# Find the values of k for which the lineis

Question:

Find the values of $k$ for which the line $(k-3) x-\left(4-k^{2}\right) y+k^{2}-7 k+6=0$ is

(a) Parallel to the $x$-axis,

(b) Parallel to the $y$-axis,

(c) Passing through the origin.

Solution:

The given equation of line is

$(k-3) x-\left(4-k^{2}\right) y+k^{2}-7 k+6=0 \ldots(1)$

(a) If the given line is parallel to the x-axis, then

Slope of the given line = Slope of the x-axis

The given line can be written as

$\left(4-k^{2}\right) y=(k-3) x+k^{2}-7 k+6=0$

$y=\frac{(k-3)}{\left(4-k^{2}\right)} x+\frac{k^{2}-7 k+6}{\left(4-k^{2}\right)}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=\frac{(k-3)}{\left(4-k^{2}\right)}$

Slope of the x-axis = 0

$\therefore \frac{(k-3)}{\left(4-k^{2}\right)}=0$

$\Rightarrow k-3=0$

$\Rightarrow k=3$

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 $\frac{(k-3)}{\left(4-k^{2}\right)}$.

Now, $\frac{(k-3)}{\left(4-k^{2}\right)}$ is undefined at $k^{2}=4$

$k^{2}=4$

$\Rightarrow k=\pm 2$

Thus, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the

given equation of line.

$(k-3)(0)-\left(4-k^{2}\right)(0)+k^{2}-7 k+6=0$

$k^{2}-7 k+6=0$

$k^{2}-6 k-k+6=0$

$(k-6)(k-1)=0$

$k=1$ or 6

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.