Browse Questions

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

$\begin {array} {1 1} (A)\;k=2 & \quad (B)\;k=-2 \\ (C)\;k=\pm 2 & \quad (D)\;\text{ none of the above } \end {array}$

Toolbox:
• If the given line is parallel to $y$ - axis, then its slope is undefined.
The given equation of the line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
It is given that the given line is parallel to $y$ axis.
The above line can be written in the form $y=mx+c$
i.e., $y = \large\frac{(k-3)}{(4-k^2)}$$x+ \large\frac{k^2-7k+6}{(4-k^2)} Hence the slope of the given line is \large\frac{k-3}{4-k^2} Since it is parallel to y - axis, \large\frac{k-3}{4-k^2}$$ = \infty$
(i.e., ) $4-k^2=0$
or $k^2=4$
$k = \pm 2$
Hence the value of $k$ is $\pm 2$