Question

Let $l_1$ be the line $4x+3y=3$ and $l_2$ be the line $y=8x$. $L_1$ 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

Answer: 57

$l_1:4x+3y=3$
$f(x)=y=\frac{3-4x}{3}$
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) $x=\frac{3(1-y)}{4};f^{-1}(y)=\frac{3(1-y)}{4}$
$\therefore f^{-1}(x)=\frac{3(1-x)}{4}$
$4y=3-3x$
$L_1=3x+4y-3=0$
$m_1=-3/4$
$\Vert y{L}_2=y=-8x\text{ with }{m}_2=-8$
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 $b=28$
$\therefore a+b=29+28=57$