Find the value of k for which the line (k-3)x | 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 $x$-axis,its slope must be $0$. Rewriting t

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$k = 3$

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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 $x$-axis,its slope must be $0$. Rewriting the equation in the form $y = mx + c$: $(4-k^2)y = (k-3)x + (k^2 - 7k + 6)$ $y = \frac{k-3}{4-k^2}x + \frac{k^2 - 7k + 6}{4-k^2}$. For the line to be parallel to the $x$-axis,the slope $m = \frac{k-3}{4-k^2}$ must be $0$. This implies $k-3 = 0$,so $k = 3$. However,we must ensure the coefficient of $y$ is not zero. If $k=3$,the coefficient of $y$ is $-(4-3^2) = -(4-9) = 5 
eq 0$. Thus,the value of $k$ is $3$.

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 $x$-axis.

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