Question

In each of the following find the value of ‘k’ for which the points are collinear (7, -2), (5, 1), (3, k)

Answer 1

Let the points be A(7, -2), B(5, 1) and C(3, k)

We know that, when the points are collinear, the area of the triangle becomes 0.

Therefore ar(∆ABC) = 0

➡ 1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0

➡ 7(1 - k) + 5(k + 2) + 3(-2 - 1) = 0

➡ 7 - 7k + 5k + 10 - 9 = 0

➡ 8 - 2k = 0

➡ -2k = -8

➡ k = -8/-2

➡ k = 4

Hence, the value of k is 4 for which the given points are collinear.