# An isosceles triangle has sides a, b and c with sides b and c being equal in length. If side a goes from (2 ,9 ) to (8 ,5 ) and the triangle's area is 64 , what are the possible coordinates of the triangle's third corner?

Jun 18, 2018

$\left(- \frac{63}{13} , - \frac{101}{13}\right) \mathmr{and} \left(\frac{193}{13} , \frac{283}{13}\right)$

#### Explanation:

Side $a$ is $\left(2 , 9\right)$ to $\left(8 , 5\right)$

Let's do this directly. We have

${a}^{2} = {\left(8 - 2\right)}^{2} + {\left(5 - 9\right)}^{2} = 36 + 16 = 52$

Area $S = \frac{1}{2} a h$ or

$h = \frac{2 S}{a} = \frac{2 \left(64\right)}{\setminus} \sqrt{52} = \frac{128}{\sqrt{52}}$

The other vertex is along the perpendicular bisector of side $a$. The direction vector for side as is $\left(8 - 2 , 5 - 9\right) = \left(6 , - 4\right)$ so the perpendicular through the midpoint is

$\left(x , y\right) = \left(\frac{2 + 8}{2} , \frac{9 + 5}{2}\right) + t \left(4 , 6\right) = \left(5 , 7\right) + t \left(4 , 6\right)$

We need $| t \left(4 , 6\right) | = h$

$t = \pm \frac{h}{|} \left(4.6\right) | = \pm \frac{\frac{128}{\sqrt{52}}}{\sqrt{52}} = \pm \frac{128}{52} = \pm \frac{32}{13}$

Two choices for the third vertex:

(x,y) = (5,7) pm (32/13) (4,6) = (-63/13, -101/13) or (193/13, 283/13)

Let's check one:

${b}^{2} = {\left(- \frac{63}{13} - 2\right)}^{2} + {\left(- \frac{101}{13} - 9\right)}^{2} = \frac{4265}{13}$

 c^2 = (-63/13 - 8)^2 + (-101/13 - 5)^2 = 4265/13 quad sqrt

Area $S$

$16 {S}^{2} = 4 \left(52\right) \left(\frac{4265}{13}\right) - {\left(52\right)}^{2}$

S = 64 quad sqrt