Question

Find the value of k of the points (k,2- 2k),(1-k,2k) ,(-4-k,6-2k) are collinear

Answer 1

let (k,2-2k),(1-k,2k),(-4-k,6-2k) be A,B&C respectively.
since,the points A,B&Care collinear
therefore, slope of AB= slope of BC
since, m=y2 -y1/x2-x1
2k-2+2k/1-k-k=6-2k-2+2k/-4-k-2+2k
-24k+12=4-8k
16k=8
k=1/2

Answer 2

if 3 points (a,b) (c,d) (e,f) are colinear
then (d-b)/(c-a)=(f-b)/(e-a)
so here in the question
the points given are colinear
so (2k--2+2k)/(1-k-k)=(6-2k-2+2k)/(-4-k-k)
or (4k-2)/(1-2k)=4/(-4-2k)
or 4-8k=-16k-8k^2+8+4k=-12k-8k^2+8
or -4k-8k^2+4=0
or 2k^2+k-1=0
or 2k^2+(2-1)k-1=0
or2k^2+2k-k-1=0
or 2k (k+1)-1 (k+1)=0
or (k+1)(2k-1)=0
or k=(-1 ) or 1/2