# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (4 ,9 ) to (8 ,5 ) and the triangle's area is 64 , what are the possible coordinates of the triangle's third corner?

May 5, 2016

The third corner will be at either $\left(- 2 , - 1\right)$ or $\left(14 , 15\right)$

#### Explanation:

Length of side A = $\sqrt{{\left(8 - 4\right)}^{2} + {\left(5 - 9\right)}^{2}} = 4 \sqrt{2}$

Height of triangle ABC relative to A $= \left(\text{area")/("length side A}\right)$
$\textcolor{w h i t e}{\text{XXX}} \frac{64}{4 \sqrt{2}} = 8 \sqrt{2}$

Note that since $\left\mid B \right\mid = \left\mid C \right\mid$ the height relative to $A$ is the length of the perpendicular bisector of $A$

Mid-point of $A$ is $\left(\frac{4 + 8}{2} , \frac{9 + 5}{2}\right) = \left(6 , 7\right)$ $A$ has a slope of $\frac{\Delta y}{\Delta x} = \frac{9 - 5}{4 - 8} = - 1$

$\Rightarrow$ the perpendicular bisector has a slope of $- \frac{1}{- 1} = 1$ (and as already noted passes through (6,7)#

A slope of $1$ implies a $\Delta x : \Delta y$ ratio of $1 : 1$ (with a corresponding hypotenuse of $\sqrt{2}$).

A point on the perpendicular bisector a distance of $8 \sqrt{2}$ from $\left(6 , 7\right)$ will be
either at $\left(6 - 8 , 7 - 8\right) = \left(- 2 , - 1\right)$
or at $\left(6 + 8 , 7 + 8\right) = \left(14 , 15\right)$ 