# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,5 ) to (8 ,2 ) and the triangle's area is 27 , what are the possible coordinates of the triangle's third corner?

Jul 1, 2017

The coordinates are $\left(23.7 , 8.9\right)$ and $\left(- 8.7 , - 1.9\right)$

#### Explanation:

The length of side $A = \sqrt{{\left(7 - 8\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{10}$

Let the height of the triangle be $= h$

The area of the triangle is

$\frac{1}{2} \cdot \sqrt{10} \cdot h = 27$

The altitude of the triangle is $h = \frac{27 \cdot 2}{\sqrt{10}} = \frac{54}{\sqrt{10}}$

The mid-point of $A$ is $\left(\frac{15}{2} , \frac{7}{2}\right)$

The gradient of $A$ is $= \frac{2 - 5}{8 - 7} = - 3$

The gradient of the altitude is $= \frac{1}{3}$

The equation of the altitude is

$y - \frac{7}{2} = \frac{1}{3} \left(x - \frac{15}{2}\right)$

$y = \frac{1}{3} x - \frac{5}{2} + \frac{7}{2} = \frac{1}{3} x + 1$

The circle with equation

${\left(x - \frac{15}{2}\right)}^{2} + {\left(y - \frac{7}{2}\right)}^{2} = {54}^{2} / 10 = 291.6$

The intersection of this circle with the altitude will give the third corner.

${\left(x - \frac{15}{2}\right)}^{2} + {\left(\frac{1}{3} x + 1 - \frac{7}{2}\right)}^{2} = 291.6$

${x}^{2} - 15 x + \frac{225}{4} + \frac{1}{9} {x}^{2} - \frac{5}{3} x + \frac{25}{4} = 291.6$

$1.11 {x}^{2} - 16.7 x - 229.1 = 0$

$x = \frac{16.7 \pm \sqrt{{16.7}^{2} + 4 \cdot 1.11 \cdot 229.1}}{2 \cdot 1.11}$

$= \frac{16.7 \pm 36}{2.22}$

${x}_{1} = 23.74$

${x}_{2} = - 8.69$

The points are $\left(23.7 , 9.65\right)$ and $\left(- 8.7 , - 1.9\right)$

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}