Question

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

• A. $(1-2\sqrt{3})y=\sqrt{3}(x-4-3\sqrt{5})$
• B. $(1-2\sqrt{3})y=(2+\sqrt{3})(x-4-3\sqrt{5})$
  
• C. $(1+2\sqrt{3})y=2-\sqrt{3}(x-4-5\sqrt{5})$
• D. none

Answer 1

The correct option is B

$(1-2\sqrt{3})y=(2+\sqrt{3})(x-4-3\sqrt{5})$

Initially:

$x+2y=4$

at x-axis $y=0;x=4$

at y-axis $x=0;y=2$

$\Rightarrow \tan \theta =\frac{2}{4}=\frac{1}{2}$ ........$\left(1\right)$

Let $AP=Z$

$\Delta AMP$

$\sin \theta =\frac{AM}{AP}$

$\sin \theta =\frac{3}{l}$

From eq $\left(1\right)$

$\sin \theta =\frac{1}{\sqrt{5}}=\frac{3}{l}$

$l=3\sqrt{5}$; now coordinate of A

$(4+l,0)$

$(4+3\sqrt{5},0)$

from figure

$\alpha =30+\theta$

$\tan \alpha =\tan (30+\theta )$

$=\frac{\tan 30+\tan \theta }{1-\tan \theta \tan 30}$

$=\frac{\frac{1}{\sqrt{3}}+\frac{1}{2}}{1-\frac{1}{2\sqrt{3}}}=\frac{2+\sqrt{3}}{2\sqrt{3}-1}$

$m=-\tan \alpha$

$m=\frac{2+\sqrt{3}}{1-2\sqrt{3}}$

now eq.

$y-0=\frac{2+\sqrt{3}}{1-2\sqrt{3}}(x-4-3\sqrt{5})$

$(1-2\sqrt{3})y=(2+\sqrt{3})(x-4-3\sqrt{5})$.