Question

A line through the point A(2, 0), which makes an angle of ${30}^{\circ }$ with the positive direction of x-axis is rotated about A in clockwise direction through an angle ${15}^{\circ }$. The equation of the straight line in the new position is

• A. $(2-\sqrt{3})x-y-4+2\sqrt{3}=0$
• B. $(2-\sqrt{3})x+y-4+2\sqrt{3}=0$
• C. $(2-\sqrt{3})x-y+4+2\sqrt{3}=0$
• D. $(2-\sqrt{3})x-9y+4+2\sqrt{3}=0$

Answer 1

The correct option is A

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

Let AB be the initial position of the line and AC be its new position.

Slope of the line

$AC=tan{15}^{\circ }=(2-\sqrt{3})$

$\therefore$ Equation of the line AC is

$(y-0)=(2-\sqrt{3})(x-2)ory=(2-\sqrt{3})x-4+2\sqrt{3}or(2-\sqrt{3})x-y-4+2\sqrt{3}=0$