Find the value of k for which the points a(6,-10)b(k,-6)c(0,-7) are collinear
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1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)) = 0
= 6×1-7k+10k-10+6 = 0
= 2+3k = 0
= 3k = -2
= k = -2/3
= 6×1-7k+10k-10+6 = 0
= 2+3k = 0
= 3k = -2
= k = -2/3
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