Question

Find the values of k for which the line$(k-3)x-(4-k^2)y+k^2-7k+6=0$ is :
(a) Parallel to the $x$-axis.
(b) Parallel to the $y$-axis.
(c) Passing through the origin.

Answer 1

The given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
(a) If the given line is parallel to the $x$-axis, then
slope of the given line$=$ slope of the $x$-axis $=0$
$\Rightarrow \frac{(k-3)}{(4-k^2)}=0$
$\Rightarrow k-3=0$
$\Rightarrow k=3$
Thus, if the given line is parallel to the $x$-axis, then the value of $k$ is $3$.
(b) If the given line is parallel to the $y$-axis, it is vertical, hence,its slope will be undefined.
The slope of the given line is $\frac{(k-3)}{(4-k^2)}$
Now,$\frac{(k-3)}{(4-k^2)}$ is undefined at $k^2=4$
$\Rightarrow k=\pm 2$
Thus, if the given line is parallel to the $y$-axis,then the value of $k$ is $\pm 2.$
(c), If the given line is passing through the origin, then point $(0,0)$ satisfies the given equation of line
$(k-3)\left(0\right)-(4-k^2)\left(0\right)+k^2-7k+6=0$
$\Rightarrow k^2-7k+6=0\Rightarrow k^2-6k-k+6=0$
$\Rightarrow (k-6)(k-1)=0\Rightarrow k=1ork=6$
Thus, if the given line is passing through the origin, then the value of $k$ is either $1$ or $6$.