Question 1

Q1: Find the distance between the following pairs of points : (i) $(2,3),(4,1)$ (ii) $(-5,7),(-1,3)$ (iii) $(a, b),(-a,-b)$

Accepted Answer

Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ (i) Points $(2,3)$ and $(4,1)$ Given: Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 1)$. Solution: Distance $= \sqrt{(4 - 2)^2 + (1 - 3)^2}$ $= \s...

Question 2

Q10: Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3,6)$ and $(-3,4)$.

Accepted Answer

Given: Let the point be $P(x,y)$. Let the given points be $A(3,6)$ and $B(-3,4)$. The point P is equidistant from A and B, which means $PA = PB$. Condition: $PA = PB \implies PA^2 = PB^2$ Formula: The square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$. So...

Question 3

Q2: Find the distance between the points $(0,0)$ and $(36,15)$. Can you now find the distance between the two towns A and B discussed in Section 7.2.

Accepted Answer

Part 1: Distance between $(0,0)$ and $(36,15)$ Given: Let the points be $O(0,0)$ and $P(36,15)$. So, $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (36, 15)$. Formula: The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For distance from the origin, this simplifies to $d = \sqrt{x^2 + y^2}$...

Question 4

Q3: Determine if the points $(1,5),(2,3)$ and $(-2,-11)$ are collinear.

Accepted Answer

Given: Let the points be $A(1,5)$, $B(2,3)$, and $C(-2,-11)$. Condition for Collinearity: Three points are collinear if the sum of the lengths of any two line segments among them is equal to the length of the remaining line segment. For example, if $AB + BC = AC$, the points are collinear. Formula:...

Question 5

Q4: Check whether $(5,-2)$, ( 6,4 ) and ( $7,-2$ ) are the vertices of an isosceles triangle.

Accepted Answer

Given: Let the vertices of the triangle be $A(5,-2)$, $B(6,4)$, and $C(7,-2)$. Condition for an Isosceles Triangle: A triangle is isosceles if at least two of its sides have equal length. Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Solution: We calculate the lengths of...

Question 6

Q5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for...

Accepted Answer

Given: From the description of the seating arrangement, which can be represented on a coordinate plane, the positions of the four friends are: $A(3,4)$, $B(6,7)$, $C(9,4)$, and $D(6,1)$. To Determine: Whether the quadrilateral ABCD formed by these points is a square. Properties of a Square: 1. All f...

Question 7

Q6: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) $(-1,-2),(1,0),(-1,2),(-3,0)$ (ii) $(-3,...

Accepted Answer

(i) Points: A(-1,-2), B(1,0), C(-1,2), D(-3,0) Solution: We find the lengths of all four sides and the two diagonals. $AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ $BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ $CD = \sqrt{(-3...

Question 8

Q7: Find the point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$.

Accepted Answer

Given: Let the point on the x-axis be $P(x, 0)$. Let the given points be $A(2,-5)$ and $B(-2,9)$. The point P is equidistant from A and B, which means $PA = PB$. Condition: $PA = PB \implies PA^2 = PB^2$ Formula: The square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y...

Question 9

Q8: Find the values of $y$ for which the distance between the points $\mathrm{P}(2,-3)$ and $\mathrm{Q}(10, y)$ is 10 units.

Accepted Answer

Given: The points are $P(2,-3)$ and $Q(10,y)$. The distance between P and Q is 10 units, i.e., $PQ = 10$. Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. Solution: We are given $PQ = 10$, so $PQ^2 = 10^2 = 100$. Using the distance...

Question 10

Q9: If $\mathrm{Q}(0,1)$ is equidistant from $\mathrm{P}(5,-3)$ and $\mathrm{R}(x, 6)$, find the values of $x$. Also find the distances QR and PR .

Accepted Answer

Part 1: Find the values of x Given: The point $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$. This means $QP = QR$, which implies $QP^2 = QR^2$. Formula: The square of the distance formula: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. Solution: $QP^2 = (5 - 0)^2 + (-3 - 1)^2 = 5^2 + (-4)^2 = 25 + 16...

Question 11

Q1: Find the coordinates of the point which divides the join of $(-1,7)$ and $(4,-3)$ in the ratio $2:3$.

Accepted Answer

Given: The line segment joins the points $A(-1,7)$ and $B(4,-3)$. The ratio of division is $m_1:m_2 = 2:3$. Let the point of division be $P(x,y)$. Here, $(x_1, y_1) = (-1, 7)$ and $(x_2, y_2) = (4, -3)$. Formula: The section formula for a point $P(x,y)$ that divides the line segment joining $A(x_1,...

Question 12

Q10: Find the area of a rhombus if its vertices are ( 3,0 ), ( 4,5 ), ( $-1,4$ ) and ( $-2,-1$ ) taken in order. [Hint : Area of a rhombus $=\frac{1}{2}$ (...

Accepted Answer

Given: Let the vertices of the rhombus be $A(3,0)$, $B(4,5)$, $C(-1,4)$, and $D(-2,-1)$. To Find: The area of the rhombus ABCD. Formula: Area of a rhombus $= \frac{1}{2} 	imes d_1 	imes d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals. The diagonals of the rhombus are AC and BD. The di...

Question 13

Q2: Find the coordinates of the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$.

Accepted Answer

Given: Let the line segment be AB, with $A(4,-1)$ and $B(-2,-3)$. The points of trisection are two points, let's call them P and Q, that divide the segment AB into three equal parts, i.e., $AP = PQ = QB$. Finding point P: Point P divides the segment AB in the ratio $1:2$. Here, $(x_1, y_1) = (4, -1)...

Question 14

Q3: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100...

Accepted Answer

Setting up the Coordinates: Let's consider the ground as a coordinate plane where the lines are parallel to the y-axis and AD is along the y-axis. The distance along AD is 100 m (since there are 100 pots at 1m distance). Position of Niharika's Green Flag (G): Niharika is on the 2nd line, so her x-co...

Question 15

Q4: Find the ratio in which the line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$.

Accepted Answer

Given: Let the line segment be AB, with $A(-3,10)$ and $B(6,-8)$. Let the point of division be $P(-1,6)$. Let the ratio in which P divides AB be $m_1:m_2$. Method 1: Using the ratio $m_1:m_2$ Formula: Section formula: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$ Solution: Using the x-coordinate: $-1 =...

Question 16

Q5: Find the ratio in which the line segment joining $\mathrm{A}(1,-5)$ and $\mathrm{B}(-4,5)$ is divided by the $x$-axis. Also find the coordinates of th...

Accepted Answer

Given: The line segment joins $A(1,-5)$ and $B(-4,5)$. It is divided by the x-axis. Understanding the problem: Any point on the x-axis has its y-coordinate equal to 0. So, let the point of division be $P(x,0)$. Let the ratio in which the x-axis divides the line segment AB be $k:1$. Formula: Using th...

Question 17

Q6: If $(1,2),(4, y),(x, 6)$ and $(3,5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Accepted Answer

Given: Let the vertices of the parallelogram be $A(1,2)$, $B(4,y)$, $C(x,6)$, and $D(3,5)$ taken in order. Property of a Parallelogram: The diagonals of a parallelogram bisect each other. This means the midpoint of diagonal AC is the same as the midpoint of diagonal BD. Formula: The midpoint of a li...

Question 18

Q7: Find the coordinates of a point A, where AB is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$.

Accepted Answer

Given: AB is the diameter of a circle. The center of the circle is $C(2,-3)$. One endpoint of the diameter is $B(1,4)$. Let the coordinates of the other endpoint A be $(x,y)$. Property of a Circle: The center of a circle is the midpoint of its diameter. Formula: The midpoint formula: $M(x_M,y_M) = (...

Question 19

Q8: If A and B are $(-2,-2)$ and $(2,-4)$, respectively, find the coordinates of P such that $\mathrm{AP}=\frac{3}{7} \mathrm{AB}$ and P lies on the line...

Accepted Answer

Given: The points are $A(-2,-2)$ and $B(2,-4)$. P is a point on the line segment AB. The condition is $AP = \frac{3}{7} AB$. To Find: The coordinates of point P. Finding the Ratio: The condition $AP = \frac{3}{7} AB$ can be written as $\frac{AP}{AB} = \frac{3}{7}$. Since P lies on AB, we have $AB =...

Question 20

Q9: Find the coordinates of the points which divide the line segment joining $\mathrm{A}(-2,2)$ and $B(2,8)$ into four equal parts.

Accepted Answer

Given: The line segment joins $A(-2,2)$ and $B(2,8)$. We need to find three points, let's call them P, Q, and R, that divide AB into four equal parts, such that $AP = PQ = QR = RB$. Strategy: 1. Point Q is the midpoint of the entire segment AB. 2. Point P is the midpoint of the segment AQ. 3. Point...

NCERT Solutions