NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

In real life, coordinate geometry can be used in navigation, GPS tracking, computer graphics, game development, and for designing various structures. It is an integral part of Mathematics. This chapter is the link between geometry and algebra. It deals with geometrical figures in coordinate systems and coordinates like the x-axis and y-axis to locate the exact position of the points in two-dimensional planes. This chapter also consists of an introductory phase into coordinate geometry, which gradually makes students aware of topics like distance formulas and section formulas.

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1. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FDF Free Download
2. Coordinate Geometry Class 10: Important Formulae
3. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Exercise)
4. NCERT Solutions for Class 10 Maths: Chapter Wise
5. NCERT Books and NCERT Syllabus
6. NCERT solutions of class 10 subject-wise
7. NCERT Exemplar solutions: Subject-wise

This chapter can be challenging for students who don't have their basic concepts clear. This article allows every student to understand all the necessary concepts and strengthen their knowledge of coordinate geometry. The NCERT solutions for Class 10 Maths Chapter 7 coordinate geometry will offer a systematic and structured approach for the exercise problems in the NCERT textbook to prepare well for your board exams by providing detailed solutions to all the exercise questions. These solutions also provide a valuable resource to the students to enhance their performance in their board exams. Follow the Coordinate Geometry Class 10 Mathematics Notes page to learn more!

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FDF Free Download

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Coordinate Geometry Class 10: Important Formulae

Distance Formulae:

• For a line defined by two points A(x₁, y₁) and B(x₂, y₂), the distance between these points can be calculated using the formula:

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

Section Formula:

• When a point P divides a line AB, with A(x₁, y₁) and B(x₂, y₂) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:

• Point P = $\{\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \}$

Midpoint Formula:

• The midpoint of a line AB, defined by A(x₁, y₁) and B(x₂, y₂), can be determined using the following formula:

• Midpoint P = $\{ \frac{x_1+x_2}2,\frac{(y_1+y_2}2\}$

Area of a Triangle:

• Consider a triangle formed by points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). The area of this triangle can be calculated using the formula:

Area ∆ABC = $\frac12$|x₁(y₂ − y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Exercise)

Q1. (i) Find the distance between the following pairs of points : (2, 3), (4, 1)

Answer:

Given points: (2, 3), (4, 1)

Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$⇒D= \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$

Q1. (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)

Answer:

Given points: (– 5, 7), (– 1, 3)

Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$⇒D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}$

Q1. (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)

Answer:

Given points: (a, b), (– a, – b)

Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$⇒D= \sqrt{(-a-a)^2+(-b-b)^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2+b^2}$

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?

Answer:

Given points: (0, 0) and (36, 15)

Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$⇒D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39$

The distance between the two towns A and B is, thus, 39 km for the given town location

$(0,0)$ and $(36,15)$.

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

Answer:

Let the points (1, 5), (2, 3), and (– 2, – 11) represent the vertices A, B, and C of the given triangle, respectively.

$A = (1,5),\ B = (2,3),\ C = (-2,-11)$

Therefore,

$AB = \sqrt{(1-2)^2+(5-3)^2} = \sqrt{5}$

$BC = \sqrt{(2-(-2))^2+(3-(-11))^2} = \sqrt{4^2+14^2} = \sqrt{16+196} = \sqrt{212}$ $CA = \sqrt{(1-(-2))^2+(5-(-11))^2} = \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265}$ Since these are not satisfied.

$AB+BC \neq CA$

$BA+AC \neq BC$

$BC+CA \neq BA$

As these cases are not satisfied.
Hence, the points are not collinear.

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

Answer:

The distance between two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is given by:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangles A, B, and C, respectively.

$AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}$

$BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}$

$CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2$

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Q5. In a classroom, 4 friends are seated at points A, B, C and D, as shown in Fig. 7.8. Champa and Chameli walk into the class, and after observing for a few minutes, Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.

Answer:

The coordinates of the points:

$A(3,4),\ B(6,7),\ C(9,4),$ and $D(6,1)$ are the positions of 4 friends.

The distance between two points $A(x_{1},y_{1}),$ and $B(x_{2},y_{2})$ is given by:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Hence,

$AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

And the lengths of diagonals:

$AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6$

$BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6$

So, here it can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square, and Champa is saying right.

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

Answer:

Let the given points $(-1,-2),\ (1,0),\ (-1,2),$ and $(-3,0)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

Finding the length of the diagonals:

$AC= \sqrt{(-1+1)^2+(-2-2)^2} =\sqrt{0+16} = 4$

$BD= \sqrt{(1+3)^2+(0-0)^2} =\sqrt{16+0} = 4$

It is clear that all sides are of the same lengths, and also, the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Q6. (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)

Answer:

Let the given points $(-3,5),\ (3,1),\ (0,3),$ and $(-1,-4)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

$BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}$

$CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2$

$AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}$

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like a square, rectangle, etc.

Q6. (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

Let the given points $(4,5),\ (7,6),\ (4,3),\ (1,2)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}$

$BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}$

$CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}$

$AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}$

And the diagonals:

$AC =\sqrt{(4-4)^2+(5-3)^2} = \sqrt{0+4} = 2$

$BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

Here, we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

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

Answer:

Let the point which is equidistant from $A(2,-5)\ and \ B(-2,9)$ be $X(x,0)$ as it lies on X-axis.

Then, we have,

Distance AX $= \sqrt{(x-2)^2+(0+5)^2}$

and Distance BX $= \sqrt{(x+2)^2+(0+9)^2}$

According to the question, these distances are equal in length.

Hence, we have,

$\sqrt{(x-2)^2+(0+5)^2}$ $= \sqrt{(x+2)^2+(0+9)^2}$

Squaring both sides, we get,

$(x-2)^2+25 = (x+2)^2+81$
$⇒x^2-4x+4+25=x^2+4x+4+81$
$⇒4x+4x=-81+25$
$⇒8x=-56$
$\therefore x=-7$

Hence, the point is $X(-7,0)$.

Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Answer:

Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.

The distance formula :

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

So, given $PQ = 10$ units

$PQ= \sqrt{(10-2)^2+(y-(-3))^2} = 10$

After squaring both sides,

$\Rightarrow (10-2)^2+(y-(-3))^2 = 100$

$\Rightarrow (y+3)^2 = 100 - 64$

$\Rightarrow y+3 = \pm 6$

$\Rightarrow y = 6 - 3$ or $y = -6-3$

Therefore, the values are $y = 3$ or $y=-9$.

Q9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Answer:

Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$ .

Then, the distances $PQ = RQ$.

Distance $PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$

Distance $RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$

$\Rightarrow \sqrt{x^2+25} = \sqrt{41}$

Squaring both sides, we get,
$⇒x^2+25=41$

$\Rightarrow x^2 = 16$

$\Rightarrow x = \pm 4$

The points are: $R(4,6)\ or\ R(-4,6.)$

CASE I: when R is $(4,6)$

The distances QR and PR.

$QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}$

CASE II: when R is $(-4,6)$

The distances QR and PR.

$QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}$

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

Answer:

Let the point $P(x,y )$ is equidistant from $A(3,6)$ and $B(-3,4)$ .

Then, the distances $AP =BP$

$AP = \sqrt{(x-3)^2+(y-6)^2}$ and $BP = \sqrt{(x-(-3))^2+(y-4)^2}$

$\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}$

Squaring both sides, we obtain

$\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2$

$\Rightarrow (2x)(-6)+(2y-10)(-2)= 0$ $\left [\because a^2-b^2 = (a+b)(a-b) \right ]$

$\Rightarrow -12x-4y+20 = 0$

$\Rightarrow 3x+y-5 = 0$

Thus, the relation is $3x+y-5 = 0$ between x and y.

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

Answer:

Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_{1}:m_{2}$ then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Substituting the values in the formula:

Here, $m_{1}:m_{2} = 2:3$

$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$

$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$

$⇒(1,3)$

Hence, the coordinate is $P \left (1, 3 \right )$.

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

Answer:

Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$

Then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

By observation point, P divides AB internally in the ratio $1:2$.

Hence, $m:n = 1:2$

Substituting the values in the equation, we get;

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

$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$

$\Rightarrow P \left (2 , \frac{-5}{3} \right )$

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n = 2:1$

Substituting the values in the equation above, we get

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

$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$

$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$

Hence, the points of trisections are $P \left (2 , \frac{-5}{3} \right )$ and $Q\left (0 , \frac{-7}{3} \right )$

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 from each other along AD, as shown in Fig. 7.12. Niharika runs $\frac14$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac15$th the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag

Answer:

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times100\ m = 25\ m$ from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$ are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times100\ m = 20\ m$ from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$.

The distance $PQ$ is given by,

$PQ = \sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61} m$

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$.

Then, by Section Formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$⇒x = \frac{2+8}{2},\ y = \frac{25+20}{2}$

$⇒x = 5,\ y = 22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m 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).

Answer:

Let the ratio be $k:1$

Then, by section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

Given point $P(x,y) = (-1,6)$

$-1 = \frac{6k-3}{k+1}$

$\Rightarrow -k-1 = 6k-3$

$\Rightarrow k = \frac{2}{7}$

Hence, the point $P$ divides the line AB in the ratio $2:7$.

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.

Answer:

Let the point on the x-axis be $P(x,0)$, and it divides it in the ratio $k:1$.

Then, we have

Section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

$⇒ \frac{ky_{2}+y_{1}}{k+1} = 0$

$⇒k =-\frac{y_{1}}{y_{2}}$

Hence, the value of k will be: $k =-\frac{-5}{5}= 1$

Therefore, the x-axis divides the line in the ratio $1:1$, and the point will be,

Putting the value of $k=1$ in the section formula.

$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$

$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$

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

Answer:

Let the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonals AC and BD also divides these diagonals.

Therefore, O is the midpoint of AC and BD.

The coordinates of the point O when it is the mid-point of AC.

$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$

The coordinates of the point O when it is the mid-point of BD.

$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$

Since both coordinates are of the same point O.

Therefore,

$\frac{x+1}{2} =\frac{7}{2}$ and $4 = \frac{5+y}{2}$

Or,

$x = 6\ and\ y = 3$

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

Answer:

As the centre point $C(2,-3)$ will be the midpoint of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$.

Given point $B(1,4)$ .

Therefore,

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

$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$

$\Rightarrow x = 3\ and\ y = -10$ .

Therefore, the coordinates of A are $(3,-10).$

Q8. If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

As $AP = \frac{3}{7}AB$

$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

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

$⇒P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$

$⇒P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$

Q9. Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Points C, D, and E.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Here, point D divides the line segment AB into two equal parts; hence

$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$

$⇒D(x_{2},y_{2})= \left (0 , 5 \right )$

Now, point C divides the line segment AD into two equal parts hence

$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$

$⇒C(x_{2},y_{2})= \left (-1 , \frac{7}{2} \right )$

Also, point E divides the line segment DB into two equal parts hence

$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$

$⇒E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$

Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

Let the vertices of the rhombus be:

$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$

Area of the rhombus ABCD is given by:

$= \frac{1}{2}\times$ Product of length of diagonals

Hence, we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of the diagonal AC:

$AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}$

Length of the diagonal BD:

$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$

Thus, the area will be,

$= \frac{1}{2}\times (AC)\times(BD)$

$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24$ square units

To ease the learning for students, the below links to the exercises can also be used. Two exercises have been segregated, and students can analyze each exercise at a time on their own will.

• Coordinate Geometry Class 10 Exercise 7.1
• Coordinate Geometry Class 10 Exercise 7.2

NCERT Solutions for Class 10 Maths: Chapter Wise

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, which is very useful for students before strategizing their study plan.
Also, links to some reference books which are important for further studies.

• NCERT Books Class 10 Maths
• NCERT Syllabus Class 10 Maths
• NCERT Books Class 10
• NCERT Syllabus Class 10

NCERT solutions of class 10 subject-wise

Students can check the following links for more in-depth learning.

• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Science

NCERT Exemplar solutions: Subject-wise

After solving all the exercises, if students want to practice more, then exemplar solutions can be practised. For that, the following links can be used.

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science