Question

find the value of $k$ for which the points $(k,-1)(2,1)$ and $(4,5)$ are collinear.

Answer 1

Consider the given points.

$(k,-1),(2,1)$ and $(4,5)$

Since, these points are collinear means that the area of triangle must me zero.

So,

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

where $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the points

Therefore,

$k(1-5)+2(5+1)+4(-1-1)=0$

$k(-4)+2\left(6\right)+4(-2)=0$

$-4k+12-8=0$

$-4k+4=0$

$4k=4$

$k=1$

Hence, this is the answer.