# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (4 ,3 ) to (8 ,9 ) and the triangle's area is 64 , what are the possible coordinates of the triangle's third corner?

Oct 19, 2016

$\left(- \frac{114}{13} , \frac{206}{13}\right)$ and $\left(\frac{270}{13} , - \frac{50}{15}\right)$

#### Explanation:

Side a is the base of the triangle its length is:

$a = \sqrt{{\left(8 - 4\right)}^{2} + {\left(9 - 3\right)}^{2}}$

$a = \sqrt{52}$

Using this and the area of the triangle when can find the altitude, h:

#64 = (1/2)sqrt(52)h

$h = 128 \frac{\sqrt{52}}{52}$

The altitude intersects the base at the midpoint of the line segment:

$\left(\frac{4 + 8}{2} , \frac{3 + 9}{2}\right) = \left(6 , 6\right)$

The slope, m, of the base is:

$m = \frac{9 - 3}{8 - 4} = \frac{6}{4} = \frac{3}{2}$

The slope, n, of the altitude is:

$n = - \frac{1}{m} = - \frac{1}{\frac{3}{2}} = - \frac{2}{3}$

The equation of the altitude is:

$y - 6 = - \frac{2}{3} \left(x - 6\right)$

$y = 10 - \frac{2}{3} x$

Using the distance formula:

$128 \frac{\sqrt{52}}{52} = \sqrt{{\left(x - 6\right)}^{2} + {\left(y - 6\right)}^{2}}$

The two points are the intersection of the distance and the line. I used WolframAlpha to solve for the two points:

$\left(- \frac{114}{13} , \frac{206}{13}\right)$ and $\left(\frac{270}{13} , - \frac{50}{15}\right)$