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Coordinate Geometry

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    Here is question which I'm stuck on:

    A triangle has vertices P(-2,2), Q(q,0) and R(5,3).
    The side PQ is twice as long as side QR.
    Find the possible values of q.

    How do I answer this question??:confused::confused:
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    (Original post by krisshP)
    Here is question which I'm stuck on:

    A triangle has vertices P(-2,2), Q(q,0) and R(5,3).
    The side PQ is twice as long as side QR.
    Find the possible values of q.

    How do I answer this question??:confused::confused:
    Try it in terms of vectors. I can post a start if you want it.
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    PQ=2QR

    What is the length of the line PQ? What is the length of the line QR?

    Post some working.
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    (Original post by notnek)
    PQ=2QR

    What is the length of the line PQ? What is the length of the line QR?

    Post some working.
    \sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

    Am I heading the right way?? Shall I continue from here?
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    (Original post by krisshP)
    \sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

    Am I heading the right way?? Shall I continue from here?
    That's correct so far. Now write down the length of PQ in terms of q.
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    \sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

    \sqrt{(q-5)^2+(-3)^2}=\frac{1}{2}PQ

    \sqrt{q^2-5q-5q+25+9}=\frac{1}{2}PQ

    \sqrt{q^2-10q+34}=\frac{1}{2}PQ

    Now what???:confused:
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    (Original post by krisshP)
    \sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

    \sqrt{(q-5)^2+(-3)^2}=\frac{1}{2}PQ

    \sqrt{q^2-5q-5q+25+9}=\frac{1}{2}PQ

    \sqrt{q^2-10q+34}=\frac{1}{2}PQ

    Now what???:confused:
    What is PQ? Use the same method as you used to find QR.
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    ok notnek I will.

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2-0)^2}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2)^2}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+4}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q  ^2+2q+2q+4+4}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q  ^2+4q+8}

    Now what do I do??
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    (Original post by krisshP)
    ok notnek I will.

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2-0)^2}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2)^2}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+4}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q  ^2+2q+2q+4+4}

    \sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q  ^2+4q+8}

    Now what do I do??
    Now square both sides so you get:

    \displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)

    Can you simplify and solve this?
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    (Original post by notnek)
    Now square both sides so you get:

    \displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)

    Can you simplify and solve this?
    I'll try to do the rest.

    Thanks a lot for your help
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    (Original post by notnek)
    Now square both sides so you get:

    \displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)

    Can you simplify and solve this?
    I end up with the following:

    -0.75q^2+11q-32=0

    Is this right?? Can I still solve it??
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    (Original post by krisshP)
    I end up with the following:

    -0.75q^2+11q-32=0

    Is this right?? Can I still solve it??
    That's correct. You can tidy the equation up and make it easier to solve by multiplying it by -4.
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    (Original post by notnek)
    That's correct. You can tidy the equation up and make it easier to solve by multiplying it by -4.
    I get the following:
    3q^2-44q+128=0

    Is it ok for me to divide both sides by 3 and use the completing the square method?
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    I got the answer:

    q=4 and q=32/3

    Thanks a lot Notnek for your help
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    (Original post by krisshP)
    I get the following:
    3q^2-44q+128=0

    Is it ok for me to divide both sides by 3 and use the completing the square method?
    You can use that method here but I've always been more of a fan of factorising:

    3q^2-44q+128 = (3q-...)(q-...)

    If you really aren't able to factorise then completing the square is OK to use.
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    (Original post by krisshP)
    I get the following:
    3q^2-44q+128=0

    Is it ok for me to divide both sides by 3 and use the completing the square method?
    That, or using the quadratic formula. (or factorising, of course, hadn't checked it to see if it did)

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Updated: July 3, 2012
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