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Find k if (2, 3), (4, k) and 6,-3) are collinear |
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Answer» If points are collinear, then area is 0 Area=(1/2)[2(k+3)+4(-3-3)+6(3-k)] 0=(1/2)[2k+6+4(-6)+18-6k] 0=(1/2)[2k+6-24+18-6k] 0=(1/2)[-4k+36] 0=-2k+18 k=18/2=9 |
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