Question

for what value of p are the points (p,2-2p) (-p+1,2p)and(-4-p,6-2p)
collinear
​

Answer 1

Answer:

p = 1/2 or -2

Step-by-step explanation:

In triangle

area = 0 as it is collinear

so

area =

1/2 { y1(x2-x3) + y2(x3-x1) + y3(x1-x2) } =0

= 1/2 { 2-2p(p+1 + 4-p) + 2p(-4-p - p)+ 6-2p(p - p+1) } = 0

= 1/2 { 5 (2-2p ) + 2p ( -4 - 2p ) + 1( 6-2p ) } = 0

= 1/2 { 10 - 10p - 8 - 4p + 6 - 2p } = 0

= 1/2 { 10 - 8 + 6 - 10p - 4p - 2p } = 0

= 1/2 ( 8 - 16p ) = 0

= 8 - 16p = 0 × 2/1 = 0

= 8 - 16p = 0

= -16p = -8

= 16p = 8

= p = 8/16

= p = 1/2

or

= p = -2