Question

Find the value of $k$ if points $(k,3),(6,-2)$ and $(-3,4)$ are collinear.

Answer 1

The given points $A(k,3),B(6,-2),$ and $C(-3,4)$ are collinear if area of the $△ABC$ formed by joining these points will be zero.

or, area $△ABC=0$

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

$\Rightarrow \frac{1}{2}|\left[k\right(-2-4)+6(4-3)+(-3\left)\right(3+2\left)\right]|=0$

$\Rightarrow \frac{1}{2}|\left[k\right(-6)+6-15]|=0$

$\Rightarrow |[-6k-9]|=0$

$\Rightarrow k=\frac{9}{-6}$

$\Rightarrow k=\frac{-3}{2}$