Show that two straight lines through the origin, which makes an angle of $\frac{\pi}{4} $ with the line $px+qy+r=0$ are given by $(p^2-q^2)(x^2-y^2)+4pqxy=0$
My Attempt: Let $y-m_1x=0$ and $y-m_2x=0$ be two straight lines passing through origin. The combined equation is $$(y-m_1x)(y-m_2x)=0$$ $$y^2-(m_1+m_2)xy+m_1.m_2x^2=0$$
In the form of equation $y=mx+c$ for a straight line, $m$ is $\tan \theta$ where $\theta$ is the angle the line makes with the $x$-axis.
Let's look at a line making an angle of $\frac {\pi}4$ with $px+qy+r=0$ - and call the angle for this line $\theta$
We have $\tan \theta =-\frac pq$ and $m= \tan (\theta \pm \frac {\pi} 4)=\frac {\tan \theta \pm \tan \frac{\pi}{4}}{1-\tan \theta \tan \frac {\pi}{4}}$
And this will give you values for $m_1$ and $m_2$ which you can substitute into your work so far.
