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

$(K-3)x-(4-K^2)y+K^2-7K+6=0$

Then $m=-\frac{(K-3)}{-(4-K^2)}=\frac{K-3}{4-K^2}$

(a) If the line is parallel to x-axis then m = 0

$\therefore \frac{K-3}{4-K^2}=0\Rightarrow K-3=0$

$\Rightarrow K=3$

(b) If the line is parallel to y-axis then $\frac{1}{m}=0$

$\therefore \frac{4-K^2}{K-3}=0\Rightarrow 4-K^2=0$

$\Rightarrow K^2=4\Rightarrow K=\pm 2$

(c) If the line passes through origin then

$(K-3)\times 0-(4-K^2)\times 0+K^2-K+6=0$

$\Rightarrow K^2-7K+6=0$

$\Rightarrow (K-1)(K-6)=0\Rightarrow K=1,6$