An equilateral triangle has two vertices at the points

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 $\frac{1-\sqrt{3}}{2}, \frac{7+5 \sqrt{3}}{2}$.