Question

find the value of p for which the points (p+1,2p-2) , (p-1,p) and ( p-3,2p-6) are collinear

Answer 1

This is how I did it.... Hope it helps :)

Answer 2

Answer:

The value of p=4

Step-by-step explanation:

Given : The points (p+1,2p-2) , (p-1,p) and ( p-3,2p-6) are collinear.

To find : The value of p?

Solution :

If the points are collinear the area of triangle is 0.

$(x_1,y_1)=(p+1,2p-2) , (x_2,y_2)=(p-1,p) ,(x_3,y_3)=(p-3,2p-6)$

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Substitute the value,

$0=\frac{1}{2}[(p+1)(p-2p+6)+(p-1)(2p-6-2p+2)+(p-3)(2p-2-p)]$

$0=\frac{1}{2}[p^2+p-2p^2+6p-2p+6+2p^2-6p-2p+6-2p^2+2p+2p-2+2p^2-2p-6p-6p+6-p^2+3p]$

$0=[p-2p-6p+3p+6+6-2+6]$

$-4p+16=0$

$p=\frac{16}{4}$

$p=4$

Therefore, The value of p=4.