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For what value of k, the points (k,-1),(5,7)and (8,11)are collinear? |
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Answer» For the points to be collinear,x1(y2-y3) + X2(y3-y1) + x3(y1-y2)=0... (1)now the points are A(1,5) B(K,1) and C(4,11) i.e.x1=1 y1=5x2=K y2=1x3=4 y3=11put these values in equation 1..1(1-11) + K(11-5) + 4(5-1) =0-10+ 6K +16 =06K+6=0.... 6K=-6i.e. K=-1... |
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