Question

if points (-4,6),(k,-2) and (5,-6) are collinear then find the value of k?

Answer 1

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

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

1/2{-4(4)+k(-12)+5(8)}=0

1/2{-16-12k+40}=0

1/2{-12k+24}=0

-12k+24÷2=0

-12k+24=0

-12k= -24

k= -24/-12

k= 2.

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