# The angle between two lines is $\large\frac{\pi}{4}$ and the slope of one of them is $\large\frac{1}{4}$ and other is $x$.Find the value of $x$.

$\begin{array}{1 1}(a)\;\large\frac{-3}{5},\large\frac{5}{3}&(b)\;\large\frac{3}{5},\frac{-5}{3}\\(c)\;\large\frac{4}{3},\frac{2}{3}&(d)\;\large\frac{7}{3},\frac{3}{5}\end{array}$

Let $\theta$ be the acute angle between the lines of slope $m_1$ and $m_2$ then
$\tan \theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
$\theta=\large\frac{\theta}{4}$
$m_1=\large\frac{1}{4},$$m_2=x \tan\large\frac{\pi}{4}=\bigg|\large\frac{\Large\frac{1}{4}-x}{1+\Large\frac{x}{4}}\bigg| \pm 1=\large\frac{\Large\frac{1}{4}-x}{1+\Large\frac{x}{4}} Taking +ve sign 1=\large\frac{\Large\frac{1}{4}-x}{1+\Large\frac{x}{4}} x+\large\frac{x}{4}=\frac{1}{4}$$-1$
$\large\frac{5x}{4}=\frac{-3}{4}$
$x=-\large\frac{3}{5}$
Taking -ve sign
$-1=\large\frac{\Large\frac{1}{4}-x}{1+\Large\frac{x}{4}}$
$\large\frac{-5}{4}=\frac{-3x}{4}$
$x=\large\frac{5}{3}$
Hence (a) is the correct answer.