Question

Find the value of $k$, so that the points are $A(-2,-3),B(3,-1)$ and $C(5,k)$ are collinear.

Answer 1

Given points $A(\underset{x_1}{-2},\underset{y_1}{-3}),B(\underset{x_2}{3},\underset{y_2}{-1})$ and $C(\underset{x_3}{5},\underset{y_3}{k})$

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

$=\frac{1}{2}|(-2)(-1-k)+3(k-(-3\left)\right)+5(-3-(-1\left)\right)|$

$=\frac{1}{2}|(+2+2k)+3(k+3)-15+5|$

$=\frac{1}{2}|2+2k+3k+9-10|$

$=\frac{1}{2}|5k+1|$

Given points are collinear

$\frac{1}{2}|5k+1|=0$

$\Rightarrow 5k+1=0$

$\Rightarrow 5k=-1$

$\Rightarrow k=-1/5$.