# 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 ,1 ) and the triangle's area is 15 , what are the possible coordinates of the triangle's third corner?

Oct 5, 2016

$\left(\frac{15}{34} , \frac{42}{34}\right)$ and $\left(\frac{495}{34} , \frac{162}{34}\right)$

#### Explanation:

We can find the length of 'a' by finding the distance between the two points:

a = sqrt((7 - 8)² + (5 - 1)²)

a = sqrt((-1)² + (4)²)

$a = \sqrt{1 + 16}$

$a = \sqrt{17}$

Let side 'a' be the base of the triangle.

Using the area, we can compute the height:

$A = \left(\frac{1}{2}\right) b h = \left(\frac{1}{2}\right) a h$

$15 = \left(\frac{1}{2}\right) \left(\sqrt{17}\right) h$

$h = 30 \frac{\sqrt{17}}{17}$

The height must lie on the line that is the perpendicular bisector of side 'a'. Let's find the equation of that line:

Side 'a' goes from left to right 1 unit and down 4 units (For later use, remember this is slope, -4), therefore, the midpoint goes from left to right $\frac{1}{2}$ unit and down 2 units.

The midpoint is $\left(\frac{15}{2} , 3\right)$

A perpendicular line will have a slope that is the negative reciprocal of -4:

$- \frac{1}{-} 4 = \frac{1}{4}$

Using the point-slope form of the equation of a line, $y - {y}_{1} = m \left(x - {x}_{1}\right)$, we write an equation of a line upon which both possible vertices must lie:

$y - 3 = \frac{1}{4} \left(x - \frac{15}{2}\right)$

$y = \frac{1}{4} x + \frac{9}{8}$

Using the equation for a circle we write an equation where the radius is $r = h = 30 \frac{\sqrt{17}}{17}$ and the center is the midpoint $\left(\frac{15}{2} , 3\right)$:

(30sqrt17/17)² = (x - 15/2)² + (y - 3)²

900/17 = x² - 15x + 225/4 + y² -6y + 9

Substitute $\frac{1}{4} x + \frac{9}{8}$ for y:

900/17 = x² - 15x + 225/4 + (1/4x + 9/8)² -6(1/4x + 9/8) + 9

I used Wolframalpha to solve this:

$x = \frac{15}{34}$ and $x = \frac{495}{34}$

The corresponding y coordinates are:

$y = \frac{42}{34}$ and $y = \frac{162}{34}$