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    A triangle is formed by the lines y=1/2x, 2x+y+5=0 and x+3y-5=0. Prove that the triangle is an isosceles.


    i knw this Questions not some challenging tricky mission but i cant get my head around some of these and really need to have them done by the end of today. thankyou sO much
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    (Original post by swirlwood)
    A triangle is formed by the lines y=1/2x, 2x+y+5=0 and x+3y-5=0. Prove that the triangle is an isosceles.


    i knw this Questions not some challenging tricky mission but i cant get my head around some of these and really need to have them done by the end of today. thankyou sO much
    Find lengths of each line, two lines will have same length, therefore triangle is isosceles.
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    (Original post by swirlwood)
    A triangle is formed by the lines y=1/2x, 2x+y+5=0 and x+3y-5=0. Prove that the triangle is an isosceles.


    i knw this Questions not some challenging tricky mission but i cant get my head around some of these and really need to have them done by the end of today. thankyou sO much
    Maybe I'm missing something but isnt y=1/2x a curve and therefore it won't form a triangle?
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    (Original post by Merlinedge)
    Maybe I'm missing something but isnt y=1/2x a curve and therefore it won't form a triangle?

    I assume it's y=0.5x, rather than y=1/(2x)


    EDIT: My explanation above is wrong, as it doesn't give coordinates. I didn't read your question properly and assumed it was the usual coordinate providing style.
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    (Original post by swirlwood)
    A triangle is formed by the lines y=1/2x, 2x+y+5=0 and x+3y-5=0. Prove that the triangle is an isosceles.


    i knw this Questions not some challenging tricky mission but i cant get my head around some of these and really need to have them done by the end of today. thankyou sO much
    Having thought about it more, draw each line to get the triangle. Then find the length of each line, and two lengths will be equal.
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    it wont be a curve because it is x^1
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    yep i thought of that too. but then i thought maybe finding th points of intersection would help...doesnt seem to be doing any good though :confused:
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    (Original post by Laith)
    it wont be a curve because it is x^1
    He thought it was a reciprocal graph (i.e. 1/x).
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    (Original post by swirlwood)
    yep i thought of that too. but then i thought maybe finding th points of intersection would help...doesnt seem to be doing any good though :confused:
    It does help.

    Draw the triangle by drawing the lines.

    You will get points of intersection. This allows you to get the length of each line.

    Length = square root of: (different in x coordinates + difference in y coordinates)

    When 2 lengths are the same, this is an isosceles.
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    Two of the lines have length root20, therefore it is isosceles.
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    (Original post by DaveJ)
    It does help.

    Draw the triangle by drawing the lines.

    You will get points of intersection. This allows you to get the length of each line.

    Length = square root of: (different in x coordinates + difference in y coordinates)

    When 2 lengths are the same, this is an isosceles.
    If you mean pythagoras, then would it not be squared root of: ((difference of x)squared)/((difference of y)squared)
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    (Original post by addylad)
    If you mean pythagoras, then would it not be squared root of: ((difference of x)squared)/((difference of y)squared)

    length^2 = (diff in x)^2 + (diff in y^2)
    Length = square root of: [ (diff in x)^2 + (diff in y)^2 ]


    I'm pretty sure I'm right with that formula.
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    (Original post by DaveJ)
    Two of the lines have length root20, therefore it is isosceles.
    OoO..I tried that yday and got root 20 for only 1..but i shall try that all again..thankyuu...:p:
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    (Original post by DaveJ)
    length^2 = (diff in x)^2 + (diff in y^2)
    Length = square root of: [ (diff in x)^2 + (diff in y)^2 ]


    I'm pretty sure I'm right with that formula.
    That's correct, just I didn't see any ^2 in your original post.
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    (Original post by addylad)
    He thought it was a reciprocal graph (i.e. 1/x).
    Ah sorry
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    ok...i just done the entire Q again...I got root 20 for only one line. the other two I got root root18 and root40..:confused: im making a mistake somewhere
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    I'm sure I did this Q... but in another book.

    How did you draw the lines... did you substitute some x values to get y, and vice versa?
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    You sure you got the right points of intersection? But I can't think of what could possibly end up in an answer of root40, root20 and root18 so I think it's more likely you calculated the lengths wrong somewhere.
    Spoiler:
    Show
    Points of intersection of the lines:
    (-4, 3)
    (-2, -1)
    (2, 1)
    Full solution:
    Spoiler:
    Show



    A = (-4, 3)
    B = (-2, -1)
    C = (2, 1)

    Length AB: \sqrt{(-1-3)^2 + (-2+4)^2} = \sqrt{16 + 4} = \sqrt{20}
    Length AC: \sqrt{(1-3)^2 + (2+4)^2} = \sqrt{4 + 36} = \sqrt{40}
    Length BC = \sqrt{1+1)^2 + (2+2)^2} = \sqrt{4 + 16} = \sqrt{20}

    Therefore the triangle ABC is isoseles with AB and BC being the 2 equal sides.
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    ahh..i love you..thankyou!

    well i took A=(-2,-1), B=(2,1) and C=-4,3)
    AB gave me root18 whilst AC gave me root20.

    anyhow..thanks a lot!
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    (Original post by addylad)
    That's correct, just I didn't see any ^2 in your original post.

    Ah yes, I forgot. :p:

    But then you put a dividing sign instead of an addition sign in your correction, so that confused me lol.
 
 
 
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