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. Find vertices of the triangle.

Answer 1

Step$-1:$ Simplification of given data
Let $ABC$ be an equilateral triangle
In equilateral triangle all sides are equal
So, $AB=BC=AC=2a$
Given that the base of equilateral triangle i.e., $BC$ lies along the y-axis
such that mid point of $BC\text{is}O(0,0)$
So, $OB=OC=a$
$\therefore$ Coordinates of $B=(0,a)$ and Coordinates of $C=(0,-a)$
Also, Since $O$ is the mid-point of $BC$
$OA$ is the median of $△ABC$
In equilateral triangle,
median & altitude coincide.
So, $OA$ is the altitude of $△ABC$
If $OA$ is the altitude of $△ABC$, then is $OA$ perpendicular to $BC$

So, point $A$ lies on $x-$axis as $x-$axis is perpendicular to $y-$axis.
Since point $A$ lie on the $x-$axis
its $y$ coordinate will be $0$
Let the $x$ coordinate be $x$
So, coordinates of point $A=(x,0)$

Step $2:$ Vertices of triangle
Now, $AB=2a$
Distance between $(x,0)\text{and}(0,a)\text{is}2a$
$\Rightarrow \sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=2a$
$\Rightarrow \sqrt{(x-0{)}^2+(0-a{)}^2}=2a$
$\Rightarrow \sqrt{(x{)}^2+(a{)}^2}=2a$
$\Rightarrow \sqrt{x^2+a^2}=2a$
Squaring both sides
$\Rightarrow {\left(\sqrt{(x^2+a^2)}\right)}^2=(2a{)}^2$
$\Rightarrow x^2+a^2=4a^2$
$\Rightarrow x^2=4a^2-a^2$
$\Rightarrow x^2=3a^2$
$\Rightarrow x=\pm \sqrt{3a^2}$
$\Rightarrow x=\pm a\sqrt{3}$
So, point $A=(x,0)$
$A=(a\sqrt{3},0)\text{or}(-a\sqrt{3},0)$

Therefore, vertices of the triangle are $(0,-a),(0,a)\text{and}(a\sqrt{3},0)\text{or}(0,-a),(0,a)\text{and}(-a\sqrt{3},0)$