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

Hi!

It is given that A (–1, 3), B (2,

*p*) and C (5, –1) are collinear.Distance between two points (

*x*_{1},*y*_{1}) and (*x*_{2},*y*_{2}) is given byAgain on squaring both the sides

(

*p*^{2}– 6*p*+ 18) (*p*^{2}+ 2*p*+ 10) = (–*p*^{2}+ 2*p*+ 12)^{2}*p*

^{4}+ 2

*p*

^{3}+10

*p*

^{2}– 6

*p*

^{3}–12

*p*

^{2}– 60

*p*+18

*p*

^{2}+ 36

*p*+180 =

*p*

^{4}+ 4

*p*

^{2}+ 144 – 4

*p*

^{3}+ 48

*p*– 24

*p*

^{2}

*p*

^{4}– 4

*p*

^{3}+16

*p*

^{2}– 24

*p*+180 =

*p*

^{4}– 4

*p*

^{3}– 20

*p*

^{2 }+ 48

*p*+ 144

36

*p*^{2}– 72*p*+ 36 = 0*p*

^{2}– 2

*p*+1 = 0

(

*P*– 1)^{2}= 0*p*– 1 = 0

*p*= 1

Cheers!

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