Coordinate Geometry

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  1. krisshP's Avatar
    1 03-07-2012  10:25
    • Peer Of The TSR Realm
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    Coordinate Geometry
    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:
  2. Oromis263's Avatar
    2 03-07-2012  10:39
    • Overlord in Training
    • Location: Alagaësia
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    Re: Coordinate Geometry
    (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.
  3. notnek's Avatar
    3 03-07-2012  10:40
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    PQ=2QR

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

    Post some working.
  4. krisshP's Avatar
    4 03-07-2012  10:47
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    Re: Coordinate Geometry
    (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?
  5. notnek's Avatar
    5 03-07-2012  10:49
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    (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.
  6. krisshP's Avatar
    6 03-07-2012  11:00
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    Re: Coordinate Geometry
    \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:
    Last edited by krisshP; 03-07-2012 at 11:02.
  7. notnek's Avatar
    7 03-07-2012  11:04
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    (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.
  8. krisshP's Avatar
    8 03-07-2012  11:17
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    Re: Coordinate Geometry
    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??
    Last edited by krisshP; 03-07-2012 at 11:23.
  9. notnek's Avatar
    9 03-07-2012  11:34
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    (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?
  10. krisshP's Avatar
    10 03-07-2012  11:36
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    Re: Coordinate Geometry
    (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
  11. krisshP's Avatar
    11 03-07-2012  11:41
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    Re: Coordinate Geometry
    (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??
    Last edited by krisshP; 03-07-2012 at 11:43.
  12. notnek's Avatar
    12 03-07-2012  11:53
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    (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.
  13. krisshP's Avatar
    13 03-07-2012  12:00
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    Re: Coordinate Geometry
    (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?
  14. krisshP's Avatar
    14 03-07-2012  12:05
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    Re: Coordinate Geometry
    I got the answer:

    q=4 and q=32/3

    Thanks a lot Notnek for your help
  15. notnek's Avatar
    15 03-07-2012  12:06
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Coordinate Geometry
    (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.
  16. Oromis263's Avatar
    16 03-07-2012  12:16
    • Overlord in Training
    • Location: Alagaësia
    • Posts: 3,492
    Re: Coordinate Geometry
    (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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