# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.3) Exercise 7.3

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.3) Exercise 7.3

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## Access NCERT solutions for Maths Chapter 7 – Coordinate Geometry

EXERCISE NO: 7.3

1. Find the area of the triangle whose vertices are: $\text{(2, 3), (- 1, 0), (2, - 4) }$

Ans: Assume A$({{x}_{1}},\text{ }{{y}_{1}})=\left( 2,\text{ }3 \right)$ , B$\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( -1\text{ },\text{ }0 \right)$ , C$\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 2,-4 \right)$

Area of a triangle =$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}$

By putting the above values of A,B,C and D

We get,

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

=$\frac{1}{2}\left\{ 8+7+6 \right\}$

=$\frac{21}{2}$ square units

2. Find the area of the triangle whose vertices are: $\text{(-5, -1), (3, -5), (5, 2) }$

Ans: Assume A$({{x}_{1}},\text{ }{{y}_{1}})=\left( -5,-1 \right)$,B$\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( \text{3,-5} \right)$ , C$\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 5,2 \right)$

Area of a triangle =$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}$

By putting the above values of A,B,C and D

We get,

=$\frac{1}{2}\left\{ (-5)((-5)-(2))+3(2-(-1))+5(-1-(-5) \right\}$

=$\frac{1}{2}\left\{ 35+9+20 \right\}$

=$32$square units

3 (i) In each of the following find the value of ‘k’, for which the points are collinear:$\text{(7, - 2), (5, 1), (3, - k})$

Ans: Assume A$({{x}_{1}},\text{ }{{y}_{1}})=\left( 7,-2 \right)$,B$\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( 5,1 \right)$ , C$\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 3,-k \right)$

Area of a triangle for collinear point$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}=0$

By putting the above values of A,B,C and D

We get,

=$\frac{1}{2}\left\{ 7(1+k)+5(2-k)+3((-2)-1) \right\}=0$

=$7+7k-5k+10-9=0$

=$2k+8=0$

=$k=-4$

As a result, the supplied points for $k=-4$ are collinear.

3 (ii) In each of the following find the value of ‘k’, for which the points are collinear:$\text{(8, 1), (k, - 4), (2, - 5)}$

Ans: Assume A$({{x}_{1}},\text{ }{{y}_{1}})=\left( 8,1 \right)$,B$\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( k,-4 \right)$ , C$\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 2,-5 \right)$

Area of a triangle for collinear point$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{1}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}=0$

By putting the above values of A,B,C and D

We get,

=$\frac{1}{2}\left\{ 8(-4-(-5)+k((-5)-(1))+2(1-(-4) \right\}=0$

=$8-6k+10=0$

=$6k=18$

=$k=3$

As a result, the supplied points for $k=3$ are collinear.

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$.Find the ratio of this area to the area of the given triangle.

Ans:

(Image Will Be Updated Soon)

Assume A$({{x}_{1}},{{y}_{1}})=(0,-1)$,B$\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( 2,1 \right)$ , C$\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 0,3 \right)$

Area of a triangle =$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{1}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}$

Let D, E, and F be the midpoints of the triangle's sides. D, E, and F coordinates are determined by :

$D=\left( \frac{0+2}{2},\frac{-1+1}{2} \right)=(1,0)$

$E=\left( \frac{0+0}{2},\frac{3-1}{2} \right)=(0,1)$

$F=\left( \frac{2+0}{2},\frac{1+3}{2} \right)=(1,2)$

By putting the above values we get area of triangle DEF

=$\frac{1}{2}\left\{ (2-1)+1(1-0)+0(0-2) \right\}$

=$\frac{1}{2}(1+1)=1$Square units

Similarly,

Area of triangle ABC=$\frac{1}{2}\left\{ 0(1-3)+2(3-(-1)+0(-1-1) \right\}$

=$\frac{1}{2}(8)=4$Square units.

As a result, Ratio of this area ∆DEF to the area of the triangle ∆ABC= $1:4$

4. Find the area of the quadrilateral whose vertices, taken in order, are $(-4,-2),(-3,-5),(3,-2)$ and $(2,3)$.

Ans:

(Image Will Be Updated Soon)

Let the quadrilateral's vertices be A $(4,2)$ , B$(3,5)$ , C $(3,-2)$, and D $(3,2)$.

Form two triangles, ABC and ACD, by joining AC.

Area of a triangle =$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{1}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}$

By putting the values of A,B,C in the above formula we get:

=$\frac{1}{2}\left\{ (-4)((-5)-(-2)+(-3)((-2)-(-2))+3((-2)-(-2) \right\}$

=$\frac{1}{2}(12+0+9)=\frac{21}{2}$Square units

By putting the values of A,C,D in the above formula we get:

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

=$\frac{1}{2}(20+15+0)=\frac{35}{2}$Square units

So, Area of  triangle ABCD= Area of triangle ABC + Area of triangle ACD

=$\left( \frac{21}{2}+\frac{35}{2} \right)=28$ Square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A $(4,-6)$ , B $(3,-2)$  and C $(5,2)$

Ans:

(Image Will Be Updated Soon)

The triangle's vertices should be A (4, 6), B (3, 2), and C (5, 2).

Assume D is the midpoint of ABC's side BC.

As a result, in ABC, AD is the median.

D point coordinates are:$\left( \frac{3+5}{2}+\frac{-2+2}{2} \right)=(4,0)$

Area of a triangle =$\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{1}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}$

By putting the values of A,B,D in the above formula we get:

Area of triangle ABD=$\frac{1}{2}\left\{ (-4)((-2)-(0)+3(0-(-6))+4((-6)-(-2) \right\}$

=$\frac{1}{2}(-8+18-16)=-3$Square units

By putting the values of A,B,D in the above formula we get:

Area of triangle ADC=$\frac{1}{2}\left\{ (4)(0-(-2)+4(2-(-6))+5((-6)-(0) \right\}$

=$\frac{1}{2}(-8+32-30)=-3$Square units

As we know area cannot be in negative. So, Area of triangle ADC is 3 Square units.

As a result, median AD has clearly separated triangle ABC into two equal-sized triangles.

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

Opting for the NCERT solutions for Ex 7.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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1. What are the topics and subtopics covered in Class 10 Maths Chapter 7?

Ans: Class 10 Maths Chapter 7 deals with Coordinate Geometry and it contains three subtopics which are given below.

• Introduction to Coordinate Geometry.

• Distance Formula.

• Section Formula.

• Area of a Triangle.

2. How many chapters are contained in the Class 10th Maths syllabus?

Ans: There are a total of 15 chapters. Take a look below.

• Chapter 1: Real Numbers.

• Chapter 2: Polynomials.

• Chapter 3: Pair of linear Equations in Two Variables.

• Chapter 5: Arithmetic Progression.

• Chapter 6: Triangles.

• Chapter 7: Coordinate Geometry.

• Chapter 8: Introduction to Trigonometry.

• Chapter 9: Some Applications of Trigonometry.

• Chapter 10: Circles.

• Chapter 11: Constructions.

• Chapter 12: Areas Related to Circles.

• Chapter 13: Surface Areas and Volumes.

• Chapter 14: Statistics.

• Chapter 15: Probability.

3. How many questions are there Class 10 Maths Chapter 7 Exercise 7.3?

Ans: Five questions are there in total in the Exercise 7.3 of Class 10 Maths Chapter 7. Solutions to all these questions are also provided in the NCERT Solutions of Class 10 Maths Chapter 7 Exercise 7.3. These solutions are created by top-notch experts from the relevant field in an accurate manner.