Question

Find the value of ${}^{\prime }{k}^{\prime }$ for which the given point are collinear
$(8,1),(k,-4),(2,-5)$

Answer 1

When the points are collinear the area of triangle formed by these points would be 0.
$\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)=0$
Area of the triangle formed by the given points $=\frac{1}{2}\left(8\right(-4-(-5))+k(-5-1)+2(1-(-4)\left)\right)$
$=\frac{1}{2}\left(8\right(1)+k(-6)+2(5\left)\right)$
$=\frac{1}{2}(8-6k+10)$
$=\frac{1}{2}(18-6k)$
Since the given points are collinear,
area of the triangle formed by the given points = 0
$\frac{1}{2}(18-6k)=0$
$18-6k=0$
$k=\frac{18}{6}=3$