Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is

Passing through the origin.


The given equation of the

(k–3) x – (4 – k2) y + 6


If the given line is passing through the origin, then point (0,0) satisfies the given equation of line.


(k -3) (0) – (4-k2)(0) + k2 -7k +6 = 0


k2 -7k +6 = 0


k2 -6k-k +6 = 0


(k-6)(k-1)=0


k = 1or 6


Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.


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