# Two vertices of an isosceles triangle are (2, 0) and (2, 5).

Question:

Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

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 isosceles triangle two sides will be of equal length.

Here two vertices of the triangle is given as (20) and (25). Let the third side of the triangle be C(x, y)

It is given that the length of the equal sides is 3 units.

Let us now find the length of the side in which both the vertices are known.

$A B=\sqrt{(2-2)^{2}+(0-5)^{2}}$

$=\sqrt{(0)^{2}+(-5)^{2}}$

$=\sqrt{0+25}$

$=\sqrt{25}$

$A B=5$

So, now we know that the side ‘AB’ is not one of the equal sides of the isosceles triangle.

So, we have

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

$B C=\sqrt{(2-x)^{2}+(5-y)^{2}}$

Equating these two equations we have,

$\sqrt{(2-x)^{2}+(0-y)^{2}}=\sqrt{(2-x)^{2}+(5-y)^{2}}$

Squaring on both sides of the equation we have,

$(2-x)^{2}+(0-y)^{2}=(2-x)^{2}+(5-y)^{2}$

$4+x^{2}-4 x+y^{2}=4+x^{2}-4 x+25+y^{2}-10 y$

$10 y=25$

$y=\frac{5}{2}$

$y=2.5$

We know that the length of the equal sides is 3 units. So substituting the value of ‘y’ in equation for either ‘AC’ or ‘BC’ we can get the value of ‘x’.

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

$3=\sqrt{(2-x)^{2}+\left(-\frac{5}{2}\right)^{2}}$

Squaring on both sides,

$9=4+x^{2}-4 x+\frac{25}{4}$

$5=x^{2}-4 x+\frac{25}{4}$

$20=4 x^{2}-16 x+25$

$-5=4 x^{2}-16 x$

We have a quadratic equation for ‘x’. Solving for roots of the above equation we have,

$4 x^{2}-16 x+5=0$

$x=\frac{16 \pm \sqrt{256-4(4)(5)}}{8}$

$=\frac{16 \pm \sqrt{176}}{8}$

$=16 \pm 4 \sqrt{11}$

$x=2 \pm \frac{\sqrt{11}}{2}$

Hence the possible co-ordinates of the third vertex of the isosceles triangle are $\left(2+\frac{\sqrt{11}}{2}, \frac{5}{2}\right)$ or $\left(2-\frac{\sqrt{11}}{2}, \frac{5}{2}\right)$.