Question

Find the value of $K$, if the points $A(2,5),B(7,0)$ and $C(9,K)$ are collinear.

Answer 1

Given points are $A(\underset{x_1}{2},\underset{y_1}{5}),B(\underset{x_2}{7},\underset{y_2}{0}),C(\underset{x_3}{9},\underset{y_3}{k})$

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

$\Rightarrow$ triangle ABC$=\frac{1}{2}\left|2\right(0-k)+7(k-5)+9(5-0\left)\right|$

$\Rightarrow$ triangle ABC$=\frac{1}{2}|-2k+7k+45-35|$

$\Rightarrow$ triangle ABC$=\frac{1}{2}|5k+10|$

$\therefore$ Given points are collinear

$\frac{1}{2}|5k+10|=0$

$\Rightarrow 5k+10=0$

$\Rightarrow 5k=-10$

$\Rightarrow k=-10/5$

$\therefore k=-2$.