**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 *A *(2*, *0) and *B *(2*, *5). 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 $\left(2-\frac{\sqrt{11}}{2}, \frac{5}{2}\right)$.

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