If $(0, -3)$ and $(0, 3)$ are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Given:

$(0, -3)$ and $(0, 3)$ are the two vertices of an equilateral triangle.

To do:

We have to find the coordinates of its third vertex.

Solution:

Let $A (0, -3)$ and $B (0, 3)$ be two vertices of the equilateral triangle and the coordinates of the third vertex be $C (x, y)$.

This implies,

$AB=BC=CA$

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( \mathrm{AC}=\mathrm{AB} \)

\( \Rightarrow (x-0)^{2}+(y+3)^{2}=(0-0)^{2}+(3+3)^{2} \)

\( \Rightarrow x^{2}+(y+3)^{2}=0+(6)^{2}=36 \)

\( \Rightarrow x^{2}+y^{2}+6 y+9=36 \)

\( \Rightarrow x^{2}+y^{2}+6 y=36-9=27 \)......(i)

\( \mathrm{BC}=\mathrm{AB} \)

\( \Rightarrow (x-0)^{2}+(y-3)^{2}=36 \)

\( \Rightarrow x^{2}+y^{2}+9-6 y=36 \)

\( \Rightarrow x^{2}+y^{2}-6 y=36-9=27 \)........(ii)

From (i) and (ii),

\( x^{2}+y^{2}+6 y=x^{2}+y^{2}-6 y \)

\( x^{2}+y^{2}+6 y-x^{2}-y^{2}+6 y=0 \)

\( 12 y=0 \)

\( \Rightarrow y=0 \)

From (i),

\( x^{2}+y^{2}+6 y=27 \)

\(  x^{2}+0+0=27 \)

\( x=\pm \sqrt{9 \times 3} \)

\( =\pm 3 \sqrt{3} \)

\( x=\pm 3 \sqrt{3} \) and  \( y=0 \).

Therefore, the coordinates of the third point is \( (3 \sqrt{3}, 0) \) or \( (-3 \sqrt{3}, 0) \).