Question

The base of an equilateral triangle with side $2a$ 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. $(\sqrt{3a},0),(0,a),(0,-a)$
• 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 $BC$ 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, $△ABC$ is equilateral, then
$AB=BC$
$A{B}^2=B{C}^2$
$\Rightarrow (h-0{)}^2+(0-a{)}^2=(2a{)}^2$
$[\because$ 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 $AB,\left(x_1,y_1\right)=(a,0)\left(x_2,y_2\right)=(0,a)$
$\Rightarrow h^2+a^2=4a^2$
$\Rightarrow h^2=3a^2$
$\Rightarrow h=\pm \sqrt{3}a$ (taking square root)
Hence, the vertices of triangle are $(\sqrt{3}a,0),(0,a),(0,a)$ or $(-\sqrt{3}a,0),(0,a),(0,-a)$