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Geometry of the line question!

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  • 03-05-2016 6:15pm
    #1
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭


    This question's been bugging me for weeks. It's why I had to stop geometry of the line.

    "Find the image of the point (-4,-5) under the axial symmetry in the line 3x+2y-4=0.

    The answer at the back of the book is (8,3)

    How do I get that answer?


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    #2
    skippy1977
    Registered Users Posts: 284 ✭✭skippy1977


    Need to find the equation of the line perpendicular to 3x+2y-4=0 through the point (-4,-5). Works out to be 2x-3y-7=0

    When you get this you can find the point of intersection of the two lines. Works out to be (2,-1)

    This is the point that you translate (-4,-5) through.

    (-4,-5) → (2,-1) → (8,3)


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    #3
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭TSMGUY


    Top-notch explanation skippy. Now that you've solved that I'll be cheeky and ask another.

    P(-1,5), Q(-2,1), R(3,-2) and S(a,b) are the four vertices of a parallelogram. with PQllRS (PQ is parallel to RS) Find two pairs of co-ordinates of the point S.


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    #4
    skippy1977
    Registered Users Posts: 284 ✭✭skippy1977


    Okay so PQ is parallel to RS. If you draw a rough sketch of the 3 points you have you can imagine a point S that lies above R to the right....
    this involves finding the image of R under the translation Q to P.

    Q(-2,1)→P(-1,5)
    R(3,-2)→S(4,2)

    But you can also visualise a point S that lies below R and to the left...
    this involves finding the image of R under the translation P to Q.

    P(-1,5)→Q(-2,1)
    R(3,-2)→S(2,-6)

    Parallelogram.jpg


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    #5
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭TSMGUY


    That makes perfect sense! The stupid book has

    (4,2) and (6,-2) as the answer for some reason. I think they meant (2,-6) like you said, it's the only one that makes sense given the equation of the line is 4x-y-14=0.

    Irish maths textbooks are so effing unreliable.


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