- Flash Back from Class IX notes
- |
- Distance formula
- |
- Section Formula
- |
- Area of triangle
- |
- How to Solve the line segment bisection ,trisection and four-section problem's
- |
- How to Prove three points are collinear
- |
- How to solve general Problems of Area in Coordinate geometry

- Coordinate Geometry Problem and Solutions
- |
- Coordinate Geometry Short questions
- |
- Coordinate Geometry 3 Marks Questions
- |
- Coordinate Geometry 5 Marks questions

In this page we have *Class 10 Maths Coordinate Geometry NCERT Solutions* for
Exercise 7.3 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

(i) (2, 3), (–1, 0), (2, – 4) (ii) (–5, –1), (3, –5), (5, 2)

(i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)

Area of triangle ABC of coordinates A(x

i) Area of the given triangle = 1/2 [2 {0- (-4)} + (-1) {(-4) - (3)} + 2 (3 - 0)]

= 1/2 {8 + 7 + 6}

= 21/2 square units.

(ii) Area of the given triangle = 1/2 [-5 {(-5)- (4)} + 3(2-(-1)) + 5{-1 - (-5)}]

= 1/2{35 + 9 + 20}

= 32 square units

Area of triangle ABC of coordinates A(x

For point A, B and C to be collinear, the value of A should be zero

(i) For collinear points, area of triangle formed by them is zero.

1/2 [7 {1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0

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

-2k + 8 = 0

k = 4

(ii) For collinear points, area of triangle formed by them is zero.

1/2 [8 { -4- (-5)} + k {(-5) -(1)} + 2{1 -(-4)}] = 0

8 - 6k + 10 = 0

6k = 18

k = 3

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle.

Coordinates of D, E, and F are given by mid-point formula

D = [(0+2)/2, (-1+1)/2] = (1,0)

E = [(0+0)/2, (-3-1)/2] = (0,1)

F = [(2+0)/2, (1+3)/2] = (1,2)

Now we know that

Area of triangle of coordinates A(x

So Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}

= 1/2 (1+1) = 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]

= 1/2 {8} = 4 square units

Therefore, the required ratio is 1:4.

Let the vertices of the quadrilateral be A (- 4, - 2), B (- 3, -5), C (3, - 2), and D (2, 3). Join AC to form two triangles ABC and ΔACD.

Now we know that

Area of triangle of coordinates (x

Area of ΔABC = 1/2 [(-4) {(-5) - (-2)} + (-3) {(-2) - (-2)} + 3{(-2) - (-5)}]

= 1/2 (12+0+9)

= 21/2 square units

Area of ΔACD = 1/2 [(-4) {(-2) - (3)} + 3{(3) - (-2)} + 2 {(-2) -(-2)}]

= 1/2 (20+15+0)

= 35/2 square units

Area of ΔABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units = 28 square units

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is he median in ΔABC.

Coordinates of point D (midpoint of B &C) = (3+5/2, -2+2/2) = (4,0)

Now we know that

Area of triangle of coordinates (x

Area of ΔABD = 1/2 [(4) {(-2) - (0)} + 3{(0) - (-6)} + (4) {(-6) - (-2)}]

= 1/2 (-8+18-16)

= -3 square units

However, area cannot be negative. Therefore, area of ΔABD s 3 square units.

Area of ΔADC 1/2 [(4) {0 - (2)} + 4{(2) - (-6)} + (5) {(-6) -(0)}]

= 1/2 (-8+32-30)

= -3 square units

However, area cannot be negative. Therefore, area of ΔADC is 3 square units.

The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas

Download Coordinate Geometry Exercise 7.3 as pdf

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

Thanks for visiting our website.

**DISCLOSURE:** THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY **DISCLOSURE** FOR MORE INFO.