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 (1 ,2 ) and the triangle's area is 36 , what are the possible coordinates of the triangle's third corner?

Nov 5, 2016

One possible coordinate would be $\left(- .8 , 13.1\right)$ and $\left(8.8 , - 6.1\right)$

Explanation:

One possible corner would be opposite to side length A if side A is the base of the isosceles triangle. We can find the length of A using the distance formula:
${A}^{2} = {6}^{2} + {3}^{2}$
$A = 3 \sqrt{5}$

The area of the triangle can be found using the formula:
$A r e a = \frac{1}{2} b h$ where b= base length and h= height
$36 = \left(\frac{1}{2}\right) \left(3 \sqrt{5}\right) \left(h\right)$
$h = \frac{24}{\sqrt{5}}$

Now we need to find the coordinates of a point that is $\frac{24}{\sqrt{5}}$ units away from the point (4,3.5) (which is the midpoint of the given line segment A) and perpendicular to line segment A.

The two possible points for the third corner are at $\left(- .8 , 13.1\right)$ and $\left(8.8 , - 6.1\right)$. I found this by using the distance formula and setting it equal to $\frac{24}{\sqrt{5}}$, and I also used the fact that the slope of the line perpendicular to A is -2.