Question

Find the values of k for which the line (k – 3) x – (4 – k2) y + k2 – 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
$(k-3) x \- (4 \- k^ 2 ) y + k ^2- 7k + 6 = 0 … (1)$

**(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
$(4- k^ 2 ) y = (k- 3) x + k^ 2 \- 7k + 6 = 0$

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

∴ Slope of the given line =$\frac{(k-3)}{(4-k^2)},$ which is of the form $y = mx + c.$
Slope of the x-axis = 0

$∴ \frac{(k-3)}{(4-k^2)}=0$

$⇒ k-3=0$

$⇒ k=3$
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 $k^2 = 4$

$k^2 = 4$
$⇒ k = ±2$
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.

$(k-3)(0)-(4-k^2)(0)+k^2-7k+6=0$

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

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

$k=1\space or\space 6$

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