Question

If (0, −3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]

Answer 1

Let the given points be A(0, −3) and B(0, 3). Suppose the coordinates of the third vertex be C(x, y).

Now, ∆ABC is an equilateral triangle.

∴ AB = BC = CA

$\sqrt{{\left(0-0\right)}^2+{\left(-3-3\right)}^2}=\sqrt{{\left(x-0\right)}^2+{\left(y-3\right)}^2}=\sqrt{{\left(x-0\right)}^2+{\left[y-\left(-3\right)\right]}^2}$ (Distance formula)

Squaring on both sides, we get

$36=x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2$

⇒ $x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2$ and $x^2+{\left(y-3\right)}^2=36$

Now,

$$\begin{gathered} x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2 \\ \Rightarrow y^2-6y+9=y^2+6y+9 \\ \Rightarrow -12y=0 \\ \Rightarrow y=0 \end{gathered}$$

Putting y = 0 in $x^2+{\left(y-3\right)}^2=36$, we get

$$\begin{gathered} x^2+{\left(0-3\right)}^2=36 \\ \Rightarrow x^2=36-9=27 \\ \Rightarrow x=\pm \sqrt{27}=\pm 3\sqrt{3} \end{gathered}$$

Thus, the coordinates of the third vertex are $\left(3\sqrt{3},0\right)$ or $\left(-3\sqrt{3},0\right)$.