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Thread: Right triangle stuff!

  1. #1
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    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?
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  2. #2
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    Re: Right triangle stuff!

    "Kirk to C, Kirk to C, we need your coordinates, over"
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  3. #3
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    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.
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  4. #4
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    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...
    Please tell me it's IMPOSSIBLE
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  5. #5
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    Re: Right triangle stuff!

    Quote Originally Posted by DenisB View Post
    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 $c_x$.
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  6. #6
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    Re: Right triangle stuff!

    Thanks Slip...looks like no EASY way
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  7. #7
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    Re: Right triangle stuff!

    Quote Originally Posted by DenisB View Post
    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.
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  8. #8
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    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 $\displaystyle y=4$
    Last edited by Idea; Apr 23rd 2018 at 09:38 AM.
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  9. #9
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    Re: Right triangle stuff!

    Quote Originally Posted by DenisB View Post
    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
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  10. #10
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    Re: Right triangle stuff!

    Don't feel like shaving right now Dan !!
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  11. #11
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    Re: Right triangle stuff!

    Quote Originally Posted by DenisB View Post
    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
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  12. #12
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    Re: Right triangle stuff!

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  13. #13
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    Re: Right triangle stuff!

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

    -Dan
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