Question

Find the value of k for which the points are collinear (7,-2) (5,1) and (3,k)

Answer 1

Answer:

1/2 [x1(y2-y3)+x2 (y3-y1) + x3(y1-y2)] = 0

[7(1-k) + 5(k+2) + 3(-2-1)]=0

7-7k +5k+10-9 = 0

8-2k=0

2k=8

k=8/2 = 4