Question

The line $x+2y=4$ is translated parallel to itself by $3$ unit in the sense of increasing $x$ and then rotated by $30^{\circ }$ in the anti-clockwise direction about the point where the shifted line cuts the $x$ -axis. The equation of the line in the new position is

• A. $y=\tan \left(\theta -30^{\circ }\right)(x-4-3\sqrt{5})$
• B. $y=\tan \left(30^{\circ }-\theta \right)(x-4-3\sqrt{5})$
• C. $y=\tan \left(\theta +30^{\circ }\right)(x+4+3\sqrt{5})$
• D. $y=\tan \left(\theta -30^{\circ }\right)(x+4+3\sqrt{5})$

Answer 1

Answer: (A)

The equation of a line parallel to $x+2y=4$ is $x+2y=k$.
Since, the distance between these two lines is 3 , therefore
$\frac{k}{\sqrt{1+4}}-\frac{4}{\sqrt{1+4}}=3$
$\Rightarrow k=4+3\sqrt{5}$
This shifted line cuts $x$ -axis at $(k,0)$.
After rotation the slope of the line is $\tan \left(\theta -30^{\circ }\right)$,
where $\tan \theta =($ slope of
$(x+2y=4)=-\frac{1}{2}$
$\therefore$ The equation of the line in the new position is
$y-0=\tan \left(\theta -30^{\circ }\right)(x-k)$
$\Rightarrow y=\tan \left(\theta -30^{\circ }\right)(x-k)$
where $k=4+3\sqrt{5}$