Class 10 Maths Coordinate Geometry NCERT Solutions Exercise 7.3
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Exercise 7.3 on page 170. Hope you like them and do not forget to like , social_share
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Formula used
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
Coordinate Geometry NCERT Solutions Exercise 7.3
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(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3})
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
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(x_{1},y_{1}) , B(x_{2},y_{2}) and C(x_{3},y_{3})
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
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(x_{1},y_{1}) , B(x_{2},y_{2}) and C(x_{3},y_{3}) is given by
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
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 (x_{1},y_{1}) , (x_{2},y_{2}) and (x_{3},y_{3}) is given by
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
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 (x_{1},y_{1}) , (x_{2},y_{2}) and (x_{3},y_{3}) is given by
$A = \frac {1}{2} [ x_1(y_2 -y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
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
Summary
NCERT Solutions for Class 10 Maths Chapter 7:Coordinate Geometry Exercise 7.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download Coordinate Geometry Class10 Exercise 7.3 as pdf
This chapter 7 has total 3 Exercise 7.1 ,7.2 and 7.3. This is the third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below