# Class 10 Maths Coordinate Geometry NCERT Solutions Exercise 7.3

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.
Question 1 Find the area of the triangle whose vertices are
(i) (2, 3), (–1, 0), (2, – 4) (ii) (–5, –1), (3, –5), (5, 2)
Question 2. In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)
Question 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.
Question 4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).
Question 5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for Δ ABC whose vertices are A (4, – 6), B (3, –2) and C (5, 2).

Solution 1
Area of triangle ABC of coordinates A(x1,y1), B(x2,y2) and C(x3,y3)

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
Solution2
Area of triangle ABC of coordinates A(x1,y1) , B(x2,y2) and C(x3,y3)

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

Solution 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(x1,y1) , B(x2,y2) and C(x3,y3) is given by

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.

Solution 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 (x1,y1) , (x2,y2) and (x3,y3) is given by

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
Solution 5
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 (x1,y1) , (x2,y2) and (x3,y3) is given by

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