# 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 48 , what are the possible coordinates of the triangle's third corner?

Dec 4, 2017

Coordinates of third corner A of the triangle is $\left(\frac{16}{5} , \frac{21}{5}\right)$

#### Explanation:

Area = 48

$a = \sqrt{{\left(4 - 8\right)}^{2} + {\left(9 - 5\right)}^{2}} = \sqrt{32} = 4 \sqrt{2}$

$A r e a = \left(\frac{1}{2}\right) a \cdot h$

$48 = \left(\frac{1}{2}\right) \cdot 4 \sqrt{2} \cdot h$

$h = \frac{48}{2 \sqrt{2}} = 12 \sqrt{2}$

$\tan \theta = \frac{h}{\frac{a}{2}} = \frac{12 \sqrt{2}}{2 \sqrt{2}} = 6$

Equation of line passing through point B and having slope 6 is
$y - 9 = 6 \cdot \left(x - 4\right)$
$6 x - y = 15$. Eqn (1)

Slope of $B C = \frac{5 - 9}{8 - 4} = - 1$
Slope of altitude h passing through mid point of BC is 1

Mid point of BC is $\left(\frac{8 + 4}{2} , \frac{9 + 5}{2}\right) = \left(6 , 7\right)$

Equation of altitude h is
$\left(y - 7\right) = 1 \cdot \left(x - 6\right)$
$x - y = - 1$ Eqn (2)

Solving Eqns (1) & (2), we get coordinates of point A.

x = 16/5, y = 21/5)

Coordinates of the third corner A is $\left(\frac{16}{5} , \frac{21}{5}\right)$