# 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 (1 ,0 ) and the triangle's area is 24 , what are the possible coordinates of the triangle's third corner?

Jan 8, 2018

Coordinates of the third corner A (4.5713, 3.8096)

#### Explanation:

$B C = \sqrt{{\left(4 - 1\right)}^{2} + {\left(9 - 0\right)}^{2}} = 9.4868$

Area of triangle ${A}_{t} = 24 = \left(\frac{1}{2}\right) \cdot B C \cdot A D = \left(\frac{1}{2}\right) \cdot 9.4868 \cdot h$

$h = \frac{2 \cdot 24}{9.4868} = 5.06$

Slope of BC ${m}_{B C} = \frac{0 - 9}{1 - 4} = 3$

Slope of AD ${m}_{A D} = - \left(\frac{1}{m} _ \left(B C\right)\right) = - \frac{1}{3}$

Coordinates of point D is
$= \frac{4 + 1}{2} , \frac{9 + 0}{2} = \left(2.5 , 4.5\right)$

Equation of AD is
$\left(y - 4.5\right) = \left(- \frac{1}{3}\right) \left(x - 2.5\right)$

$3 y + x = 16$ Eqn (1)

$\tan C = {m}_{C A} = \frac{A D}{C D} = \frac{5.06}{\frac{9.4868}{2}} = 1.0667$

Equation of CA is
$y - 0 = 1.0667 \left(x - 1\right)$

$y - 1.0667 x = - 1.0667$ Eqn (2)

Solving Equations (1) & (2), we get the coordinates of A

$x = \left(4.5713\right) , y = \left(3.8096\right)$