The equation of the line common to the pair of lines $\left(\mathrm{p}^2-\mathrm{q}^2\right)…
The equation of the line common to the pair of lines $\left(\mathrm{p}^2-\mathrm{q}^2\right) \mathrm{x}^2+\left(\mathrm{q}^2-\mathrm{r}^2\right) \mathrm{xy}+\left(\mathrm{r}^2-\mathrm{p}^2\right) \mathrm{y}^2=0$ and $(l-\mathrm{m}) \mathrm{x}^2+$ $(\mathrm{m}-\mathrm{n}) \mathrm{xy}+(\mathrm{n}-l) \mathrm{y}^2=0$ is
$x+y=0$
$x-y=0$
$x+y=p q r$
$x-y=p q r$
Solution
We have given equations of lines are:
$\left(p^2-q^2\right) x^2+\left(q^2-r^2\right) x y+\left(r^2-p^2\right) y^2=0$ ...(i)
And $(\ell-m) x^2+(m-n) x y+(n-\ell) y^2=0$ ...(ii)
On comparing these two equations, we get
$p^2-q^2=\ell-m, q^2-r^2=m-n \& r^2-p^2=n-\ell$
Note that points $(1,1)$ and $(0,0)$ satisfy the equation (i) \& (ii). So, the line passing through $(1,1)$ and $(0,0)$ will be the common equation of line.
$(y-0)=\frac{1-0}{1-0}(x-0) \Rightarrow y=x \Rightarrow y-x=0$