Question

Find the acute angle between the pair of lines represented by the following equation.
$x^2-7xy+12y^2=0$.

Answer 1

Given:Pair of lines $x^2-7xy+12y^2=0$

$x^2-3xy-4xy+12y^2=0$

$x(x-3y)-4y(x-3y)=0$

$(x-3y)(x-4y)=0$

$3y-x=0,4y-x=0$

Slope of the line $3y=x$ is $m_1=\frac{1}{3}$

Slope of the line $4y=x$ is $m_2=\frac{1}{4}$

Angle between the lines$=\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{\frac{1}{3}-\frac{1}{4}}{1+\frac{1}{3}\times \frac{1}{4}}\right|$

$=\left|\frac{\frac{1}{3}-\frac{1}{4}}{1+\frac{1}{12}}\right|$

$=\left|\frac{4-3}{12+1}\right|$

$=\left|\frac{1}{13}\right|$

$\therefore \theta ={\tan }^{-1}\left(\frac{1}{13}\right)$