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    AS MATHS - COORDINATE GEOMETRY
    The point A has coordinates (7,4). The straight lines with equations x+3y+1=0 and 2x+5y=0 intersect at the point B. Find the coordinates of B and hence show that one of the two lines is perpendicular to AB.

    I solved point B as (5,-2) simultaeneously. Now i'm stuck.
    :confused: :woo:





    Rep to the first one that makes sense! :o:
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    Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
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    (Original post by FranticMind)
    Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
    I tried doing this.
    I assumed that coordinate A(7,4) and coordinate B(5,-2) joined and found the gradient of the line AB to be 3.
    Do i re-arrange any one of the 2 formulae to find the gradient?
    If so, what is the formula linking m1 and m2?
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    (Original post by FranticMind)
    Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
    oh okay so after that I tried doing this:
    I assumed that coordinate A(7,4) and coordinate B(5,-2) joined and found the gradient of the line AB to be 3.
    Do i re-arrange any one of the 2 equations to find the gradient?
    If so, what is the formula linking m1 and m2?
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    kay so you found the gradient of AB to be 3, then pick one of the other equations, ie x + 3y + 1 = 0

    rearrange to y = -1/3 x - 1/3

    and you can see that the gradient is -1/3.

    the gradient of a line perpendicular to AB would be -1/m > -1/3, and you've shown that it does have that gradient.
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    Find the gradient for line AB

    Rearrange both the equations you have to get them in the form y=mx+c

    Compare gradient for line AB to the equations.
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    Any two perpendicular lines will have product gradient -1, i.e. m1 x m2 = -1

    So if a line has a gradient say -4, then any line perpendicular to this will have gradient 1/4. This is the same as flipping round the fraction (think of 4 as 4/1) and sticking a minus sign in front
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    (Original post by BrightGirl)
    kay so you found the gradient of AB to be 3, then pick one of the other equations, ie x + 3y + 1 = 0

    rearrange to y = -1/3 x - 1/3

    and you can see that the gradient is -1/3.

    the gradient of a line perpendicular to AB would be -1/m > -1/3, and you've shown that it does have that gradient.
    omg thanks so much.
    +rep
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    hehe okay GOT IT.
    I owe rep to Goldfishy, sexuali, Brightgirl and FranticMind :P
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    (Original post by Goldfishy)
    Any two perpendicular lines will have product gradient -1, i.e. m1 x m2 = -1

    So if a line has a gradient say -4, then any line perpendicular to this will have gradient 1/4. This is the same as flipping round the fraction (think of 4 as 4/1) and sticking a minus sign in front
    thaaankyouu
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    (Original post by Glee)
    omg thanks so much.
    +rep
    you're welcome, don't worry about the rep
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    (Original post by sexuali)
    Find the gradient for line AB

    Rearrange both the equations you have to get them in the form y=mx+c

    Compare gradient for line AB to the equations.
    thankss
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    The perpendicular gradient will simply be 1/m and of inverse sign.

    The gradient of the line AB is

    (4- -2)/(7-5) = 3

    The gradient of a perpendicular would be

    -1/3

    And x + 3y +1 =0

    rearranges to y = (-1-x)/3

    which is the same as y = -1/3(x+1) or y = -1/3x-1/3

    thus the gradient is -1/3 and so it is the perpendicular to the first equation.

    Hope that helped!
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    (Original post by BrightGirl)
    you're welcome, don't worry about the rep
    nahh i insist
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    oops I see brightgirl beat me to it. Whilst i was typing lol
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    (Original post by FranticMind)
    The perpendicular gradient will simply be 1/m and of inverse sign.

    The gradient of the line AB is

    (4- -2)/(7-5) = 3

    The gradient of a perpendicular would be

    -1/3

    And x + 3y +1 =0

    rearranges to y = (-1-x)/3

    which is the same as y = -1/3(x+1) or y = -1/3x-1/3

    thus the gradient is -1/3 and so it is the perpendicular to the first equation.

    Hope that helped!
    :rolleyes: :yes:
 
 
 
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