Question

If one vertex of an equilateral triangle of side ${}^{\prime }{a}^{\prime }$ lies at the origin and the other lies on the line $x-\sqrt{3}y=0$, then the co-ordinates of the third vertex are :

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

Answer 1

Answer: (A), (C)

The correct options are
A $(0,a)$
C $(0,-a)$

$x-\sqrt{3}y=0$

$(x_2,y_2)$ lies on $x-\sqrt{3}y=0$

$\Rightarrow x_2=y_2\sqrt{3}$

Dist. b/w $(0,0)\&(x_2,y_2)$

$a=\sqrt{x_2^2+y_2^2}$

$\Rightarrow y_2=\pm a/2$

So, $x_2=\pm a\sqrt{3}/2$

Dist. b/w $(x_1,y_1),(0,0)$

$x_1^2+y_1^2=a^2$

Also,

${(x_1-\frac{a\sqrt{3}}{2})}^2+{(y_1-\frac{a}{2})}^2=a^2$

which gives $x_1=0y_1=\pm a$

$(0,a)(0,-a)$