Question

Equation of a straight line passing through the point $(2,3)$ and inclined at an angle of ${\tan }^{-1}\left(\frac{1}{2}\right)$ with the line $y+2x=5$ is

• A. $y=3$
• B. $x=2$
• C. $3x+4y-18-0$
• D. $4x+3y-17=0$

Answer 1

Answer: (B), (C)

The correct options are
B $x=2$
C $3x+4y-18-0$

let the slope of required line be m
we know that, angle$\left(\theta \right)$ between two line with slopes $m_{1}and{m}_2$ is given by
$\theta ={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

then ${\tan }^{-1}\frac{1}{2}={\tan }^{-1}|\frac{-2-m}{1-2m}|$
$⟹-4-2m=1-2m⟹$ m is not defined
also$-4-2m=2m-1⟹4m=-3⟹m=\frac{-3}{4}$
let the lines be $x=c_1$ and $3x+4y=c$
these lines pass through $(2,3)$
Thus, $c_1=2$ and $6+12=c=18$
Therefore, lines are $x=2$ and $3x+4y-18=0$