Question

Find the value (s) of k for which the points $(3k-1,k-2)$, $(k,k-7)$ and $(k-1,-k-2)$ are collinear.

Answer 1

If the area of a triangle formed by three points is zero, Hence the points are collinear.

Area of a triangle whose sides are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given as,

Area of triangle = $\frac{1}{2}[(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_1-x_1{y}_3)]$

$0=\frac{1}{2}[(3k-1)(k-7)-k(k-2)+k(-k-2)-(k-1)(k-7)+(k-1)(k-2)-(-k-2)(3k-1)]$
$\Rightarrow$ $3k^2-21k-k+7-k^2+2k-k^2-2k-k^2+8k-7+(k^2-3k+2)+(3k^2+6k-k-2)=0$
$\Rightarrow$ $4k^2-12k=0$
$\Rightarrow$ $4k(k-3)=0$
$\Rightarrow$ $k=0,3$
$\therefore$The points are collinear when $k=0$ or $k=3$