For what values of k is the line (k-3)x-(5-k^{2})y+k^{2}-7k+6=0 parallel to the line x+y=1?

Explanation

The slope of the line (k-3)x-(5-k^2)y+C=0 is m_1 = \frac{k-3}{5-k^2}. The slope of the line x+y=1 is m_2 = -1. Since parallel lines have equal slopes, \frac{k-3}{5-k^2} = -1. This gives k-3 = k^2-5 \implies k^2-k-2=0. Factoring the quadratic yields (k-2)(k+1)=0, so k=2 or k=-1.

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