Question

Find the value of k whose (7,-2),(5,1)and (3,k) ate collinear

Answer 1

Since the given points are in collinear then the area of triangle is zero

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

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

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

0 = 7-7k +5k +20 - 9

0 = 7+20 - 9 - 7k +5k

0 = 18 - 3k

- 18 = - 3k

- 18/-3 = k

6 = k

Therefore the value of k = 6

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