# 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 ,5 ) and the triangle's area is 32 , what are the possible coordinates of the triangle's third corner?

Feb 4, 2018

Coordinates of vertex A $\textcolor{b r o w n}{12.74 , - 0.68}$

#### Explanation:

Given : B (4,9), C (1,5), ${A}_{t} = 32$, AB = AC, $A N \left(h\right) \bot$ bisector of BC

To find A coordinates.

Coordinates of N (4+1)/2, (9+5)/2 = N (5/2, 7)

$B C = a = \sqrt{{\left(1 - 4\right)}^{2} + {\left(5 - 9\right)}^{2}} = 5$

${A}_{t} = 32 = \left(\frac{1}{2}\right) a \cdot h = \left(\frac{1}{2}\right) \cdot 5 \cdot h$

$h = 12.8$

Slope of BC ${m}_{a} = \frac{5 - 9}{1 - 4} = \frac{4}{3}$

Slope of $A N \bot B C = {m}_{h} = - \frac{1}{m} _ a = - \frac{3}{4}$

Equation of AN

$y - 7 = - \left(\frac{3}{4}\right) \left(x - \left(\frac{5}{2}\right)\right)$

$y - 7 = - \left(\frac{3}{8}\right) \left(2 x - 5\right)$

$8 y - 56 = - 6 x + 15$

$8 y + 6 x = 71$ Eqn (1)

AN = h = 12.8 = sqrt((x-(5/2))^2 + (y-7)^2)#

${\left(y - 7\right)}^{2} + {\left(x - \left(\frac{5}{2}\right)\right)}^{2} = {12.8}^{2}$ Eqn (2)

Solving equations (1), (2), we get the coordinates of vertex A

$A \left(12.74 , - 0.68\right)$

Verification :

$A B = c = \sqrt{{\left(4 - 12.74\right)}^{2} + {\left(9 + 0.68\right)}^{2}} = 13.0419$

$A C = b = \sqrt{{\left(1 - 12.74\right)}^{2} + {\left(5 + 0.68\right)}^{2}} = 13.0419$

$b = c = 13.0419$ since it is an isosceles triangle