# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (2 ,5 ) to (8 ,1 ) and the triangle's area is 15 , what are the possible coordinates of the triangle's third corner?

Dec 19, 2017

Coordinates of the third vertex is $\left(\frac{187}{9}\right) , \left(\frac{80}{3}\right)$

#### Explanation:

Area = 15

$a = \sqrt{{\left(8 - 2\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{52} = 2 \sqrt{13}$

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

$15 = \left(\frac{1}{2}\right) \cdot 2 \sqrt{13} \cdot h$

$h = \frac{15}{\sqrt{13}}$

$\tan \theta = \frac{h}{\frac{a}{2}} = \frac{\frac{15}{\sqrt{13}}}{\frac{2 \sqrt{13}}{2}} = \frac{15}{13}$

Slope of line AB is $\tan \theta = \frac{15}{13}$
Equation of line AB is given by
$y - 5 = \left(\frac{15}{13}\right) \cdot \left(x - 2\right)$
$13 y - 65 = 15 x - 30$
15x - 13y = —35. Eqn (1)

Slope of base $m = \frac{1 - 5}{8 - 2} = - \frac{2}{3}$
Slope of altitude ${m}_{1} = - \left(\frac{1}{m}\right) = \frac{3}{2}$

Midpoint of base a is $\frac{8 + 2}{2} , \frac{1 + 5}{2} = \left(5 , 3\right)$

Equation of altitude and passing through point A and having slope 3/2 is
$y - 3 = \left(\frac{3}{2}\right) \cdot \left(x - 5\right)$
$2 y - 6 = 3 x - 15$
$3 x - 2 y = 9$ Eqn (2)

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

$x = \left(\frac{187}{9}\right) , y = \left(\frac{80}{3}\right)$

Coordinates of the third corner A is $\left(\frac{187}{9} , \frac{80}{3}\right)$