Question

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

Answer 1

(k,3),(6,-2),(-3,4) are

collinear

here $x_1=k,y_1=3$

$x_2=6,y_2=-2$

$x_3=-3,y_3=4$

So, as given that the points are collinear, then,

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

$k(-2-4)+6(4-3)+3(3+2)=0$

$k(-6)+6\left(1\right)+(-3)\times 5=0$

$-6k+6-15=0$

$-6k-9=0$

$k=\frac{9}{6}=\frac{3}{2}$