If the lines represented by $\left(\mathrm{k}^2+2\right) \mathrm{x}^2+3 \mathrm{xy}-6 \mathrm{y}^2=0$ are…
If the lines represented by $\left(\mathrm{k}^2+2\right) \mathrm{x}^2+3 \mathrm{xy}-6 \mathrm{y}^2=0$ are perpendicular to each other, then the values of $\mathrm{K}$ are
$\pm 3$
$\pm 4$
$\pm 1$
$\pm 2$
Solution
The lines $\left(k^2+2\right) x^2+3 x y-6 y^2=0$ are perpendicular to each other.
$\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2$