2)(4,−2).

Solution

The slope of a line is double of the slope of another line. If tangent of the angle between them is 13\frac{1}{3}31​, find the slopes of the lines.

Solution

A line passes through (x1,y1)(x_1, y_1)(x1​,y1​) and (h,k)(h, k)(h,k). If slope of the line is mmm, show that k−y1=m(h−x1)k-y_1=m(h-x_1)k−y1​=m(h−x1​).

Solution

Write the equations for the xxx-and yyy-axes.

Solution

Passing through the point (−4,3)(-4, 3)(−4,3) with slope 12\frac{1}{2}21​.

Solution

Passing through (0,0)(0, 0)(0,0) with slope mmm.

Solution

Passing through (2,23)(2, 2\sqrt{3})(2,23​) and inclined with the xxx-axis at an angle of 75∘75^{\circ}75∘.

Solution

Intersecting the xxx-axis at a distance of 3 units to the left of origin with slope -2.

Solution

Intersecting the yyy-axis at a distance of 2 units above the origin and making an angle of 30∘30^{\circ}30∘ with positive direction of the xxx-axis.

Solution

Passing through the points (−1,1)(-1, 1)(−1,1) and (2,−4)(2, -4)(2,−4).

Solution

The vertices of △PQR\triangle \mathrm{PQR}△PQR are P(2,1),Q(−2,3)\mathrm{P}(2, 1), \mathrm{Q}(-2, 3)P(2,1),Q(−2,3) and R(4,5)\mathrm{R}(4, 5)R(4,5). Find equation of the median through the vertex R.

Solution

Find the equation of the line passing through (−3,5)(-3, 5)(−3,5) and perpendicular to the line through the points (2,5)(2, 5)(2,5) and (−3,6)(-3, 6)(−3,6).

Solution

A line perpendicular to the line segment joining the points (1,0)(1, 0)(1,0) and (2,3)(2, 3)(2,3) divides it in the ratio 1:n1:n1:n. Find the equation of the line.

Solution

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3)(2, 3)(2,3).

Solution

Find equation of the line passing through the point (2,2)(2, 2)(2,2) and cutting off intercepts on the axes whose sum is 9.

Solution

Find equation of the line through the point (0,2)(0, 2)(0,2) making an angle 2π3\frac{2 \pi}{3}32π​ with the positive xxx-axis. Also, find the equation of line parallel to it and crossing the yyy-axis at a distance of 2 units below the origin.

Solution

The perpendicular from the origin to a line meets it at the point (−2,9)(-2, 9)(−2,9), find the equation of the line.

Solution

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L=124.942\mathrm{L}=124.942L=124.942 when C=20\mathrm{C}=20C=20 and L=125.134\mathrm{L}=125.134L=125.134 when C=110\mathrm{C}=110C=110, express L in terms of C.

Solution

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Solution

P(a,b)\mathrm{P}(a, b)P(a,b) is the mid-point of a line segment between axes. Show that equation of the line is xa+yb=2\frac{x}{a}+\frac{y}{b}=2ax​+by​=2.

Solution

Point R(h,k)\mathrm{R}(h, k)R(h,k) divides a line segment between the axes in the ratio 1:21:21:2. Find equation of the line.

Solution

By using the concept of equation of a line, prove that the three points (3,0),(−2,−2)(3, 0), (-2, -2)(3,0),(−2,−2) and (8,2)(8, 2)(8,2) are collinear.

Solution

Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

x+7y=0x + 7y = 0x+7y=0,

6x+3y−5=06x + 3y - 5 = 06x+3y−5=0,

y=0y = 0y=0.

Solution

Reduce the following equations into intercept form and find their intercepts on the axes.

3x+2y−12=03x + 2y - 12 = 03x+2y−12=0,

4x−3y=64x - 3y = 64x−3y=6,

3y+2=03y + 2 = 03y+2=0.

Solution

Find the distance of the point (−1,1)(-1, 1)(−1,1) from the line 12(x+6)=5(y−2)12(x + 6) = 5(y - 2)12(x+6)=5(y−2).

Solution

Find the points on the xxx-axis, whose distances from the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 13x​+4y​=1 are 4 units.

Solution

Find the distance between parallel lines

15x+8y−34=015x + 8y - 34 = 015x+8y−34=0 and 15x+8y+31=015x + 8y + 31 = 015x+8y+31=0

l(x+y)+p=0l(x + y) + p = 0l(x+y)+p=0 and l(x+y)−r=0l(x + y) - r = 0l(x+y)−r=0.

Solution

Find equation of the line parallel to the line 3x−4y+2=03x - 4y + 2 = 03x−4y+2=0 and passing through the point (−2,3)(-2, 3)(−2,3).

Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4, 5), (0, 7), (5, -5)$ and $(-4, -2)$ . Also, find its area.

The base of an equilateral triangle with side $2a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Find the distance between $\mathrm{P}(x_{1}, y_{1})$ and $\mathrm{Q}(x_{2}, y_{2})$ when : (i) PQ is parallel to the $y$ -axis, (ii) PQ is parallel to the $x$ -axis.

Find a point on the $x$ -axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$ .

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $\mathrm{P}(0,-4)$ and $\mathrm{B}(8,0)$ .

Without using the Pythagoras theorem, show that the points $(4, 4), (3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

Find the slope of the line, which makes an angle of $30^{\circ}$ with the positive direction of $y$ -axis measured anticlockwise.

Without using distance formula, show that points $(-2, -1), (4, 0), (3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

Find the angle between the $x$ -axis and the line joining the points $(3, -1)$ and $(4, -2)$ .

The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$ , find the slopes of the lines.

A line passes through $(x_1, y_1)$ and $(h, k)$ . If slope of the line is $m$ , show that $k-y_1=m(h-x_1)$ .

Write the equations for the $x$ -and $y$ -axes.

Passing through the point $(-4, 3)$ with slope $\frac{1}{2}$ .

Passing through $(0, 0)$ with slope $m$ .

Passing through $(2, 2\sqrt{3})$ and inclined with the $x$ -axis at an angle of $75^{\circ}$ .

Intersecting the $x$ -axis at a distance of 3 units to the left of origin with slope -2.

Intersecting the $y$ -axis at a distance of 2 units above the origin and making an angle of $30^{\circ}$ with positive direction of the $x$ -axis.

Passing through the points $(-1, 1)$ and $(2, -4)$ .

The vertices of $\triangle \mathrm{PQR}$ are $\mathrm{P}(2, 1), \mathrm{Q}(-2, 3)$ and $\mathrm{R}(4, 5)$ . Find equation of the median through the vertex R.

Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points $(2, 5)$ and $(-3, 6)$ .

A line perpendicular to the line segment joining the points $(1, 0)$ and $(2, 3)$ divides it in the ratio $1:n$ . Find the equation of the line.

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2, 3)$ .

Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is 9.

Find equation of the line through the point $(0, 2)$ making an angle $\frac{2 \pi}{3}$ with the positive $x$ -axis. Also, find the equation of line parallel to it and crossing the $y$ -axis at a distance of 2 units below the origin.

The perpendicular from the origin to a line meets it at the point $(-2, 9)$ , find the equation of the line.

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if $\mathrm{L}=124.942$ when $\mathrm{C}=20$ and $\mathrm{L}=125.134$ when $\mathrm{C}=110$ , express L in terms of C.

$\mathrm{P}(a, b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a}+\frac{y}{b}=2$ .

Point $\mathrm{R}(h, k)$ divides a line segment between the axes in the ratio $1:2$ . Find equation of the line.

By using the concept of equation of a line, prove that the three points $(3, 0), (-2, -2)$ and $(8, 2)$ are collinear.

$x + 7y = 0$ ,

$6x + 3y - 5 = 0$ ,

$y = 0$ .

$3x + 2y - 12 = 0$ ,

$4x - 3y = 6$ ,

$3y + 2 = 0$ .

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$ .

Find the points on the $x$ -axis, whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are 4 units.

$15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$

$l(x + y) + p = 0$ and $l(x + y) - r = 0$ .

Find equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(-2, 3)$ .

Find equation of the line perpendicular to the line $x - 7y + 5 = 0$ and having $x$ intercept 3.

Find angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$ .

The line through the points $(h, 3)$ and $(4, 1)$ intersects the line $7x - 9y - 19 = 0$ . at right angle. Find the value of $h$ .

Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax + By + C = 0$ is $A(x - x_1) + B(y - y_1) = 0$ .

Two lines passing through the point $(2, 3)$ intersects each other at an angle of $60^{\circ}$ . If slope of one line is 2, find equation of the other line.

Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1, 2)$ .

Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$ .

The perpendicular from the origin to the line $y = mx + c$ meets it at the point $(-1, 2)$ . Find the values of $m$ and $c$ .

In the triangle ABC with vertices $\mathrm{A}(2, 3), \mathrm{B}(4, -1)$ and $\mathrm{C}(1, 2)$ , find the equation and length of altitude from the vertex A.

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$ , then show that $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$ .

Find the values of $k$ for which the line $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is

What are the points on the $y$ -axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

Find perpendicular distance from the origin to the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ .

Find the equation of the line parallel to $y$ -axis and drawn through the point of intersection of the lines $x - 7y + 5 = 0$ and $3x + y = 0$ .

Find the equation of a line drawn perpendicular to the line $\frac{x}{4} + \frac{y}{6} = 1$ through the point, where it meets the $y$ -axis.

Find the area of the triangle formed by the lines $y - x = 0, x + y = 0$ and $x - k = 0$ .

Find the value of $p$ so that the three lines $3x + y - 2 = 0, px + 2y - 3 = 0$ and $2x - y - 3 = 0$ may intersect at one point.

Find the equation of the lines through the point $(3, 2)$ which make an angle of $45^{\circ}$ with the line $x - 2y = 3$ .

Find the equation of the line passing through the point of intersection of the lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ that has equal intercepts on the axes.

Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is $\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}$ .

In what ratio, the line joining $(-1, 1)$ and $(5, 7)$ is divided by the line $x + y = 4$ ?

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$ .

Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of 3 units from this point.

The hypotenuse of a right angled triangle has its ends at the points $(1, 3)$ and $(-4, 1)$ . Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Find the image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ assuming the line to be a plane mirror.

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$ , find the value of $m$ .

If sum of the perpendicular distances of a variable point $\mathrm{P}(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always 10. Show that P must move on a line.

Find equation of the line which is equidistant from parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$ .

A ray of light passing through the point $(1, 2)$ reflects on the $x$ -axis at point A and the reflected ray passes through the point $(5, 3)$ . Find the coordinates of A.