Question

The line joining two points $A(2,0)$ and $B(3,1)$ ir rotated about $A$ about in the anticlock wise direction through an angle of ${15}^o$. The equation of the line in the new position is:

• A. $x-\sqrt{3}y-2=0$
• B. $x-2y-2=0$
• C. $\sqrt{3}x-y-2\sqrt{3}=0$
• D. $\sqrt{2}x-y-2\sqrt{2}=0$

Answer 1

The correct option is C $\sqrt{3}x-y-2\sqrt{3}=0$

Consider the diagram shown below.

The slope of the line joining the points $A(2,0)$ and $B(3,1)$ is,

$\Rightarrow \frac{1-0}{3-2}=1$

Slope of $AB=\tan \theta$

$\Rightarrow \tan \theta =1$

$\Rightarrow \theta =45{}^{\circ }$

Thus, the line $AB$ makes an angle of $45{}^{\circ }$ with the $x-$axis. Therefore, line $AC$ makes an angle of $45{}^{\circ }+15{}^{\circ }=60{}^{\circ }$ with the $x-$axis. Thus,

Slope of line $AC=\tan 60{}^{\circ }=\sqrt{3}$

Therefore, the equation of new line be,

$y-0=\sqrt{3}(x-2)$

$\sqrt{3}x-y-2\sqrt{3}=0$