Question

Let origin is one vertex of an equilateral triangle of side length $a$ units. If other vertex lies on the line $x-\sqrt{3}y=0$ in the first quadrant, then the co-ordinates of third vertex is/are

• A. $(0,-a)$
• B. $(0,a)$
• C. $(\frac{\sqrt{3}a}{2},-\frac{a}{2})$
• D. $(-\frac{\sqrt{3}a}{2},\frac{a}{2})$

Answer 1

Answer: (B), (C)

The correct options are
B $(0,a)$
C $(\frac{\sqrt{3}a}{2},-\frac{a}{2})$

Given line: $x=\sqrt{3}y$
Slope $m=\frac{1}{\sqrt{3}}$
So it makes an angle of ${30}^{\circ }$ with $x-$axis in anticlockwise direction.
Now the two possible choices for third vertex make an angle of ${90}^{\circ }$ and $-{30}^{\circ }$ with $x-$ axis.
Now, using parametric coordinates, we get
$(0+a\cos \theta ,0+a\sin \theta )$

Therefore, the possible coordinates third vertex are
$(a\cos {90}^{\circ },a\sin {90}^{\circ })=(0,a)\left(a\cos \right(-{30}^{\circ }),a\sin (-{30}^{\circ }\left)\right)=(\frac{\sqrt{3}a}{2},-\frac{a}{2})$