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Thread: Hi, quick distance between points ?

  1. #1
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    Hi, quick distance between points ?

    Hello everyone! I can't believe I actually found a forum for math! Mathletes unite, Well it's my last class to finish up my degree and I can't believe I waited till last minute to take it! now its beating me up!

    I have a few questions, but will post one at a time in this thread so I can work them out to to understand them, thank you. The first one is


    **Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint. (Hint: let (x,y) be the unknown endpoint. Apply the midpoint formula, and solve the 2 equations for x and y.)

    Midpoint (27,1), endpoint (19,-1)


    thank you so much its due before midnight!!
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  2. #2
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    $\displaystyle \left(\frac{x+19}{2} \, , \frac{y + (-1)}{2} \right) = (27,1)$

    solve for x and y.
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  3. #3
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    I dont know how?

    and thank you for replying?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dillon658 View Post
    I dont know how?

    and thank you for replying?
    $\displaystyle \left(\frac{x+19}{2} \, , \frac{y -1}{2} \right) = (27,1)$

    Compare the x and y coordinates.

    $\displaystyle \frac{x+19}{2}=27\implies x=......$ and $\displaystyle \frac{y-1}{2}=1\implies y=......$

    This shouldn't be much of a problem now. I'm sure you can take it from here.

    --Chris
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  5. #5
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    I guess im looking for the answer....

    Theres so many different ways that I have in my head right now it all looks japanese...
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dillon658 View Post
    I guess im looking for the answer....

    Theres so many different ways that I have in my head right now it all looks japanese...
    I'll solve for y, and leave it for you to solve for x:

    $\displaystyle \frac{y-1}{2}=1$

    multiply both sides by 2 to get $\displaystyle y-1=2$

    To isolate y, add 1 to both sides. We should now see that $\displaystyle \color{red}\boxed{y=3}$

    Does this make sense? Try to find x in a similar way.

    --Chris
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  7. #7
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    is it 11?
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dillon658 View Post
    is it 11?
    No. Did you first multiply both sides by 2? You should get $\displaystyle x+19=54\implies x=......$

    I hope this clarifies things!

    --Chris
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  9. #9
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    I'm lost...

    I suck at algebra it's due at midpoint

    Id pay you if I had money bro

    I just need the answer

    why I go online because the tutors suck and I dont get it at school
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    36x^2+25y^2=900
    5y+6x=30


    the solutions are????

    please! need some answers before midnight, thank you!
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dillon658 View Post
    I'm lost...

    I suck at algebra it's due at midpoint

    Id pay you if I had money bro

    I just need the answer

    why I go online because the tutors suck and I dont get it at school
    MHF is a community where you get FREE help. We will not take money from you to help you.

    I will work them out step by step and hopefully then, it will make sense.

    You were asked to find the coordinate $\displaystyle (x,y)$, given the midpoint and one endpoint.

    We came up with the formula $\displaystyle \left(\frac{x+19}{2},\frac{y-1}{2}\right)=(27,1)$

    We then compare the coordinates. We have an equation for x and y.

    $\displaystyle \frac{x+19}{2}=27$

    Multiply both sides by 2:

    $\displaystyle {\color{red}2}\cdot\frac{x+19}{2}={\color{red}2}\c dot 27\implies {\not 2}\cdot\frac{x+19}{\not 2}=54\implies x+19=54$

    Subtract 19 from both sides to solve for x:

    $\displaystyle x+19{\color{red}-19}=54{\color{red}-19}\implies x+\not 19-\not 19=35\implies \color{red}\boxed{x=35}$

    $\displaystyle \frac{y-1}{2}=1$

    Multiply both sides by 2:

    $\displaystyle {\color{red}2}\cdot\frac{y-1}{2}={\color{red}2}\cdot 1\implies {\not 2}\cdot\frac{y-1}{\not 2}=2\implies y-1=2$

    Add 1 to both sides to solve for y:

    $\displaystyle y-1{\color{red}+1}=1{\color{red}+2}\implies x-\not 1+\not 1=3\implies \color{red}\boxed{y=3}$

    Does this make sense?

    --Chris
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  12. #12
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dillon658 View Post
    36x^2+25y^2=900
    5y+6x=30


    the solutions are????

    please! need some answers before midnight, thank you!
    $\displaystyle 36x^2+25y^2=900$

    $\displaystyle 6x+5y=30$

    Note that $\displaystyle 36x^2+25y^2=900\implies (6x)^2+(5y)^2=900$

    To find what x and y are, we need to substitute $\displaystyle 6x+5y=30$ into $\displaystyle (6x)^2+(5y)^2=900$...but how? We need to solve for either $\displaystyle 6x$ or $\displaystyle 5y$.

    I'll solve for 5y:

    $\displaystyle 6x+5y=30\implies 5y=30-6x$

    Substitute this into the other equation:

    $\displaystyle (6x)^2+(5y)^2=900\implies 36x^2+(\underbrace{30-6x}_{5y})^2=900$

    Now expand and simplify.

    $\displaystyle 36x^2+(900-180x+36x^2)=900\implies 72x^2-180x=0\implies 18x(4x-10)=0$$\displaystyle \implies x=\tfrac{10}{4} \text{ or }x=0$.

    Now find y:

    $\displaystyle 5y=30-6x\implies 5y=30\text{ or }5y=30-15\implies y=6\text{ or } y=3$

    The correct paired solutions would be $\displaystyle \left(\tfrac{10}{4},3\right)$ and $\displaystyle \left(0,6\right)$.

    Does this make sense?

    --Chris
    Last edited by Chris L T521; Aug 12th 2008 at 09:06 PM.
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