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

Let the given collinear points are,

$A=(3k-1,k-2),B=(k,k-7)$ and $C=(k-1,-k-2)$

Since, if $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear then the area of triangle formed by these vertices is zero.

$\Rightarrow Ar\left(\Delta ABC\right)=0$

$\Rightarrow \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0$

$\Rightarrow x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

$\Rightarrow (3k-1)(k-7-(-k-2\left)\right)+k\left(\right(-k-2)-(k-2\left)\right)+(k-1)(k-2-(k-7\left)\right)=0$

$\Rightarrow (3k-1)(k-7+k+2))+k(-k-2-k+2)+(k-1\left)\right(k-2-k+7)=0$

$\Rightarrow (3k-1)(2k-5)+k(-2k)+(k-1)\left(5\right)=0$

$\Rightarrow 6k^2-15k-2k+5-2k^2+5k-5=0$

$\Rightarrow 4k^2-12k=0$

$\Rightarrow 4k(k-3)=0$

$\Rightarrow 4k=0$ or $k-3=0$

$\Rightarrow k=0$ or $k=3$.

Hence, the value of $k$ is either $0$ or $3$.