Question

3.
The value of k for which the points A (0,1), B (2 ,k) and C(4, -5) are
collinear is
​

Answer 1

-2

x1*[y2-y3]+x2*[y3-y1]+x3*[y1-y2]=0

0*k-5+2*-5-1+4*1-k=0

0+2*-6+4-4k=0

-12+4-4k=0

-8-4k=0

-8=4k

-8/4=k

-2=k

Therefore

k=-2