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

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

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