Question

Let $l_1$ be the line $4 x +3 y =3$ and $l_2$ be the line $y =8 x$. $L _1$ is the line formed by reflecting $l_1$ across the line $y = x$ and $L _2$ is the line formed by reflecting $l_2$ across the $x$-axis. If $\theta$ is the acute angle between $L_1$ and $L_2$ such that $\tan \theta=a / b$, where a and $b$ are coprime then find $(a+b)$.

Answer 1

$ l_1: 4 x +3 y =3 $
$f ( x )= y =\frac{3-4 x }{3}$
since $f(x)$ and $f^{-1}(x)$ are the mirror images of each other in the line $y=x$ hence we find $f^{-1}(x)$.
$\text { now } y=f(x) \Rightarrow f^{-1}(y)=x$
from(1) $ x =\frac{3(1- y )}{4} ; f ^{-1}( y )=\frac{3(1- y )}{4}$
$\therefore f ^{-1}( x )=\frac{3(1- x )}{4} $
$4 y =3-3 x$
$L _1=3 x +4 y -3=0$
$m _1=-3 / 4 $
$\| y L _2= y =-8 x \text { with } m _2=-8$
if $\theta$ is the acute angle between the lines
$\tan \theta=\left|\frac{ m _2- m _1}{1+ m _1 m _2}\right|=\left|\frac{-8+\frac{3}{4}}{1+(-8)\left(-\frac{3}{4}\right)}\right|=\left|\frac{-29}{28}\right| \Rightarrow \frac{29}{28} \Rightarrow a =29$ and $b =28$
$\therefore a+b=29+28=57$