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)

Asked by Pragya Singh | 1 year ago |  87

##### Solution :-

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

Let points (7, -2) (5, 1), and (3, k) are vertices of a triangle.

Area of triangle = $$\dfrac{1}{2}$$ [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.

Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0

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

8 – 6k + 10 = 0

6k = 18

k = 3

Answered by Abhisek | 1 year ago

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