NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2 are designed to help students apply the distance formula and midpoint theorem to solve real-world problems. This exercise builds on the foundational concepts introduced in Exercise 7.1 and explores the relationship between midpoints of line segments and coordinates of points.

Exercise 7.2 covers:

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2

Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{To conduct Sports Day activities, in your rectangular }\\ \text{shaped school ground ABCD, lines have been drawn}\\ \text{with chalk powder at a distance of 1m each}\text{. 100 flower}\\ \text{pots have been placed at a distance of 1m from each}\\ \text{other along AD, as shown in the following figure}\text{. Niharika}\\ \text{runs }\frac{1}{4}\text{th the distance AD on the 2nd line and posts a}\\ \text{green flag}\text{. Preet runs }\frac{\text{1}}{5}\text{th the distance AD on the eighth}\\ \text{line and posts a red flag}\text{. What is the distance between}\\ \text{both the flags? If Rashmi has to post a blue flag exactly}\\ \text{halfway between the line segment joining the two flags, }\\ \text{where should she post her flag?}\end{array}$

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Coordinate Geometry

Medium

Q.

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

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Coordinate Geometry

Medium

Q.

Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

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Coordinate Geometry

Medium

Q.

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

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Coordinate Geometry

Medium

Q.

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).

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{If A and B are (}-\text{2, }-\text{2) and (2, }-\text{4), respectively,}\\ \text{find the coordinates of P such that AP}=\frac{3}{7}\text{AB and P}\\ \text{lies on the line segment AB.}\end{array}$

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Find the coordinates of the points which divide the}\\ \text{line segment joining A(}-\text{2, 2) and B(2, 8) into four}\\ \text{equal parts.}\end{array}$

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Find the area of a rhombus if its vertices are (3, 0),}\\ \text{(4, 5), (}-\text{1, 4) and (}-\text{2, }-\text{1) taken in order.}\\ \text{[Hint : Area of a rhombus}=\frac{1}{2}\text{(product of its diagonals)]}\end{array}$

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Coordinate Geometry

Medium

Q.

The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{The vertices of a }\mathrm{\Delta }\text{\hspace{0.17em}ABC are A(4, 6), B(1, 5) and C(7, 2). }\\ \text{A line is drawn to intersect sides AB and AC at D and E }\\ \text{respectively, such that }\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{1}}{4}.\text{ Calculate the area }\\ \text{of the }\mathrm{\Delta }\text{\hspace{0.17em}ADE and compare it with the area of }\mathrm{\Delta }\text{\hspace{0.17em}ABC.}\\ \text{(Recall converse of basic proportionality theorem and }\\ \text{Theorem 6.6 related to the ratio of the areas of two}\\ \text{similar triangles).}\end{array}$

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Coordinate Geometry

Medium

Q.

The coordinates of the centre of a circle are (2a, a – 7). Find the value(s) of 'a' if the circle passes through the point (11, –9) and has diameter $$10 \sqrt{2}$$​ units.

[CBSE - 2025]

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Coordinate Geometry

Medium

Q.

Find the point on y-axis which is equidistant form the points (5, –2) and (–3, 2).

[CBSE - 2019]

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Coordinate Geometry

Easy

Q.

Find the coordinates of the mid-point of the line segment joining the points P(–4, 5) and Q(4, 6). Justify on which axis this point lies.

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• Midpoint Theorem: Finding the midpoint of a line segment with given endpoints.

• Application of distance formula to determine the length of line segments.

• Real-life problems that involve finding midpoints and distances between points.

The solutions are explained in step-by-step detail, helping students apply coordinate geometry concepts to practical problems and enhance their problem-solving skills for Class 10 exams.

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

Solution:

Given,
Points A(−1, 7) and B(4, −3) and the ratio

$2:3$

2:3.
The formula for the coordinates of the point

$P(x, y)$

P(x,y) that divides the line segment joining two points

$A(x_1, y_1)$

A(x1​,y1​) and

$B(x_2, y_2)$

B(x2​,y2​) in the ratio

$m:n$

m:n is:

$$P(x, y) = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$$

P(x,y)=(m+nmx2​+nx1​​,m+nmy2​+ny1​​)

Substituting the values into the formula:

$$P(x, y) = \left( \frac{2(4) + 3(-1)}{2 + 3}, \frac{2(-3) + 3(7)}{2 + 3} \right)$$

P(x,y)=(2+32(4)+3(−1)​,2+32(−3)+3(7)​)

$$P(x, y) = \left( \frac{8 - 3}{5}, \frac{-6 + 21}{5} \right) = \left( \frac{5}{5}, \frac{15}{5} \right)$$

P(x,y)=(58−3​,5−6+21​)=(55​,515​)

$$P(x, y) = (1, 3)$$

P(x,y)=(1,3)

Thus, the coordinates of the point are

$(1, 3)$

(1,3).

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

Solution:

Given,
Points A(4, −1) and B(−2, −3).
We need to find the points of trisection, i.e., points that divide the line segment in the ratio 1:2 and 2:1.

First Point (P) (Divides AB in the ratio 1:2):

Using the section formula:

$$P(x, y) = \left( \frac{1(-2) + 2(4)}{1 + 2}, \frac{1(-3) + 2(-1)}{1 + 2} \right)$$

P(x,y)=(1+21(−2)+2(4)​,1+21(−3)+2(−1)​)

$$P(x, y) = \left( \frac{-2 + 8}{3}, \frac{-3 - 2}{3} \right) = \left( \frac{6}{3}, \frac{-5}{3} \right)$$

P(x,y)=(3−2+8​,3−3−2​)=(36​,3−5​)

$$P(x, y) = (2, −\frac{5}{3})$$

P(x,y)=(2,−35​)

Second Point (Q) (Divides AB in the ratio 2:1):

Using the section formula:

$$Q(x, y) = \left( \frac{2(-2) + 1(4)}{2 + 1}, \frac{2(-3) + 1(-1)}{2 + 1} \right)$$

Q(x,y)=(2+12(−2)+1(4)​,2+12(−3)+1(−1)​)

$$Q(x, y) = \left( \frac{-4 + 4}{3}, \frac{-6 - 1}{3} \right) = \left( \frac{0}{3}, \frac{-7}{3} \right)$$

Q(x,y)=(3−4+4​,3−6−1​)=(30​,3−7​)

$$Q(x, y) = (0, −\frac{7}{3})$$

Q(x,y)=(0,−37​)

Thus, the points of trisection are

$(2, −\frac{5}{3})$

(2,−35​) and

$(0, −\frac{7}{3})$

(0,−37​).

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 flower pots have been placed at a distance of 1m each along AD. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?

Solution:

From the figure,
Niharika posts the green flag at

$\frac{1}{4}$

41​th the distance of AD, i.e., at a distance of 25m from the starting point of the 2nd line. Therefore, the coordinates of the point P are

$(2, 25)$

(2,25).
Preet posts the red flag at

$\frac{1}{5}$

51​th the distance of AD, i.e., at a distance of 20m from the starting point of the 8th line. Therefore, the coordinates of the point Q are

$(8, 20)$

(8,20).

Now, the distance between the two flags is:

$$d = \sqrt{(8 - 2)^2 + (25 - 20)^2} = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \, \text{m}$$

d=(8−2)2+(25−20)2​=62+52​=36+25​=61​m

The coordinates of the midpoint M (where Rashmi should post her blue flag) are:

$$M(x, y) = \left( \frac{2 + 8}{2}, \frac{25 + 20}{2} \right) = (5, 22.5)$$

M(x,y)=(22+8​,225+20​)=(5,22.5)

Thus, Rashmi should post her blue flag at

$(5, 22.5)$

(5,22.5) on the 5th line.

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

Solution:

Let the ratio be

$k:1$

k:1.
Using the section formula, the coordinates of the point dividing the line segment in the ratio

$k:1$

k:1 are:

$$P(x, y) = \left( \frac{k \cdot 6 + 1 \cdot (-3)}{k + 1}, \frac{k \cdot (-8) + 1 \cdot 10}{k + 1} \right)$$

P(x,y)=(k+1k⋅6+1⋅(−3)​,k+1k⋅(−8)+1⋅10​)

Substitute the values for

$P(-1, 6)$

P(−1,6):

$$-1 = \frac{k \cdot 6 - 3}{k + 1}, \quad 6 = \frac{k \cdot (-8) + 10}{k + 1}$$

−1=k+1k⋅6−3​,6=k+1k⋅(−8)+10​

Solving the equations:

$$-1 = \frac{6k - 3}{k + 1} \quad \text{and} \quad 6 = \frac{-8k + 10}{k + 1}$$

−1=k+16k−3​and6=k+1−8k+10​

By solving these two equations, we find

$k = 2$

k=2.
Thus, the line segment is divided in the ratio

$2:7$

2:7.

Q5. Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Let the ratio be

$k:1$

k:1.
The x-axis divides the line segment at

$y = 0$

y=0, so applying the section formula:

$$P(x, y) = \left( \frac{k \cdot (-4) + 1 \cdot 1}{k + 1}, \frac{k \cdot 5 + 1 \cdot (-5)}{k + 1} \right)$$

P(x,y)=(k+1k⋅(−4)+1⋅1​,k+1k⋅5+1⋅(−5)​)

Since

$y = 0$

y=0, we substitute this into the second equation:

$$0 = \frac{k \cdot 5 - 5}{k + 1}$$

0=k+1k⋅5−5​

Solving this gives

$k = 1$

k=1.

Thus, the point of division is at

$(x, 0) = (-\frac{3}{2}, 0)$

(x,0)=(−23​,0).
So, the ratio is 1:1 and the coordinates of the point of division are

$(-\frac{3}{2}, 0)$

(−23​,0).

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

Solution:

Using the midpoint property of parallelograms (diagonals bisect each other), we equate the midpoints of diagonals:

$$\text{Midpoint of AC} = \text{Midpoint of BD}$$

Midpoint of AC=Midpoint of BD

From the midpoint formula:

$$\left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right)$$

(21+x​,22+6​)=(24+3​,2y+5​)

Solving the system of equations:

$$\frac{1 + x}{2} = \frac{7}{2} \quad \text{and} \quad \frac{2 + 6}{2} = \frac{y + 5}{2}$$

21+x​=27​and22+6​=2y+5​

We get:

$$x = 6 \quad \text{and} \quad y = 1$$

x=6andy=1

Thus, the values of

$x$

x and

$y$

y are 6 and 1, respectively.

FAQs: Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2

Q1. What is the focus of Exercise 7.2?
Answer:
Exercise 7.2 focuses on solving problems involving the distance formula and midpoint theorem, including finding the midpoint of a line segment and calculating distances in coordinate geometry.

Q2. What is the Midpoint Theorem?
Answer:
The Midpoint Theorem states that the midpoint of a line segment joining two points

$(x_1, y_1)$

(x1​,y1​) and

$(x_2, y_2)$

(x2​,y2​) is given by the formula:

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

M=(2x1​+x2​​,2y1​+y2​​)

where

$M$

M is the midpoint of the line segment.

Q3. How do I find the midpoint of a line segment using the formula?
Answer:
To find the midpoint of a line segment, simply:

1. Take the average of the x-coordinates of the two points.

2. Take the average of the y-coordinates of the two points.

This gives the coordinates of the midpoint.

Q4. How do I apply the distance formula in this exercise?
Answer:
To apply the distance formula, use the following steps:

1. Identify the coordinates of the two points

   $(x_1, y_1)$(x1​,y1​) and

   $(x_2, y_2)$(x2​,y2​).

2. Substitute these values into the distance formula:

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

d=(x2​−x1​)2+(y2​−y1​)2​

3. Simplify the expression to find the distance

   $d$d.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step guidance for solving problems involving the distance formula and midpoint theorem, ensuring students can apply the concepts correctly in the exams. By practicing these solutions, students will improve their problem-solving skills and boost their confidence for Class 10 exams.