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 (2 ,9 ) and the triangle's area is 32 , what are the possible coordinates of the triangle's third corner?

Apr 25, 2018

 (1825/178, 765/89) or( -223/178, 125/89)

Explanation:

We relabel in standard notation: $b = c$, $A \left(x , y\right)$, $B \left(7 , 1\right) ,$ $C \left(2 , 9\right)$. We have $\textrm{a r e a} = 32$.

The base of our isosceles triangle is $B C$. We have

$a = | B C | = \sqrt{{5}^{2} + {8}^{2}} = \sqrt{89}$

The midpoint of $B C$ is $D = \left(\frac{7 + 2}{2} , \frac{1 + 9}{2}\right) = \left(\frac{9}{2} , 5\right)$. $B C$'s perpendicular bisector goes through $D$ and vertex $A$.

$h = A D$ is an altitude, which we get from the area:

$32 = \setminus \frac{1}{2} a h = \frac{1}{2} \setminus \sqrt{89} \setminus h$

$h = \frac{64}{\sqrt{89}}$

The direction vector from $B$ to $C$ is

$C - B = \left(2 - 7 , 9 - 1\right) = \left(- 5 , 8\right)$.

The direction vector of its perpendiculars is $P = \left(8 , 5\right)$, swapping the coordinates and negating one. Its magnitude must also be $| P | = \sqrt{89}$.

We need to go $h$ in either direction. The idea is:

$A = D \setminus \pm h \frac{P}{|} P |$

$A = \left(\frac{9}{2} , 5\right) \setminus \pm \left(\frac{64}{\sqrt{89}}\right) \frac{\left(8 , 5\right)}{\setminus} \sqrt{89}$

$A = \left(\frac{9}{2} , 5\right) \setminus \pm \frac{64}{89} \left(8 , 5\right)$

$A = \left(\frac{9}{2} + \frac{8 \left(64\right)}{89} , 5 + \frac{5 \left(64\right)}{89}\right) \mathmr{and}$$A = \left(\frac{9}{2} - \frac{8 \left(64\right)}{89} , 5 - \frac{5 \left(64\right)}{89}\right)$

 A=(1825/178, 765/89) or A=( -223/178, 125/89)

That's a bit messy. Is it right? Let's ask Alpha.

Great! Alpha verifies its isosceles and the area is $32.$ The other $A$ is right too.