Find the value of k for which the line (k-3) | vedclass

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(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$. If a line is parallel to the $y$-axis,it is a vertical line of the for

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$\pm 2$

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(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$. If a line is parallel to the $y$-axis,it is a vertical line of the form $x = c$. For the equation to represent a vertical line,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero. Setting the coefficient of $y$ to zero: $-(4-k^2) = 0$ $\Rightarrow 4-k^2 = 0$ $\Rightarrow k^2 = 4$ $\Rightarrow k = \pm 2$. Now,check the coefficient of $x$ for these values: If $k = 2$,the coefficient of $x$ is $(2-3) = -1 
eq 0$. If $k = -2$,the coefficient of $x$ is $(-2-3) = -5 
eq 0$. Since both values result in a valid vertical line,the value of $k$ is $\pm 2$.

Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ is parallel to the $y$-axis.

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