Q.Prove that the equation to the straight lines through the origin, each of which makes an angle $\alpha$ with the straight line $y = x$, is $x^2 - 2xy\sec2\alpha + y^2 = 0$.

Pair of Straight Lines

A second-degree equation $ax^2+2hxy+by^2=0$ represents two straight lines through the origin; their slopes $m_1,m_2$ satisfy $m_1+m_2=-\dfrac{2h}{b}$ and

$m_1m_2=\dfrac{a}{b}$. The lines are real and distinct when $h^2>ab$, coincident when

$h^2=ab$, and the angle between them is $\tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}$;

they are perpendicular iff $a+b=0$ and coincident/parallel-behaving iff $h^2=ab$.

The general second-degree equation

$ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of lines exactly when

$\Delta=abc+2fgh-af^2-bg^2-ch^2=0$. When it does, the lines meet at the point

found from $\dfrac{\partial}{\partial x}=0,\ \dfrac{\partial}{\partial y}=0$, i.e.

$\Big(\dfrac{hf-bg}{ab-h^2},\dfrac{gh-af}{ab-h^2}\Big)$, and they are parallel when …