1. ## Right triangle stuff!

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?

2. ## Re: Right triangle stuff!

"Kirk to C, Kirk to C, we need your coordinates, over"

3. ## Re: Right triangle stuff!

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.

4. ## Re: Right triangle stuff!

Agree Halls. There are 2 solutions of course: (2,4) and (38/5,44/5)

Was trying to solve WITHOUT involving a circle...

5. ## Re: Right triangle stuff!

Originally Posted by DenisB
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?
$\sqrt{17} = a = BC = \sqrt{(c_x-6)^2+(c_y-5)^2}$

$\sqrt{68} = b = AC = \sqrt{(c_x-0)^2+(c_y-12)^2}$

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 $c_x$.

6. ## Re: Right triangle stuff!

Thanks Slip...looks like no EASY way

7. ## Re: Right triangle stuff!

Originally Posted by DenisB
Thanks Slip...looks like no EASY way
The OP didn't say easy. The OP asked for easiest. That is a subjective and relative superlative. I gave the OP the method that is easiest for me.

8. ## Re: Right triangle stuff!

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 $\displaystyle y=4$

9. ## Re: Right triangle stuff!

Originally Posted by DenisB
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?
Get out a straight edge and a protractor...

-Dan

10. ## Re: Right triangle stuff!

Don't feel like shaving right now Dan !!

11. ## Re: Right triangle stuff!

Originally Posted by DenisB
Don't feel like shaving right now Dan !!
Then get over here and shave me. I've got a summer tan line to think about.

-Dan

13. ## Re: Right triangle stuff!

Originally Posted by DenisB
You Canadians. Just pack up the dog sled and go... eh?

-Dan