Question

find the value of 'K' for which the points are collinear
1) (K,K) (2,3) (4,-1)​

Answer 1

We know that for collinear points, the area of the triangle formed by them is 0.

Therefore, Area of triangle = 1/2(x1(y2-y3) + x2(y3-y1) + x3(y1-y2) = 0

$\bold{Given}\begin{cases}\sf{x_1=k}\sf{x_2=2}\sf{x_3=4}\\ \sf{y_1=k}\sf{y_2=3}\sf{y_3=-1}\end{cases}$

Hence,

1/2(k(3-(-1)) + 2((-1)-k) + 2(k-3)) = 0

1/2(4k-2-2k+2k-6)= 0

(4k-8) = 0

4k = 8

k = 2

Therefore the value of (k,k) = (2,2)

Hope this helps !