Question

If a line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in anticlockwise direction through an angle ${15}^{\circ }$, then the equation of the line in the new position is

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

Answer 1

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

Given:

$A(2,0)\equiv (x_1,y_1)$ and $B(3,1)\equiv (x_2,y_2)$

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}$

Slope of $AB=\frac{1-0}{3-2}=1=\tan {45}^{\circ }$

$\Rightarrow \angle BAX={45}^{\circ }$ ....(From figure)

If $AC$ is the new position of the line $AB$, then $\angle CAX={45}^{\circ }+{15}^{\circ }={60}^{\circ }$.

And thus its equation is $y=\tan {60}^{\circ }(x-2)$

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

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