# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,1 ) to (8 ,5 ) and the triangle's area is 27 , what are the possible coordinates of the triangle's third corner?

Dec 15, 2016

$\left(6.55882 , 3.23529\right) \mathmr{and} \left(8.44118 , 2.76471\right)$

#### Explanation:

Let's begin by finding the length of side "a":

$a = \sqrt{{\left(8 - 7\right)}^{2} + {\left(5 - 1\right)}^{2}}$

$a = \sqrt{{1}^{2} + {4}^{2}}$

$a = \sqrt{17}$

Use side "a" as the base of the triangle, find the height:

$A r e a = \frac{1}{2} b a s e \times h e i g h t$

$2 = \frac{1}{2} \sqrt{17} \times h e i g h t$

$h e i g h t = \frac{4 \sqrt{17}}{17}$

The height must bisect side "a" at the point $\left(7.5 , 3\right)$

We can used the distance formula to write one equation for the two possible points:

$\frac{4 \sqrt{17}}{17} = \sqrt{{\left(x - 7.5\right)}^{2} + {\left(y - 3\right)}^{2}}$

Square both sides:

$\frac{16}{17} = {\left(x - 7.5\right)}^{2} + {\left(y - 3\right)}^{2} \text{ [1]}$

We need one more equation. Find the slope of side "a":

$m = \frac{5 - 1}{8 - 7} = 4$

Because the height is perpendicular to side "a", its slope is: $- \frac{1}{4}$

We can use the slope and the point to find the equation of line for the height:

$3 = - \frac{1}{4} \left(7.5\right) + b$

$b = \frac{39}{8}$

The equation for the height is:

$y = - \frac{1}{4} x + \frac{39}{8} \text{ [2]}$

Solving equations [1] and [2] is very long and tedious process. I gave them to WolframAlpha. Here are the points:

$\left(6.55882 , 3.23529\right) \mathmr{and} \left(8.44118 , 2.76471\right)$