Question

Find the value 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

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

a) parallel to x axis

$\therefore y=p$ p is constant

$\therefore$ no x term

so k-3=0

$\Rightarrow k=3$

b) parallel to y axis $\therefore x=p\Rightarrow$ no y terms

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

$\Rightarrow -(4-k^2)y=0$

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

c) passing through origin (0,0)

$\therefore x=0$ y = 0

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

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

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

$\Rightarrow k^2-6k-k+6=0$

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

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

$\therefore k=6,1$