# An equilateral triangle has two vertices at the points $(3, 4)$, and $(-2, 3)$, find the coordinates of the third vertex.

Given:

An equilateral triangle has two vertices at the points $(3, 4)$, and $(-2, 3)$.

To do:

We have to find the coordinates of the third vertex.
Solution:

Let the two vertices of the equilateral triangle be $A (3,4)$ and $B (-2,3)$ and 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,

$AB=\sqrt{(-2-3)^2+(3-4)^2}$

$=\sqrt{(-5)^2+(-1)^2}$

$=\sqrt{25+1}$

$=\sqrt{26}$

$BC=\sqrt{(x+2)^2+(y-3)^2}$

$C A=\sqrt{(3-x)^{2}+(4-y)^{2}}$
$\mathrm{BC}=\mathrm{AB}$
$\Rightarrow \sqrt{(x+2)^{2}+(y-3)^{2}}=\sqrt{26}$

Squaring on both sides, we get,
$\Rightarrow(x+2)^{2}+(y-3)^{2}=26$
$\Rightarrow x^{2}+4 x+4+y^{2}-6 y+9=26$
$\Rightarrow x^{2}+y^{2}+4 x-6 y+13=26$

$\Rightarrow x^{2}+y^{2}+4 x-6 y=26-13=13$.........(i)

$\mathrm{CA}=\mathrm{AB}$

$\Rightarrow \sqrt{(3-x)^{2}+(4-y)^{2}}=\sqrt{26}$

Squaring on both sides, we get,
$\Rightarrow (3-x)^{2}+(4-y)^{2}=26$

$\Rightarrow 9+x^{2}-6 x+16+y^{2}-8 y=26$
$\Rightarrow x^{2}+y^{2}-6 x-8 y+25=26$
$\Rightarrow x^{2}+y^{2}-6 x-8 y=26-25=1$..........(ii)
Subtracting (ii) from (i), we get,
$10 x+2 y=12$

$5 x+y=6$
$y=6-5 x$
Substituting $y=6-5x$ in (i), we get,

$x^{2}+(6-5 x)^{2}+4 x-6(6-5 x)=13$
$\Rightarrow x^{2}+36+25 x^{2}-60 x+4 x-36+30 x-13=0$
$\Rightarrow 26 x^{2}-26 x-13=0$

$\Rightarrow 2 x^{2}-2 x-1=0$

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 2 \times(-1)}}{2 \times 2}$
$=\frac{2 \pm \sqrt{4+8}}{4}$
$=\frac{2 \pm \sqrt{12}}{4}$
$=\frac{2 \pm \sqrt{4 \times 3}}{4}$
$=\frac{2 \pm 2 \sqrt{3}}{4}$

$=\frac{1 \pm \sqrt{3}}{2}$
$x=\frac{1+\sqrt{3}}{2}$ or $x=\frac{1-\sqrt{3}}{2}$
If $x=\frac{1+\sqrt{3}}{2},$ then,

$y=6-5 x$
$=6-\frac{5(1+\sqrt{3})}{2}$
$=\frac{12-5-5 \sqrt{3}}{2}$

$=\frac{7-5 \sqrt{3}}{2}$
If $x=\frac{1-\sqrt{3}}{2},$ then,
$y=6-5 x=6-5 \frac{(1-\sqrt{3})}{2}$
$=\frac{12-5+5 \sqrt{3}}{2}$

$=\frac{7+5 \sqrt{3}}{2}$
Hence, the coordinates of the third vertex are $\left(\frac{1+\sqrt{3}}{2}, \frac{7-5 \sqrt{3}}{2}\right)$ or $\left(\frac{1-\sqrt{3}}{2}, \frac{7+5 \sqrt{3}}{2}\right)$.

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Updated on: 10-Oct-2022

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