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Question

Find the value of k, if the points (k,3),(6,2) and (3,4) are collinear.

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Solution

Points A(K,3); B(6,2); C(3,4) are coplanar.
Hence, area =0

12x1x2y1y2x1x3y1y3=0

k63+2k+334=0

(k6)(1)(5)(k+3)=0
k+65k15=0
6k=6
k=32

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