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Math Help - 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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    \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?
    \left(\frac{x+19}{2} \, , \frac{y -1}{2} \right) = (27,1)

    Compare the x and y coordinates.

    \frac{x+19}{2}=27\implies x=...... and \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:

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

    multiply both sides by 2 to get y-1=2

    To isolate y, add 1 to both sides. We should now see that \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 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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  10. #10
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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 (x,y), given the midpoint and one endpoint.

    We came up with the formula \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.

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

    Multiply both sides by 2:

    {\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:

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

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

    Multiply both sides by 2:

    {\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:

    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!
    36x^2+25y^2=900

    6x+5y=30

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

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

    I'll solve for 5y:

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

    Substitute this into the other equation:

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

    Now expand and simplify.

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

    Now find y:

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

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

    Does this make sense?

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