Right triangle ABC.
BC = a = sqrt(17)
AC = b = sqrt(68)
AB = c = sqrt(85)
A's coordinates: 0,12
B's coordinates: 6,5
What's EASIEST way to get C's coordinates?
Knowing the coordinates of A and B, you can find the slope of the line AB. Since angle ABC is a right angle, if AB has slope m, BC has slope -1/m. So you can write the equation of line BC. Write the equation of the circle with center B and radius sqrt(17). To find C's coordinates, find the point where the line BC and that circle intersect.
$\sqrt{17} = a = BC = \sqrt{(c_x-6)^2+(c_y-5)^2}$
$\sqrt{68} = b = AC = \sqrt{(c_x-0)^2+(c_y-12)^2}$
From the first equation:
$17 = (c_x-6)^2+(c_y-5)^2 \Longrightarrow c_x = 6 \pm \sqrt{17-(c_y-5)^2}$
Plugging into the second equation:
$\displaystyle 68 = c_x^2 + (c_y-12)^2 = \left( 6\pm \sqrt{17-(c_y-5)^2} \right)^2 + (c_y-12)^2$
https://www.wolframalpha.com/input/?...%2B+(y-12)%5E2
https://www.wolframalpha.com/input/?...%2B+(y-12)%5E2
You wind up with $c_y=4$ or $c_y = \dfrac{44}{5}$
Plugging into the first equation, calculate $c_x$.
Let $\displaystyle C(x,y)$ be the solution closer to the origin
Draw the line (L) through A and parallel to the x-axis
CM and BK perpendicular to (L), M and K both on (L)
Triangles KMC and BCA are similar, angle AKC = angle ABC since quad AKBC is inscribed
We get
$\displaystyle \frac{\text{MC}}{\text{KM}}=\frac{\text{CA}}{\text {CB}}=2$
$\displaystyle \frac{12-y}{6-x}=2$
This gives $\displaystyle y=2x$
We have $\displaystyle A(0,12)$, $\displaystyle B(6,5)$, and $\displaystyle C(x,2x)$
AC is perpendicular to BC therefore
$\displaystyle \left(\frac{5-2x}{6-x}\right)\left(\frac{12-2x}{-x}\right)=-1$
$\displaystyle 2(5-2x)=x$
Therefore $\displaystyle x=2$ and $\displaystyle y=4$
I'll think about it:
https://www.google.ca/search?ei=C-Pf....0.jblQhsqG7Qs