# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (4 ,1 ) to (8 ,9 ) and the triangle's area is 64 , what are the possible coordinates of the triangle's third corner?

Apr 22, 2018

Let $P R = A$ be the side of the isosceles triangle having coordinates of its end points as follows

$P \to \left(4 , 1\right)$ and $R \to \left(8 , 9\right)$

Let the coordinates of the third point of the triangle be $\left(x , y\right)$.

As $\left(x , y\right)$ is equidistant from P and R we can write

${\left(x - 4\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(x - 8\right)}^{2} + {\left(y - 9\right)}^{2}$

$\implies {x}^{2} - 8 x + 16 + {y}^{2} - 2 y + 1 = {x}^{2} - 16 x + 64 + {y}^{2} - 18 y + 81$

$\implies 8 x + 16 y = 128$

$\implies x + 2 y = 16$

$\implies x = 16 - 2 y \ldots \ldots \left[1\right]$

Again $\left(x , y\right)$ being equidistant from P and R, the perpendicular dropped from $\left(x , y\right)$ to $P R$ must bisect it, Let this foot of the perpendicular or mid point of $P R$ be $T$

So coordinates of $T \to \left(6 , 5\right)$

Now height of the isosceles triangle

$H = \sqrt{{\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2}}$

And the base of the isosceles triangle

$P R = A = \sqrt{{\left(4 - 8\right)}^{2} + {\left(1 - 9\right)}^{2}} = 4 \sqrt{5}$

So by the problem its area

$\frac{1}{2} \times A \times H = 64$

$\implies H = \frac{128}{A} = \frac{128}{4 \sqrt{5}} = \frac{32}{\sqrt{5}}$

$\sqrt{{\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2}} = \frac{32}{\sqrt{5}}$

$\implies {\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2} = \frac{1024}{5} = 204.8 \ldots . \left[2\right]$

By [2] and [1] we get

${\left(16 - 2 y - 6\right)}^{2} + {\left(y - 5\right)}^{2} = 204.8$

$\implies {\left(10 - 2 y\right)}^{2} + {\left(y - 5\right)}^{2} = 204.8$

$\implies 5 {y}^{2} - 50 y + 125 = = 204.8$

$\implies {y}^{2} - 10 y + 25 = = 40.96$

$\implies {\left(y - 5\right)}^{2} = {6.4}^{2}$

$\implies y = 5 \pm 6.4$

So $y = 11.4 \mathmr{and} y = - 1.4$

when $y = 11.4$

$x = 16 - 2 \times 11.4 = - 6.8$

when $y = - 1.4$

$x = 16 - 2 \times \left(- 1.4\right) = 18.8$

So the coordinates of third point will be

$\left(- 6.8 , 11.4\right) \to \text{Q in figure}$

OR

$\left(18.8 , - 1.4\right) \to \text{S in figure}$