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.