Question

If one vertex of an equilateral triangle of side ${}^{\prime }{a}^{\prime }$ is origin and the other lies on the line $x-\sqrt{3}y=0$, then the coordinates of third vertex 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: (A), (B), (C), (D)

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

Given line $x-\sqrt{3}y=0$
Slope of this line is $\frac{1}{\sqrt{3}}$
$\tan \theta =\frac{1}{\sqrt{3}}$
$\theta ={30}^0$
Since, this line passes through origin and makes an angle of ${30}^0$ with x-axis.
So, the third vertex lies on the line passing through origin and makes an angle of
$({30}^0+{60}^0)$or$({30}^0-{60}^0)$i.e.${90}^0$or${-30}^0$ with x-axis at a distance $a$ from the origin
Hence, equation of these lines are
$\frac{x}{\cos {90}^0}=\frac{y}{\sin {90}^0}=a$

$\Rightarrow x=0,y=a$

$\frac{x}{\cos {(-30}^0)}=\frac{y}{\sin {(-30}^0)}=a$

$\Rightarrow x=\frac{\sqrt{3}a}{2},y=-\frac{a}{2}$
Coordinates of points at a distance $a$ from origin are
$(0,a),(0,-a),(\frac{\sqrt{3}a}{2},-\frac{a}{2}),(-\frac{\sqrt{3}a}{2},\frac{a}{2})$