Question

If one vertex of an equilateral triangle of side $a$ is origin and the other vertex lies on the line $x-\sqrt{3}y=0$, then the possible coordinates of the other two vertices of the triangle 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 $(-\frac{\sqrt{3}a}{2},-\frac{a}{2})$

Assume vertices of the triangle are $A(0,0),B(\sqrt{3}{y}_1,y_1)$ and $C(x_2,y_2)$
Now triangle ABC is equilateral with side $a$
thus using $|AB|=|BC|=|CA|=a$
we get $y_1=\pm \frac{a}{2},x_2=0$ and $y_2=\pm a$
Hence all choices are correct.