Question

For what value of k the points a(1 5) b(k 1) and c(4 11) are collinear

Answer 1

Answer:

if the points are collinear then area of triangle=0

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

0×2=[1(1-11)+k(11-5)+4(5-1)]

0=(-10+6k+16)

0=6k+6

k=-1

Answer 2

Answer:

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