Question

Find the value of 'k', for which the following points are collinear.
(7, -2), (5, 1), (3, k)

Answer 1

For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1) and (3, k), area = 0
Let $(x_1,y_1)=(7,-2)$, $(x_2,y_2)=(5,1)$ and $(x_3,y_3)=(3,k)$,
Area of a triangle
$=\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right\}$

=$\frac{1}{2}[7\{1-k\}+5(k-(-2))+3\left\{\right(-2)-1\}]=0$
=$7-7k+5k+10-9=0$
=$-2k+8=0$
=$k=4$