Question

if(k,2-2k),(-k+1,2k),(-4-k,6-2k) are collinear then k=

Answer 1

For the three points to be collinear, the area of triangle formed by them must be equal to 0.
By equating triangle formula with 0, we get

X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)=0
k(2k-6+2k)-k+1(6-2k-2+2k)-4-k(2-2k-2k)=0
k(4k-6)-k+1(4)-4-k(2-4k)=0
4k^2-6k-4k+4-8+16k-2k+4k^2=0
8k^2+4k-4=0
4(2k^2+k-1)=0
2k^2-2k+k-1=0
2k(k-1)+1(k-1)=0
So, 2k+1=0 or k-1=0
k=-1/2 or k=1