Question

If a vertex of an equilateral triangle is on origin and second vertex is $(4,0)$, then its third vertex is

• A. $(2,\pm \sqrt{3})$
• B. $(3,\pm \sqrt{2})$
• C. $(2,\pm 2\sqrt{3})$
• D. $(3,\pm 2\sqrt{2})$

Answer 1

The correct option is C $(2,\pm 2\sqrt{3})$

Let $(x,y)$ be the co-ordinate of the third vertex.

Then distance between $(x,y)$ and $(0,0)$ is same as the distance between $(x,y)$ and $(4,0)$ is same as the distance between $(0,0)$ and $\left(4.0\right)$.

Then we've

$\sqrt{x^2+y^2}=\sqrt{(x-4{)}^2+y^2}=\sqrt{4^2+0^2}$

or, $x^2+y^2=(x-4{)}^2+y^2=4^2$ [ Squaring we get]

Taking the first two equality we get,

$x^2=(x-4{)}^2$

or, $x=4-x$

or, $x=2$.

Also $x^2+y^2=16$

or, $y^2=12$ [ Using $x=2$]

or, $y=\pm 2\sqrt{3}$.

So the third vertex is $(2,\pm 2\sqrt{3})$.