Question

The base of an equilateral triangle with side $2 a$ lies along the $Y$-axis such that the mid-point of the base is at the origin. Then, which of the following are not the vertices of triangle?

• A. $(\sqrt{3 a}, 0),(0, a),(0,-a)$
• B. $(\sqrt{3} a, 0),(-\sqrt{3} a, 0),(0, a)$
• C. $(-\sqrt{3} a, 0),(0, a),(0,-a)$
• D. None of these

Answer 1

Answer: (B)

Let $B C$ be the base of a triangle which lies on $Y$-axis and third vertex may be $A(h, 0)$ or $A^{\prime}(-h, 0)$.

Since, $\triangle A B C$ is equilateral, then
$A B=B C$
$A B^2=B C^2$
$\Rightarrow (h-0)^2+(0-a)^2=(2 a)^2$
$[\because $ distance between two points $\left(x_1, y_1\right)]$
and $\left(x_2, y_2\right)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}]$
For distance $A B,\left(x_1, y_1\right)=(a, 0)\left(x_2, y_2\right)=(0, a)$
$\Rightarrow h^2+a^2 =4 a^2 $
$ \Rightarrow h^2 =3 a^2 $
$ \Rightarrow h =\pm \sqrt{3} a $ (taking square root)
Hence, the vertices of triangle are $(\sqrt{3} a, 0),(0, a),(0, a)$ or $(-\sqrt{3} a, 0),(0, a),(0,-a)$