Question

Find the angle between the following lines:
$x-y\sqrt{3}-5=0;x\sqrt{3}+y-7=0$

Answer 1

angle between $x-y\sqrt{3}-5=0$ and $x\sqrt{3}+y-7=0$

Slope of $x-y\sqrt{3}-5=0=\frac{-a}{b}=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}=m_1$

Slope of $x\sqrt{3}+y-7=\frac{-a}{b}=\frac{-\sqrt{3}}{1}=-\sqrt{3}=m_2$

$m_1\times m_2=\frac{1}{\sqrt{3}}\times -\sqrt{3}=-1$

$\therefore$ lines are perpendicular

angle between $x-\sqrt{3}y-5=0$ and $x\sqrt{3}+y-7=0$