An isosceles triangle has sides a, b, and c with sides a and c being equal in length. If side b goes from (5 ,1 ) to (3 ,2 ) and the triangle's area is 8 , what are the possible coordinates of the triangle's third vertex?

Jun 9, 2018

$\left(\frac{23}{2} , \frac{19}{2}\right) \mathmr{and} \left(- \frac{9}{2} , - \frac{13}{2}\right)$

Explanation:

We have $A \left(4 , 1\right)$ and $C \left(3 , 2\right)$ and unknown third vertex $B \left(x , y\right) ,$

For AC we have direction vector $C - A = \left(3 - 4 , 2 - 1\right) = \left(- 1 , 1\right) .$ The perpendicular has direction vector given by swapping the coordinates and negating one of them, $\left(1 , 1\right)$.

The foot of the bisector is the midpoint of $A C$, $F \left(\frac{4 + 3}{2} , \frac{1 + 2}{2}\right) = F \left(\frac{7}{2} , \frac{3}{2}\right)$

$B$ lies along the perpendicular bisector of AC, so through F.

$\left(x , y\right) = F + t \left(1 , 1\right) = \left(t + \frac{7}{2} , t + \frac{3}{2}\right)$

The area is $S = 8.$ We can use the Shoelace Theorem:

$S = 8 = \frac{1}{2} | 4 \left(t + \frac{3}{2}\right) - \left(t + \frac{7}{2}\right) + 3 \left(1\right) - 2 \left(4\right) + \left(t + \frac{7}{2}\right) \left(2\right) - \left(t + \frac{3}{2}\right) \left(3\right) |$

$\pm 16 = 2 t$

$t = \pm 8$

$\left(x , y\right) = \left(t + \frac{7}{2} , t + \frac{3}{2}\right)$

$\left(x , y\right) = \left(\frac{23}{2} , \frac{19}{2}\right) \mathmr{and} \left(- \frac{9}{2} , - \frac{13}{2}\right)$