Question

The base $BC$of an equilateral triangle $ABC$ lies on $y-axis$. The coordinates of point$C$ are $(0,-3)$. If the origin is the mid-point of the base$BC$, find the coordinates of the points $A$ and$B.$

Answer 1

Step 1: Find the coordinates of point $B.$

Given $(0,0)$is the mid-point of the base $BC.$

$C=(0,-3)$

So, according to question, the points of equilateral triangle $ABC$ is as shown in the below figure.

It is given that origin is mid-point of the base $BC.$

So, the coordinate of point $B$ is given by

$\left(\frac{0+0}{2},\frac{3+3}{2}\right)\equiv \left(0,3\right)$

So $B=(0,3)$

Step 2: Use distance formula and find the coordinate of point $A$

Let $A=(x,0).$

Since, $△ABC$is an equilateral triangle

Therefore,$BC=AB$

Using distance formula, we get

$\sqrt{[{(0-0)}^2+{(3+3)}^2]}=\sqrt{[{(x-0)}^2+{(0-3)}^2]}$

Squaring both sides we get,

$$\begin{gathered} 6^2=x^2+9 \\ \Rightarrow x^2=36-9 \\ \Rightarrow x^2=27 \\ \Rightarrow x=\sqrt{27} \\ \Rightarrow x=\pm 3\sqrt{3} \end{gathered}$$

So$A=(3\sqrt{3},0)$

Hence, the coordinates of the points $A=(3\sqrt{3},0)$ and $B=(0,3)$