**Question:**

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

**Solution:**

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In an equilateral triangle all the sides are of equal length.

Here we are given that *A *(3*, *4) and *B *(*−*2*, *3) are two vertices of an equilateral triangle. Let *C*(*x, y*) be the third vertex of the equilateral triangle.

First let us find out the length of the side of the equilateral triangle.

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

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

$=\sqrt{25+1}$

$A B=\sqrt{26}$

Hence the side of the equilateral triangle measures $\sqrt{26}$ units.

Now, since it is an equilateral triangle, all the sides need to measure the same length.

Hence we have $B C=A C$

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

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

Equating both these equations we have,

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

Squaring on both sides we have,

$(-2-x)^{2}+(3-y)^{2}=(3-x)^{2}+(4-y)^{2}$

$4+x^{2}+4 x+9+y^{2}-6 y=9+x^{2}-6 x+16+y^{2}-8 y$

$10 x+2 y=12$

$5 x+y=6$

From the above equation we have, $y=6-5 x$

Substituting this and the value of the side of the triangle in the equation for one of the sides we have,

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

$\sqrt{26}=\sqrt{(-2-x)^{2}+(3-6+5 x)^{2}}$

Squaring on both sides,

$26=(-2-x)^{2}+(-3+5 x)^{2}$

$26=4+x^{2}+4 x+9+25 x^{2}-30 x$

$13=26 x^{2}-26 x$

$1=2 x^{2}-2 x$

Now we have a quadratic equation for ‘*x*’. Solving for the roots of this equation,

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

$x=\frac{2 \pm \sqrt{4-4(2)(-1)}}{4}$

$=\frac{2 \pm \sqrt{12}}{4}$

$x=\frac{1 \pm \sqrt{3}}{2}$

We know that $y=6-5 x$. Substituting the value of ' $x$ ' we have,

$y=6-5\left(\frac{1 \pm \sqrt{3}}{2}\right)$

$=\frac{12-5 \mp 5 \sqrt{3}}{2}$

$y=\frac{7 \mp 5 \sqrt{3}}{2}$

Hence the two possible values of the third vertex are $\frac{1+\sqrt{3}}{2}, \frac{7-5 \sqrt{3}}{2}$ and $\frac{1-\sqrt{3}}{2}, \frac{7+5 \sqrt{3}}{2}$.

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