Question

Find the value of p for which the points $(-1,3),(2,p)and(5,-1)$are colinear.

Answer 1

Step-1 Given data

Points $(-1,3),(2,p)and(5,-1)$ are colinear.

Then area from by these points is equal to zero.

We know the area of triangle of vertices are $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is given by the formula.

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

Step-2 Finding area of a triangle.

$∆=\frac{1}{2}\left|\left\{-1\left[p-(-1)\right]+2(-1-3)+5(3-p)\right\}\right|$

$\frac{1}{2}\left|-p-1-8+15-5p\right|=0$

$\left|-6p+6\right|=0$

$p=1$

Hence, the value of $p$ is $1$.