Question

Find the joint equation of lines passing through the origin, each of which making angle of measure ${150}^o$ with the line $x-y=0$.

Answer 1

Slope of the first line $=m_1=\tan {150}^{\circ }=\tan ({180}^{\circ }-{30}^{\circ })=-\tan {30}^{\circ }=\frac{-1}{\sqrt{3}}$

The equation of the first line is $y=\frac{-1}{\sqrt{3}}x$

or $\sqrt{3}y+x=0$

$\therefore$ equations of the lines are $\sqrt{3}y+x=0$ and $x-y=0$

The joint equation is $(x+\sqrt{3}y)(x-y)=0$

$=x^2-xy+\sqrt{3}xy-\sqrt{3}{y}^2=0$

or $x^2+(\sqrt{3}-1)xy-\sqrt{3}{y}^2=0$ is the required joint equation.