Question

Line x + 2y = 4 is translated by $\sqrt{5}$ units closer to the origin and then rotated by angle $ta{n}^{-1}\left(\frac{1}{2}\right)$ in the clockwise direction about the point where the shifted line cuts the x-axis. Find the distance of new line from point M(3, 3).___

Answer 1

Perpendicular distance of line $x+2y–4=0$

from (0, 0) is $\frac{4}{\sqrt{5}}$

Let the equation of line parallel to $x+2y–4=0$ be $x+2y+k=0$

As $\frac{|k-4|}{\sqrt{5}}=\sqrt{5}\Rightarrow k=9,-1$

As the line is translated closer to the origin, so k = –1.

$\therefore$ Equation of translated line is $x+2y+1=0$

Now, translated line is rotated through an angle $\theta =ta{n}^{-1}\left(\frac{1}{2}\right)$ in clockwise direction.

So, slope of new line $=tan(\varphi -\theta )=\frac{tan\varphi -tan\theta }{1+tan\varphi tan\theta }=m\left(say\right)$

$\Rightarrow m=\frac{-\frac{1}{2}-\frac{1}{2}}{1+(-\frac{1}{2})\left(\frac{1}{2}\right)}=\frac{-1}{1-\frac{1}{4}}=-\frac{4}{3}$

$\therefore$ Equation of new line is $(y-0)=-\frac{4}{3}(x+1)$

$\Rightarrow 4x+3y+4=0$

Clearly, distance of new line from M(3, 3)

$=|\frac{4\left(3\right)+3\left(3\right)+4}{\sqrt{4^2+3^2}}|=5units$