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.
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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