Q15 (ii): whether the two circles intersect each other.

Given: Screen dimensions: 800 pixels wide, 600 pixels high. Origin (0,0) at bottom-left. Screen boundaries are from x=0 to x=800 and y=0 to y=600. Circle 1: Center A(100, 150), radius r_A = 80 pixels. Circle 2: Center B(250, 230), radius r_B = 100 pixels. To Determine: (i) If either circle goes…

Orienting Yourself: The Use of Coordinates Class 9 Mathematics NCERT Solutions

Q1End-of-Chapter Exercises

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution

Given: The Cartesian coordinate system with two perpendicular axes, the x-axis and the y-axis.

To Find: The coordinates of the point of intersection of the axes.

Solution: The point of intersection of the x-axis and the y-axis is called the origin. By definition, the origin is the reference point from which all distances are measured. It has a value of 0 on the x-axis and a value of 0 on the y-axis. Therefore, the x-coordinate is 0 and the y-coordinate is 0.

Final Answer: The x-coordinate is 0 and the y-coordinate is 0. The coordinates of the point of intersection are (0, 0).

Q2End-of-Chapter Exercises

Point W has x-coordinate equal to -5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Solution

Given:

Point W has an x-coordinate of -5.
Point H is on a line that passes through W and is parallel to the y-axis.

To Find:

The general form of the coordinates of point H.
The possible quadrants for point H.

Solution: A line parallel to the y-axis is a vertical line. All points on a vertical line have the same x-coordinate. Since the line passes through point W, which has an x-coordinate of -5, every point on this line must also have an x-coordinate of -5. The equation of this line is

x = -5

.

Therefore, the coordinates of point H must be of the form

(-5, y)

, where

y

can be any real number.

The quadrants H can lie in depend on the value of its y-coordinate:

If $y > 0$ (e.g., H(-5, 2)), point H lies in Quadrant II (where x is negative and y is positive).
If $y < 0$ (e.g., H(-5, -3)), point H lies in Quadrant III (where x is negative and y is negative).
If $y = 0$ , point H(-5, 0) lies on the negative x-axis and is not in any quadrant.

Final Answer: The coordinates of point H can be predicted to be of the form

(-5, y)

for any real number

y

. Point H can lie in Quadrant II (if

y > 0

) or Quadrant III (if

y < 0

).

Q3End-of-Chapter Exercises

Consider the points R (3, 0), A (0, -2), M (-5, -2) and P (-5, 2). If they are joined in the same order, predict:

(i)

Two sides of RAMP that are perpendicular to each other.

(ii)

One side of RAMP that is parallel to one of the axes.

(iii)

Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

Solution

Given: Points R(3, 0), A(0, -2), M(-5, -2), and P(-5, 2).

Predictions:

(i)

Perpendicular sides: A side is perpendicular to another if one is horizontal and the other is vertical.

Side AM connects A(0, -2) and M(-5, -2). Since the y-coordinates are the same (-2), this is a horizontal line segment.
Side MP connects M(-5, -2) and P(-5, 2). Since the x-coordinates are the same (-5), this is a vertical line segment. A horizontal line is perpendicular to a vertical line. Therefore, sides AM and MP are perpendicular to each other.

(ii)

Side parallel to an axis:

As determined above, side AM has constant y-coordinate, so it is parallel to the x-axis.
Side MP has constant x-coordinate, so it is parallel to the y-axis. One such side is AM (or MP).

(iii)

Mirror images: Two points are mirror images in an axis if one coordinate is the same and the other is the opposite.

Consider points M(-5, -2) and P(-5, 2).
Their x-coordinates are the same (-5).
Their y-coordinates are opposites (-2 and 2). Therefore, M and P are mirror images of each other in the x-axis.

Verification by Plotting:

Plot the points R(3, 0), A(0, -2), M(-5, -2), and P(-5, 2) on a Cartesian plane.
Join the points in order: R to A, A to M, M to P, and P to R.

(i)

Observing the plot, the line segment AM is horizontal and the line segment MP is vertical. They meet at point M, forming a right angle. Thus, AM is perpendicular to MP. The prediction is correct.

(ii)

The segment AM is a horizontal line, confirming it is parallel to the x-axis. The prediction is correct.

(iii)

The point P(-5, 2) is 2 units above the x-axis, and the point M(-5, -2) is 2 units below the x-axis, both at the same horizontal position

x=-5

. This confirms they are mirror images of each other with respect to the x-axis. The prediction is correct.

Final Answer:

(i)

Sides AM and MP are perpendicular to each other.

(ii)

Side AM is parallel to the x-axis (and side MP is parallel to the y-axis).

(iii)

Points M(-5, -2) and P(-5, 2) are mirror images of each other in the x-axis.

Q4End-of-Chapter Exercises

Plot point Z (5,-6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

Solution

Given: Point Z(5, -6).

To Do: Construct a right-angled triangle IZN and find the lengths of its sides.

Solution: This is an open-ended question. We can construct a right-angled triangle in many ways. A simple method is to create two sides that are parallel to the coordinate axes.

Plot Point Z: Locate the point (5, -6) on the Cartesian plane. This point is in Quadrant IV.
Construct the Triangle: Let's form a right angle at a new point, N. To do this, we can draw a horizontal line and a vertical line that intersect at N. Let's place N such that ZN is vertical and IN is horizontal.
- Let's choose point N to have the same x-coordinate as Z, so ZN is vertical. Let N be (5, -2).
- Let's choose point I to have the same y-coordinate as N, so IN is horizontal. Let I be (1, -2).
- The vertices of our triangle are I(1, -2), Z(5, -6), and N(5, -2).
- The angle at N is a right angle because ZN is a vertical line ( $x=5$ ) and IN is a horizontal line ( $y=-2$ ).
Find the lengths of the sides: We use the distance formula, or simply count units for horizontal and vertical segments.
- Length of IN: Since it is a horizontal segment, the length is the difference in x-coordinates. $IN = |5 - 1| = 4 \text{ units.}$
- Length of ZN: Since it is a vertical segment, the length is the difference in y-coordinates. $ZN = |-6 - (-2)| = |-4| = 4 \text{ units.}$
- Length of IZ (the hypotenuse): We use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ . $IZ = \sqrt{(5-1)^2 + (-6 - (-2))^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$ $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \approx 5.66 \text{ units.}$

Final Answer: One possible right-angled triangle IZN has vertices I(1, -2), Z(5, -6), and N(5, -2), with the right angle at N. The lengths of the sides are:

IN = 4 units
ZN = 4 units
IZ = $4\sqrt{2}$ units

Q5End-of-Chapter Exercises

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Solution

To consider: A coordinate system without negative numbers and its ability to locate all points in a 2-D plane.

Solution: If we did not have negative numbers, our coordinates

(x, y)

would be restricted to values where

x \geq 0

and

y \geq 0

. This system would consist of two perpendicular rays starting from an origin, extending infinitely in one direction (e.g., to the right and upwards).

This coordinate system would look like only the first quadrant of the standard Cartesian plane, including the positive x-axis and the positive y-axis.

This system would not allow us to locate all the points on a 2-D plane. A complete 2-D plane extends infinitely in all directions. The standard Cartesian plane is divided into four quadrants by the x and y axes.

Quadrant I: $(+, +)$ - Points here could be located.
Quadrant II: $(-, +)$ - Points with negative x-coordinates could not be located.
Quadrant III: $(-, -)$ - Points with both negative x and y-coordinates could not be located.
Quadrant IV: $(+, -)$ - Points with negative y-coordinates could not be located.

Brahmagupta's formalization of negative numbers was crucial for the development of the full four-quadrant Cartesian plane, which is necessary to describe the locations of all points in a plane.

Final Answer: A coordinate system without negative numbers would only cover the first quadrant of the Cartesian plane (where both x and y are non-negative). No, this system would not allow us to locate all the points on a 2-D plane; it would be impossible to describe the location of any point in Quadrants II, III, or IV.

Q6End-of-Chapter Exercises

*6. Are the points $\mathrm{M}(-3,-4), \mathrm{A}(0,0)$ and $\mathrm{G}(6,8)$ on the same straight line? Suggest a method to check this without plotting and joining the points.

Solution

Given: Points M(-3, -4), A(0, 0), and G(6, 8).

To Check: Whether the points are collinear (on the same straight line).

Method: Three points are collinear if the slope of the line segment joining the first two points is equal to the slope of the line segment joining the second and third points. The formula for the slope

(m)

between two points

(x_1, y_1)

and

(x_2, y_2)

is

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

.

Solution:

Calculate the slope of the segment MA: Let M be $(x_1, y_1) = (-3, -4)$ and A be $(x_2, y_2) = (0, 0)$ . $m_{MA} = \frac{0 - (-4)}{0 - (-3)} = \frac{4}{3}$
Calculate the slope of the segment AG: Let A be $(x_1, y_1) = (0, 0)$ and G be $(x_2, y_2) = (6, 8)$ . $m_{AG} = \frac{8 - 0}{6 - 0} = \frac{8}{6} = \frac{4}{3}$
Compare the slopes: The slope of MA is $\frac{4}{3}$ and the slope of AG is $\frac{4}{3}$ . Since $m_{MA} = m_{AG}$ and they share a common point A, the points M, A, and G must lie on the same straight line.

Alternative Method (Distance Formula): Three points A, B, C are collinear if the sum of the lengths of two segments equals the length of the third (e.g., AB + BC = AC). We can calculate the distances MA, AG, and MG and check if MA + AG = MG.

$MA = \sqrt{(0 - (-3))^2 + (0 - (-4))^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$ .
$AG = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$ .
$MG = \sqrt{(6 - (-3))^2 + (8 - (-4))^2} = \sqrt{9^2 + 12^2} = \sqrt{81+144} = \sqrt{225} = 15$ . Since $MA + AG = 5 + 10 = 15$ , and $MG = 15$ , we have $MA + AG = MG$ . This confirms the points are collinear.

Final Answer: Yes, the points M(-3, -4), A(0, 0), and G(6, 8) are on the same straight line. A method to check this is to calculate the slopes of the segments MA and AG. If the slopes are equal, the points are collinear.

Q7End-of-Chapter Exercises

*7. Use your method (from Problem 6) to check if the points R (-5, -1), B (-2, -5) and C (4, -12) are on the same straight line. Now plot both sets of points and check your answers.

Solution

Given: Points R(-5, -1), B(-2, -5), and C(4, -12).

To Check: Whether the points are collinear using the slope method.

Method: We will calculate the slope of the segment RB and the slope of the segment BC. If the slopes are equal, the points are collinear.

Solution:

Calculate the slope of the segment RB: Let R be $(x_1, y_1) = (-5, -1)$ and B be $(x_2, y_2) = (-2, -5)$ . $m_{RB} = \frac{-5 - (-1)}{-2 - (-5)} = \frac{-5 + 1}{-2 + 5} = \frac{-4}{3}$
Calculate the slope of the segment BC: Let B be $(x_1, y_1) = (-2, -5)$ and C be $(x_2, y_2) = (4, -12)$ . $m_{BC} = \frac{-12 - (-5)}{4 - (-2)} = \frac{-12 + 5}{4 + 2} = \frac{-7}{6}$
Compare the slopes: The slope of RB is $-\frac{4}{3}$ and the slope of BC is $-\frac{7}{6}$ . Since $-\frac{4}{3} \neq -\frac{7}{6}$ , the slopes are not equal. Therefore, the points R, B, and C do not lie on the same straight line.

Verification by Plotting:

Plot the points R(-5, -1), B(-2, -5), and C(4, -12) on a Cartesian plane.
When you try to draw a straight line through R and B, you will observe that point C does not fall on this line. The line segment BC has a different steepness (slope) than the line segment RB.
This visual check confirms that the three points are not collinear.

Final Answer: No, the points R(-5, -1), B(-2, -5), and C(4, -12) are not on the same straight line. The slope of segment RB is

-\frac{4}{3}

while the slope of segment BC is

-\frac{7}{6}

. Since the slopes are not equal, the points are not collinear.

Q8End-of-Chapter Exercises

*8. Using the origin as one vertex, plot the vertices of:

(i)

A right-angled isosceles triangle.

(ii)

An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Solution

Given: One vertex of each triangle is the origin O(0, 0).

To Construct:

(i)

A right-angled isosceles triangle.

(ii)

An isosceles triangle with one vertex in Q III and another in Q IV.

Solution:

(i) Right-angled isosceles triangle:

Let the vertices be O(0, 0), A, and B.
For the triangle to be right-angled at the origin O, the other two vertices A and B must lie on the coordinate axes.
Let vertex A be on the x-axis, so its coordinates are $(a, 0)$ for some $a \neq 0$ .
Let vertex B be on the y-axis, so its coordinates are $(0, b)$ for some $b \neq 0$ .
For the triangle to be isosceles, the lengths of the sides adjacent to the right angle must be equal, i.e., $OA = OB$ .
The length $OA = \sqrt{(a-0)^2 + (0-0)^2} = |a|$ .
The length $OB = \sqrt{(0-0)^2 + (b-0)^2} = |b|$ .
So we need $|a| = |b|$ . Let's choose $a = 4$ . Then $b$ can be 4 or -4. Let's choose $b = 4$ .
The vertices are O(0, 0), A(4, 0), and B(0, 4).

(ii) Isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV:

Let the vertices be O(0, 0), P, and Q.
Let P be in Quadrant III. Its coordinates $(x_P, y_P)$ must satisfy $x_P < 0$ and $y_P < 0$ .
Let Q be in Quadrant IV. Its coordinates $(x_Q, y_Q)$ must satisfy $x_Q > 0$ and $y_Q < 0$ .
For the triangle OPQ to be isosceles, we can have $OP = OQ$ , or $OP = PQ$ , or $OQ = PQ$ . The simplest case is to make $OP = OQ$ .
$OP^2 = x_P^2 + y_P^2$ and $OQ^2 = x_Q^2 + y_Q^2$ .
To make $OP = OQ$ , we need $x_P^2 + y_P^2 = x_Q^2 + y_Q^2$ .
A simple way to satisfy this is to choose the same y-coordinate for P and Q, and have their x-coordinates be opposites. Let $y_P = y_Q = -3$ . For the x-coordinates, let $x_P = -5$ and $x_Q = 5$ .
The vertices are O(0, 0), P(-5, -3), and Q(5, -3).
P(-5, -3) is in Quadrant III.
Q(5, -3) is in Quadrant IV.
$OP = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}$ .
$OQ = \sqrt{(5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}$ .
Since $OP = OQ$ , the triangle is isosceles.

Final Answer:

(i)

An example of a right-angled isosceles triangle is one with vertices O(0, 0), A(4, 0), and B(0, 4).

(ii)

An example of an isosceles triangle with one vertex in Q III and one in Q IV is one with vertices O(0, 0), P(-5, -3), and Q(5, -3).

Q9End-of-Chapter Exercises

*9. The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.
S M T Is M the midpoint of ST? Yes or No Reason for your answer
(-3, 0) $(0,0)$ $(3,0)$
$(2,3)$ $(3,4)$ $(4,5)$
$(0,0)$ $(0,5)$ (0, -10)
$(-8,7)$ (0, -2) (6, -3)
When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

S	M	T
(-3, 0)	$(0,0)$	$(3,0)$
$(2,3)$	$(3,4)$	$(4,5)$
$(0,0)$	$(0,5)$	(0, -10)
$(-8,7)$	(0, -2)	(6, -3)

Solution

Method: The midpoint of a line segment with endpoints S

(x_S, y_S)

and T

(x_T, y_T)

is given by the formula:

\text{Midpoint} = \left( \frac{x_S + x_T}{2}, \frac{y_S + y_T}{2} \right)

We will calculate the midpoint of ST for each case and check if it matches the coordinates of M.

Completed Table:

S	M	T	Is M the midpoint of ST? Yes or No	Reason for your answer
(-3, 0)	(0, 0)	(3, 0)	Yes	The calculated midpoint is $(\frac{-3+3}{2}, \frac{0+0}{2}) = (0, 0)$ , which matches M.
(2, 3)	(3, 4)	(4, 5)	Yes	The calculated midpoint is $(\frac{2+4}{2}, \frac{3+5}{2}) = (3, 4)$ , which matches M.
(0, 0)	(0, 5)	(0, -10)	No	The calculated midpoint is $(\frac{0+0}{2}, \frac{0+(-10)}{2}) = (0, -5)$ , which does not match M(0, 5).
(-8, 7)	(0, -2)	(6, -3)	No	The calculated midpoint is $(\frac{-8+6}{2}, \frac{7+(-3)}{2}) = (-1, 2)$ , which does not match M(0, -2).

Connection between coordinates: When M

(x_M, y_M)

is the midpoint of ST, where S is

(x_S, y_S)

and T is

(x_T, y_T)

, the coordinates of M are the average of the coordinates of S and T.

x_M = \frac{x_S + x_T}{2}

y_M = \frac{y_S + y_T}{2}

Final Answer: The completed table is provided above. The connection is that the coordinates of the midpoint M are the arithmetic mean (average) of the corresponding coordinates of the endpoints S and T.

Q10End-of-Chapter Exercises

*10. Use the connection you found to find the coordinates of B given that M (-7, 1) is the midpoint of A (3, -4) and $\mathrm{B}(x, y)$ .

Solution

Given:

M(-7, 1) is the midpoint of the segment AB.
The coordinates of point A are (3, -4).
The coordinates of point B are $(x, y)$ .

To Find: The coordinates of point B.

Formula: The midpoint M

(x_M, y_M)

of a segment with endpoints A

(x_A, y_A)

and B

(x_B, y_B)

has coordinates:

x_M = \frac{x_A + x_B}{2} \quad \text{and} \quad y_M = \frac{y_A + y_B}{2}

Solution: We can set up two separate equations, one for the x-coordinates and one for the y-coordinates.

For the x-coordinate: $-7 = \frac{3 + x}{2}$ Multiply both sides by 2: $-14 = 3 + x$ Subtract 3 from both sides: $x = -14 - 3 = -17$
For the y-coordinate: $1 = \frac{-4 + y}{2}$ Multiply both sides by 2: $2 = -4 + y$ Add 4 to both sides: $y = 2 + 4 = 6$

So, the coordinates of point B are (-17, 6).

Final Answer: The coordinates of B are (-17, 6).

Q11End-of-Chapter Exercises

*11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A $(4,7)$ and B (16, -2).

Solution

Given:

Points A(4, 7) and B(16, -2).
Points P and Q trisect the segment AB, meaning they divide it into three equal parts: AP = PQ = QB.
P is closer to A, and Q is closer to B.

Method using Midpoint Formula: The trisection points can be related using the midpoint concept:

Point P is the midpoint of the segment AQ.
Point Q is the midpoint of the segment PB.

Let P have coordinates

(x_P, y_P)

and Q have coordinates

(x_Q, y_Q)

. We can set up a system of equations based on the midpoint formula.

From (1),

P = \text{midpoint of AQ}

:

x_P = \frac{x_A + x_Q}{2} = \frac{4 + x_Q}{2} \quad \implies \quad 2x_P - x_Q = 4 \quad ...(\text{i})

y_P = \frac{y_A + y_Q}{2} = \frac{7 + y_Q}{2} \quad \implies \quad 2y_P - y_Q = 7 \quad ...(\text{ii})

From (2),

Q = \text{midpoint of PB}

:

x_Q = \frac{x_P + x_B}{2} = \frac{x_P + 16}{2} \quad \implies \quad 2x_Q - x_P = 16 \quad \implies \quad -x_P + 2x_Q = 16 \quad ...(\text{iii})

y_Q = \frac{y_P + y_B}{2} = \frac{y_P - 2}{2} \quad \implies \quad 2y_Q - y_P = -2 \quad \implies \quad -y_P + 2y_Q = -2 \quad ...(\text{iv})

Solving for x-coordinates: Multiply equation (i) by 2:

4x_P - 2x_Q = 8

. Add this to equation (iii):

(4x_P - 2x_Q) + (-x_P + 2x_Q) = 8 + 16

3x_P = 24 \implies x_P = 8

Substitute

x_P = 8

into (i):

2(8) - x_Q = 4 \implies 16 - x_Q = 4 \implies x_Q = 12

.

Solving for y-coordinates: Multiply equation (ii) by 2:

4y_P - 2y_Q = 14

. Add this to equation (iv):

(4y_P - 2y_Q) + (-y_P + 2y_Q) = 14 + (-2)

3y_P = 12 \implies y_P = 4

Substitute

y_P = 4

into (ii):

2(4) - y_Q = 7 \implies 8 - y_Q = 7 \implies y_Q = 1

.

So, the coordinates are P(8, 4) and Q(12, 1).

(Note: A more direct method is the section formula. P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1. This gives the same result.)

Final Answer: Using the midpoint relationship between the points, we solve a system of equations. The coordinates of the trisection points are P(8, 4) and Q(12, 1).

Q12End-of-Chapter Exercises

*12. (i) Given the points A (1, -8), B (-4, 7) and C (-7, -4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K? (ii) Given the points D (-5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

Solution

Given:

Points A(1, -8), B(-4, 7), C(-7, -4), D(-5, 6), E(0, 9).
A circle K with center at the origin O(0, 0).

Formula: The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

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

.

Part (i):

To Show: Points A, B, and C lie on circle K. To do this, we must show that their distances from the center O are equal. This common distance will be the radius.

Solution:

Calculate distance OA: $OA = \sqrt{(1-0)^2 + (-8-0)^2} = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$
Calculate distance OB: $OB = \sqrt{(-4-0)^2 + (7-0)^2} = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$
Calculate distance OC: $OC = \sqrt{(-7-0)^2 + (-4-0)^2} = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$

Since

OA = OB = OC = \sqrt{65}

, all three points are equidistant from the origin. Therefore, they lie on a circle K with center O(0, 0). The radius of circle K is this common distance,

r = \sqrt{65}

units.

Part (ii):

To Check: The position of points D and E relative to circle K.

Solution: We compare the distance of each point from the origin with the radius

r = \sqrt{65}

. Let

r^2 = 65

.

Check point D(-5, 6): Calculate the square of the distance OD: $OD^2 = (-5-0)^2 + (6-0)^2 = (-5)^2 + 6^2 = 25 + 36 = 61$ Since $OD^2 = 61$ and $r^2 = 65$ , we have $OD^2 < r^2$ . This implies $OD < r$ . Therefore, point D lies within the circle K.
Check point E(0, 9): Calculate the square of the distance OE: $OE^2 = (0-0)^2 + (9-0)^2 = 0^2 + 9^2 = 0 + 81 = 81$ Since $OE^2 = 81$ and $r^2 = 65$ , we have $OE^2 > r^2$ . This implies $OE > r$ . Therefore, point E lies outside the circle K.

Final Answer:

(i)

Yes, the points A, B, and C lie on a circle with center at the origin because their distances to the origin are all equal to

\sqrt{65}

. The radius of circle K is

\sqrt{65}

units.

(ii)

Point D lies within the circle K. Point E lies outside the circle K.

Q13End-of-Chapter Exercises

*13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and ( 0,3 ), respectively, find the coordinates of A, B and C .

Solution

Given:

D, E, and F are the midpoints of the sides of triangle ABC.
D = (5, 1), E = (6, 5), F = (0, 3).
Let A = $(x_A, y_A)$ , B = $(x_B, y_B)$ , C = $(x_C, y_C)$ .
Let D be the midpoint of BC, E be the midpoint of AC, and F be the midpoint of AB.

To Find: The coordinates of vertices A, B, and C.

Formula: Midpoint of a segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

is

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

.

Solution: We can set up systems of equations for the x and y coordinates.

For x-coordinates:

D is midpoint of BC: $\frac{x_B + x_C}{2} = 5 \implies x_B + x_C = 10$ ...(i)
E is midpoint of AC: $\frac{x_A + x_C}{2} = 6 \implies x_A + x_C = 12$ ...(ii)
F is midpoint of AB: $\frac{x_A + x_B}{2} = 0 \implies x_A + x_B = 0$ ...(iii)

Add the three equations:

(x_B + x_C) + (x_A + x_C) + (x_A + x_B) = 10 + 12 + 0

2x_A + 2x_B + 2x_C = 22

x_A + x_B + x_C = 11

...(iv)

Now, subtract each of the first three equations from (iv):

(iv) - (i): $(x_A + x_B + x_C) - (x_B + x_C) = 11 - 10 \implies x_A = 1$ .
(iv) - (ii): $(x_A + x_B + x_C) - (x_A + x_C) = 11 - 12 \implies x_B = -1$ .
(iv) - (iii): $(x_A + x_B + x_C) - (x_A + x_B) = 11 - 0 \implies x_C = 11$ .

For y-coordinates:

D is midpoint of BC: $\frac{y_B + y_C}{2} = 1 \implies y_B + y_C = 2$ ...(v)
E is midpoint of AC: $\frac{y_A + y_C}{2} = 5 \implies y_A + y_C = 10$ ...(vi)
F is midpoint of AB: $\frac{y_A + y_B}{2} = 3 \implies y_A + y_B = 6$ ...(vii)

Add the three equations:

(y_B + y_C) + (y_A + y_C) + (y_A + y_B) = 2 + 10 + 6

2y_A + 2y_B + 2y_C = 18

y_A + y_B + y_C = 9

...(viii)

Now, subtract each of the equations (v), (vi), (vii) from (viii):

(viii) - (v): $(y_A + y_B + y_C) - (y_B + y_C) = 9 - 2 \implies y_A = 7$ .
(viii) - (vi): $(y_A + y_B + y_C) - (y_A + y_C) = 9 - 10 \implies y_B = -1$ .
(viii) - (vii): $(y_A + y_B + y_C) - (y_A + y_B) = 9 - 6 \implies y_C = 3$ .

Combining the coordinates: A = (1, 7) B = (-1, -1) C = (11, 3)

Final Answer: The coordinates of the vertices are A(1, 7), B(-1, -1), and C(11, 3).

Q14End-of-Chapter Exercises

A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South (N-S) direction and East-West (E-W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(i)

Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

(ii)

There are street intersections in the model. Each street intersection is formed by two streets-one running in the N-S direction and another in the E-W direction. Each street intersection is referred to in the following manner: If the second street running in the N-S direction and 5th street in the E-W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find:

(a)

how many street intersections can be referred to as (4, 3).

(b)

how many street intersections can be referred to as (3, 4).

Solution

Given:

A city grid with a central N-S road and a central E-W road.
10 streets parallel to the N-S road and 10 streets parallel to the E-W road.
A naming convention for intersections: (N-S street number, E-W street number).

To Find: (a) The number of street intersections referred to as (4, 3). (b) The number of street intersections referred to as (3, 4).

Solution:

(i)

Model of the city: To draw the model, one would draw a horizontal line (E-W main road) and a vertical line (N-S main road) intersecting at the center. Then, draw 10 vertical lines parallel to the N-S road and 10 horizontal lines parallel to the E-W road. With a scale of 1 cm = 200 m, these parallel lines would be 1 cm apart.

(ii)

Analyzing the intersection convention: The convention (N-S street number, E-W street number) is analogous to the Cartesian coordinate system, where each pair of coordinates

(x, y)

refers to a unique point in the plane. Here, 'N-S street number' acts like the x-coordinate and 'E-W street number' acts like the y-coordinate.

(a) How many street intersections can be referred to as (4, 3)? This refers to the specific, unique intersection point where the 4th North-South street crosses the 3rd East-West street. Just like there is only one point with coordinates (4, 3) on a graph, there is only one such intersection in the city grid.

(b) How many street intersections can be referred to as (3, 4)? Similarly, this refers to the unique intersection point where the 3rd North-South street crosses the 4th East-West street. This is a different intersection from (4, 3). There is only one such intersection.

Final Answer: (a) There is only one street intersection that can be referred to as (4, 3). (b) There is only one street intersection that can be referred to as (3, 4).

Q15End-of-Chapter Exercises

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:

(i)

whether any part of either circle lies outside the screen.

(ii)

whether the two circles intersect each other.

Solution

Given:

Screen dimensions: 800 pixels wide, 600 pixels high. Origin (0,0) at bottom-left.
Screen boundaries are from $x=0$ to $x=800$ and $y=0$ to $y=600$ .
Circle 1: Center A(100, 150), radius $r_A = 80$ pixels.
Circle 2: Center B(250, 230), radius $r_B = 100$ pixels.

To Determine:

(i)

If either circle goes off-screen.

(ii)

If the two circles intersect.

Solution:

(i) Check if circles are on-screen: We need to check the minimum and maximum x and y coordinates reached by each circle and see if they are within the screen boundaries [0, 800] for x and [0, 600] for y.

For Circle 1 (Center A, radius 80):

Min x: $100 - 80 = 20$ . ( $0 \leq 20 \leq 800$ ) -> OK
Max x: $100 + 80 = 180$ . ( $0 \leq 180 \leq 800$ ) -> OK
Min y: $150 - 80 = 70$ . ( $0 \leq 70 \leq 600$ ) -> OK
Max y: $150 + 80 = 230$ . ( $0 \leq 230 \leq 600$ ) -> OK Conclusion: All parts of Circle 1 lie within the screen.

For Circle 2 (Center B, radius 100):

Min x: $250 - 100 = 150$ . ( $0 \leq 150 \leq 800$ ) -> OK
Max x: $250 + 100 = 350$ . ( $0 \leq 350 \leq 800$ ) -> OK
Min y: $230 - 100 = 130$ . ( $0 \leq 130 \leq 600$ ) -> OK
Max y: $230 + 100 = 330$ . ( $0 \leq 330 \leq 600$ ) -> OK Conclusion: All parts of Circle 2 lie within the screen.

(ii) Check for intersection: Two circles intersect if the distance between their centers is less than the sum of their radii and greater than the absolute difference of their radii.

Calculate the distance between centers A and B: $d = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$ $d = \sqrt{(250 - 100)^2 + (230 - 150)^2}$ $d = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170 \text{ pixels}$
Calculate the sum of the radii: $r_A + r_B = 80 + 100 = 180 \text{ pixels}$
Compare the distance to the sum of radii: Distance between centers ( $d=170$ ) is less than the sum of radii ( $r_A + r_B = 180$ ). Since $d < r_A + r_B$ , the circles intersect.

Final Answer:

(i)

No, no part of either circle lies outside the screen.

(ii)

Yes, the two circles intersect each other because the distance between their centers (170 pixels) is less than the sum of their radii (180 pixels).

Q16End-of-Chapter Exercises

Plot the points A (2, 1), B (-1, 2), C (-2, -1), and D (1, -2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Solution

Given: Points A(2, 1), B(-1, 2), C(-2, -1), and D(1, -2).

To Find:

Whether the quadrilateral ABCD is a square.
The area of ABCD if it is a square.

Solution: To determine if ABCD is a square, we need to check two properties:

All four sides are equal in length.
The diagonals are equal in length (or, adjacent sides are perpendicular).

Step 1: Calculate the lengths of the four sides.

AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{9+1} = \sqrt{10}

BC = \sqrt{(-2 - (-1))^2 + (-1-2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}

CD = \sqrt{(1 - (-2))^2 + (-2 - (-1))^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}

DA = \sqrt{(2-1)^2 + (1 - (-2))^2} = \sqrt{1^2 + 3^2} = \sqrt{1+9} = \sqrt{10}

Since

AB = BC = CD = DA = \sqrt{10}

, all four sides are equal. This means ABCD is at least a rhombus.

Step 2: Calculate the lengths of the diagonals.

AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20}

BD = \sqrt{(1 - (-1))^2 + (-2-2)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}

Since the lengths of the diagonals AC and BD are equal, the rhombus ABCD is a square.

Alternative for Step 2: Check for a right angle. We can check if adjacent sides are perpendicular by comparing their slopes. The product of slopes of perpendicular lines is -1.

Slope of AB: $m_{AB} = \frac{2-1}{-1-2} = \frac{1}{-3} = -\frac{1}{3}$
Slope of BC: $m_{BC} = \frac{-1-2}{-2 - (-1)} = \frac{-3}{-1} = 3$
Product of slopes: $m_{AB} \times m_{BC} = -\frac{1}{3} \times 3 = -1$ . Since the product is -1, AB is perpendicular to BC. A rhombus with one right angle is a square.

Step 3: Calculate the area. The area of a square is given by the formula Area = (side)

^2

.

Side length = $\sqrt{10}$ units.
Area = $(\sqrt{10})^2 = 10$ square units.

Final Answer: Yes, ABCD is a square. This is because all four sides have an equal length of

\sqrt{10}

units, and the two diagonals are also equal in length (

\sqrt{20}

units). The area of the square is 10 square units.

Q1Exercise Set 1.1

If $\mathrm{D}_{1} \mathrm{R}_{1}$ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

Solution

Given: The layout of Reiaan's room is represented on a coordinate plane. The left wall of the room is along the y-axis. The door is represented by the line segment

\mathrm{D}_{1} \mathrm{R}_{1}

. Point

\mathrm{D}_{1}

has coordinates

(8.5, 0)

and point

\mathrm{R}_{1}

has coordinates

(11.5, 0)

.

To Find:

The distance of the door from the left wall (y-axis).
The distance of the door from the x-axis.

Solution:

The distance of any point from the y-axis is given by its x-coordinate. The door segment $\mathrm{D}_{1} \mathrm{R}_{1}$ lies on the x-axis, extending from $x=8.5$ to $x=11.5$ . The point on the door closest to the y-axis is $\mathrm{D}_{1}(8.5, 0)$ . Therefore, the distance of the door from the left wall is 8.5 units. Since the scale is 1 unit : 1 foot, the distance is 8.5 feet.
The distance of any point from the x-axis is given by the absolute value of its y-coordinate. Both points $\mathrm{D}_{1}$ and $\mathrm{R}_{1}$ have a y-coordinate of 0. This means the entire door segment lies on the x-axis. Therefore, the distance of the door from the x-axis is 0 units, or 0 feet.

Final Answer: The door is 8.5 feet away from the left wall (the y-axis). The door is 0 feet away from the x-axis (it lies on the x-axis).

Q2Exercise Set 1.1

What are the coordinates of $\mathrm{D}_{1}$ ?

Solution

Given: The layout of Reiaan's room is shown on a coordinate plane. The door is represented by

\mathrm{D}_{1} \mathrm{R}_{1}

. The point

\mathrm{D}_{1}

is on the x-axis.

To Find: The coordinates of point

\mathrm{D}_{1}

.

Solution: By observing the provided coordinate plane for Reiaan's room, we can locate point

\mathrm{D}_{1}

. It lies on the x-axis between the marks for 8 and 9, specifically at the midpoint, which is 8.5. Since the point is on the x-axis, its y-coordinate is 0.

Therefore, the coordinates of

\mathrm{D}_{1}

are

(8.5, 0)

.

Final Answer: The coordinates of

\mathrm{D}_{1}

are

(8.5, 0)

.

Q3Exercise Set 1.1

If $\mathrm{R}_{1}$ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

Solution

Given: The door is represented by the line segment

\mathrm{D}_{1} \mathrm{R}_{1}

. The coordinates of

\mathrm{D}_{1}

are

(8.5, 0)

. The coordinates of

\mathrm{R}_{1}

are

(11.5, 0)

. The scale is 1 unit : 1 foot.

To Find:

The width of the door.
An opinion on whether the width is comfortable and suitable for wheelchair access.

Solution:

The width of the door is the distance between points $\mathrm{D}_{1}$ and $\mathrm{R}_{1}$ . Since both points lie on the x-axis, the distance is the absolute difference of their x-coordinates. $\text{Width} = |x_{R_1} - x_{D_1}| = |11.5 - 8.5| = 3 \text{ units}$ Since the scale is 1 unit : 1 foot, the width of the door is 3 feet.
A standard interior door is typically 30 to 32 inches wide. A 3-foot door is 36 inches wide, which is quite generous and comfortable for a room door. For wheelchair accessibility, the minimum recommended clear width is 32 inches. Since the door is 36 inches wide, a person in a wheelchair will be able to enter the room easily.

Final Answer: The door is 3 feet wide. Yes, this is a comfortable width. Yes, a person in a wheelchair will be able to enter the room easily because the 36-inch width is greater than the minimum required 32-inch clearance for a wheelchair.

Q4Exercise Set 1.1

If $\mathrm{B}_{1}(0,1.5)$ and $\mathrm{B}_{2}(0,4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Solution

Given: The ends of the bathroom door are at points

\mathrm{B}_{1}(0, 1.5)

and

\mathrm{B}_{2}(0, 4)

. The width of the main room door is 3 feet (from the previous question). The scale is 1 unit : 1 foot.

To Find: Whether the bathroom door is narrower or wider than the room door.

Solution: The width of the bathroom door is the distance between points

\mathrm{B}_{1}

and

\mathrm{B}_{2}

. Since both points lie on the y-axis, the distance is the absolute difference of their y-coordinates.

\text{Bathroom door width} = |y_{B_2} - y_{B_1}| = |4 - 1.5| = 2.5 \text{ units}

Using the scale, the width of the bathroom door is 2.5 feet.

The main room door has a width of 3 feet.

Comparing the widths:

2.5 \text{ feet} < 3 \text{ feet}

The bathroom door is narrower than the room door.

Final Answer: The bathroom door is narrower than the room door.

Q1Exercise Set 1.2

Place Reiaan's rectangular study table with three of its feet at the points (8, 9), $(11,9)$ and (11, 7).

(i)

Where will the fourth foot of the table be?

(ii)

Is this a good spot for the table?

(iii)

What is the width of the table? The length? Can you make out the height of the table?

Solution

Given: A rectangular study table has three of its feet at the points A(8, 9), B(11, 9), and C(11, 7).

To Find:

(i)

The coordinates of the fourth foot.

(ii)

An opinion on the placement of the table.

(iii)

The width, length, and height of the table.

Solution:

(i)

Let the four feet of the rectangular table be at points A, B, C, and D. Let D be the fourth foot with coordinates

(x, y)

. For the table to be rectangular, the sides must be parallel to the coordinate axes (since AB is horizontal and BC is vertical). The side AB connects (8, 9) and (11, 9), which is a horizontal line segment of length

11 - 8 = 3

. The side BC connects (11, 9) and (11, 7), which is a vertical line segment of length

9 - 7 = 2

. The fourth point D must form a rectangle with A, B, and C. The side CD must be parallel to AB and the side DA must be parallel to BC. Therefore, the x-coordinate of D must be the same as A, which is 8. The y-coordinate of D must be the same as C, which is 7. So, the coordinates of the fourth foot are (8, 7).

(ii)

The room's dimensions are roughly 12 ft by 15 ft. The table is placed between x=8 and x=11, and y=7 and y=9. This spot is in the upper right section of the room, away from the main door (at x=8.5 to 11.5) and the bathroom door (on the y-axis). It does not block major pathways. Thus, it seems to be a good spot for the table.

(iii)

The side connecting (8, 9) and (11, 9) has length

11 - 8 = 3

units. This can be considered the width. The side connecting (11, 9) and (11, 7) has length

9 - 7 = 2

units. This can be considered the length. So, the table's dimensions are 3 feet by 2 feet. The height of the table cannot be determined from a 2-D floor plan, as the floor plan only shows the x and y coordinates, not the z coordinate (height).

Final Answer:

(i)

The fourth foot of the table will be at the point (8, 7).

(ii)

Yes, this is a good spot as it does not obstruct the doors or main pathways.

(iii)

The width of the table is 3 feet and the length is 2 feet. The height of the table cannot be determined from the 2-D map.

Q2Exercise Set 1.2

If the bathroom door has a hinge at $\mathrm{B}_{1}$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Solution

Given: The bathroom door has a hinge at

\mathrm{B}_{1}(0, 1.5)

. The door opens into the bedroom (the main room area). The width of the bathroom door is 2.5 feet (from Exercise 1.1, Question 4). The wardrobe is a rectangular object with corners at

\mathrm{S}_{1}(6, 15)

,

\mathrm{S}_{2}(10, 15)

, and other corners likely at (6, 13) and (10, 13). The front of the wardrobe is along the line segment from (6, 13) to (10, 13).

To Find:

Whether the bathroom door will hit the wardrobe.
Suggestions for changes if the door is made wider.

Solution:

The hinge of the bathroom door is at $\mathrm{B}_{1}(0, 1.5)$ on the y-axis. The door has a width of 2.5 feet. When it opens into the bedroom, it will swing in a circular arc with the center at (0, 1.5) and a radius of 2.5 feet. The path of the outer edge of the door will be a quarter-circle from point $\mathrm{B}_{2}(0, 4)$ to a point (2.5, 1.5). The wardrobe is located between x-coordinates 6 and 10. The closest the wardrobe gets to the y-axis is at $x=6$ . Since the swinging door only reaches a maximum x-coordinate of 2.5, and the wardrobe starts at $x=6$ , the door will not hit the wardrobe. There is a clearance of $6 - 2.5 = 3.5$ feet.
If the door is made wider, it will swing further into the room. For the door to hit the wardrobe, its width would have to be greater than 6 feet (the x-coordinate of the wardrobe's edge). A door width of 6 feet is highly impractical for a bathroom. However, if the door were made significantly wider, say 4 feet, it would still not hit the wardrobe, but it would take up more space in the room and might obstruct furniture like the bed or the study table if they are placed near the bathroom door. A suggested change if the door is made wider might be to use a sliding door instead of a swinging door to save space within the bedroom.

Final Answer: No, the bathroom door will not hit the wardrobe. If the door is made wider, it would still not hit the wardrobe, but it could interfere with other furniture. A possible suggestion would be to install a sliding door to conserve space.

Q3Exercise Set 1.2

Look at Reiaan's bathroom.

(i)

What are the coordinates of the four corners O, F, R, and P of the bathroom?

(ii)

What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.

(iii)

Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

Solution

Given: The layout of Reiaan's bathroom on a coordinate plane.

To Find:

(i)

Coordinates of the bathroom corners O, F, R, and P.

(ii)

The shape and corner coordinates of the showering area (SHWR).

(iii)

Possible coordinates for a 3 ft × 2 ft washbasin and a 2 ft × 3 ft toilet.

Solution:

(i)

By observing the coordinate plane (Fig. 1.5 in the source), we can identify the coordinates of the corners of the bathroom:

O is the origin: (0, 0)
F is on the y-axis: (0, 7)
R is the corner opposite to O: (5, 7)
P is on the x-axis: (5, 0) So the coordinates are O(0, 0), F(0, 7), R(5, 7), and P(5, 0).

(ii)

The showering area SHWR is shown in the top-right corner of the bathroom. It appears to be a square. Its corners can be identified from the grid:

One corner is at R(5, 7).
The corner along the wall FR is at (2, 7).
The corner along the wall RP is at (5, 4).
The fourth corner is at (2, 4). The side lengths are $5-2=3$ ft and $7-4=3$ ft. Since the sides are equal and parallel to the axes, the shape is a square. The coordinates of the four corners are (2, 4), (5, 4), (5, 7), and (2, 7).

(iii)

This is an open-ended question asking to place items. We need to find space for a 3 ft × 2 ft washbasin and a 2 ft × 3 ft toilet within the remaining bathroom area. The total bathroom area is 5 ft x 7 ft. The shower takes up a 3 ft x 3 ft area. Let's place the washbasin and toilet. (Note: Other valid placements are possible).

Washbasin (3 ft × 2 ft): We can place it along the wall OF, below the shower area. Let its corners be (0, 4), (2, 4), (2, 1), and (0, 1). This space is 2 ft wide (along x-axis) and 3 ft long (along y-axis), which matches one possible orientation of a 3 ft x 2 ft space.
Toilet (2 ft × 3 ft): We can place it along the wall OP. Let its corners be (3, 3), (5, 3), (5, 0), and (3, 0). This space is 2 ft wide (along x-axis) and 3 ft long (along y-axis).

Final Answer:

(i)

The coordinates of the corners are O(0, 0), F(0, 7), R(5, 7), and P(5, 0).

(ii)

The showering area is a square. Its corners have coordinates (2, 4), (5, 4), (5, 7), and (2, 7).

(iii)

A possible placement is:

Washbasin corners: (0, 1), (2, 1), (2, 4), (0, 4).
Toilet corners: (3, 0), (5, 0), (5, 3), (3, 3).

Q4Exercise Set 1.2

Other rooms in the house:

(i)

Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

(ii)

Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Solution

Given:

A dining room with length 18 ft and width 15 ft.
The description 'The length of the dining room extends from point P to point A' appears to be an error in the source text, as P(5,0) and A(12,0) are only 7 ft apart. We will assume the dining room is a rectangular room adjacent to Reiaan's room.
Let's assume the 15 ft width of the dining room runs along the wall AB of Reiaan's room, which is from (12,0) to (12,15). The 18 ft length would then extend outwards from this wall.

To Find:

(i)

A sketch and coordinates of the dining room corners.

(ii)

The coordinates of the feet of a 5 ft × 3 ft table centered in the dining room.

Solution:

(i)

Let's define the dining room based on our assumption. Reiaan's room has a corner at A(12, 0) and B(12, 15). We can place the dining room adjacent to the wall AB.

Let one corner of the dining room be at A(12, 0).
The wall shared with Reiaan's room is from A(12, 0) to B(12, 15). This side has length 15 ft, which we can take as the width of the dining room.
The length of 18 ft will extend from this wall. Let's extend it in the positive x-direction.
The four corners of the dining room, let's call them A, B, B', A', would be:
- A = (12, 0)
- B = (12, 15)
- A' = (12 + 18, 0) = (30, 0)
- B' = (12 + 18, 15) = (30, 15) The dining room is a rectangle defined by the vertices (12, 0), (30, 0), (30, 15), and (12, 15).

(ii)

To place a 5 ft × 3 ft table precisely in the centre, we first find the centre of the dining room.

The x-coordinate of the center is the average of the x-coordinates of the corners: $\frac{12 + 30}{2} = \frac{42}{2} = 21$
The y-coordinate of the center is the average of the y-coordinates of the corners: $\frac{0 + 15}{2} = 7.5$
The center of the dining room (and the table) is at (21, 7.5).

The table is 5 ft long and 3 ft wide. Let's align its length with the x-axis. The four feet of the table will be positioned symmetrically around the center (21, 7.5).

The x-coordinates of the feet will be $21 \pm \frac{5}{2} = 21 \pm 2.5$ , which are 18.5 and 23.5.
The y-coordinates of the feet will be $7.5 \pm \frac{3}{2} = 7.5 \pm 1.5$ , which are 6 and 9.
The coordinates of the four feet are: (18.5, 6), (23.5, 6), (23.5, 9), and (18.5, 9).

Final Answer:

(i)

Assuming the dining room is adjacent to wall AB of Reiaan's room, its corners are at (12, 0), (30, 0), (30, 15), and (12, 15).

(ii)

The coordinates of the feet of the centrally placed dining table are (18.5, 6), (23.5, 6), (23.5, 9), and (18.5, 9).

Orienting Yourself: The Use of CoordinatesClass 9 Mathematics NCERT Solutions

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution

Point W has x-coordinate equal to -5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Solution

Consider the points R (3, 0), A (0, -2), M (-5, -2) and P (-5, 2). If they are joined in the same order, predict:

Two sides of RAMP that are perpendicular to each other.

One side of RAMP that is parallel to one of the axes.

Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

Solution

Plot point Z (5,-6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

Solution

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Solution

*6. Are the points M(−3,−4),A(0,0)\mathrm{M}(-3,-4), \mathrm{A}(0,0)M(−3,−4),A(0,0) and G(6,8)\mathrm{G}(6,8)G(6,8) on the same straight line? Suggest a method to check this without plotting and joining the points.

Solution

*7. Use your method (from Problem 6) to check if the points R (-5, -1), B (-2, -5) and C (4, -12) are on the same straight line. Now plot both sets of points and check your answers.

Solution

*8. Using the origin as one vertex, plot the vertices of:

A right-angled isosceles triangle.

An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Solution

Solution

*10. Use the connection you found to find the coordinates of B given that M (-7, 1) is the midpoint of A (3, -4) and B(x,y)\mathrm{B}(x, y)B(x,y).

Solution

*11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4,7)(4,7)(4,7) and B (16, -2).

Solution

*12. (i) Given the points A (1, -8), B (-4, 7) and C (-7, -4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K? (ii) Given the points D (-5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

Solution

*13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and ( 0,3 ), respectively, find the coordinates of A, B and C .

Solution

A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South (N-S) direction and East-West (E-W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

Solution

whether any part of either circle lies outside the screen.

whether the two circles intersect each other.

Solution

Plot the points A (2, 1), B (-1, 2), C (-2, -1), and D (1, -2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Solution

If D1R1\mathrm{D}_{1} \mathrm{R}_{1}D1​R1​ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

Solution

What are the coordinates of D1\mathrm{D}_{1}D1​?

Solution

If R1\mathrm{R}_{1}R1​ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

Solution

If B1(0,1.5)\mathrm{B}_{1}(0,1.5)B1​(0,1.5) and B2(0,4)\mathrm{B}_{2}(0,4)B2​(0,4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Solution

Place Reiaan's rectangular study table with three of its feet at the points (8, 9), (11,9)(11,9)(11,9) and (11, 7).

Where will the fourth foot of the table be?

Is this a good spot for the table?

What is the width of the table? The length? Can you make out the height of the table?

Solution

If the bathroom door has a hinge at B1\mathrm{B}_{1}B1​ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Solution

Look at Reiaan's bathroom.

What are the coordinates of the four corners O, F, R, and P of the bathroom?

What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.

Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

Solution

Other rooms in the house:

Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Solution

*6. Are the points $\mathrm{M}(-3,-4), \mathrm{A}(0,0)$ and $\mathrm{G}(6,8)$ on the same straight line? Suggest a method to check this without plotting and joining the points.

*10. Use the connection you found to find the coordinates of B given that M (-7, 1) is the midpoint of A (3, -4) and $\mathrm{B}(x, y)$ .

*11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A $(4,7)$ and B (16, -2).

If $\mathrm{D}_{1} \mathrm{R}_{1}$ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

What are the coordinates of $\mathrm{D}_{1}$ ?

If $\mathrm{R}_{1}$ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

If $\mathrm{B}_{1}(0,1.5)$ and $\mathrm{B}_{2}(0,4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Place Reiaan's rectangular study table with three of its feet at the points (8, 9), $(11,9)$ and (11, 7).

If the bathroom door has a hinge at $\mathrm{B}_{1}$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?