Alg6.19

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Find the value of k,if the points (k,2-2k),(-k+1,2k),(-4-k,6-2k) are collinear

If we find the area of the triangle formed as by the vertices given,and should make equal to zero,since they are collinear.

\frac{1}{2}|[k(2k-6+2k)+(-k+1)(6-2k-2+2k)+(-4-k)(2-2k-2k)=0]|\,

|4k^2+2k-2=0\,

k={\frac{1}{2}}\,


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