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Math Help - Linear functions

  1. #1
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    Linear functions

    If A = (-4, 6) and B = (6, -7) find:
    I) the coordinates of P, where P AB and AP:PB = 3:1
    ii) the coordinates of P, where P AB and AP:AB = 3:1
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  2. #2
    Member integral's Avatar
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    P \in AB, AP:PB=3:1
    p is in AB, The ratio of AP to PB is 3 to 1.... (right?)
    a=(-4,6) b=(6,-7)
    I.
    y=-1.3x+0.8
    AB:
    \Delta y=13
    \Delta x=10

    AB=\sqrt{a^{2}+b^{2}}
    AB=\sqrt{\Delta x^{2}+\Delta y^{2}}
    AB=\sqrt{13^{2}+10^{2}}
    AB=\sqrt{269}

    AP:

    \frac{3\sqrt{269}}{4}


    PB:

    \frac{\sqrt{269}}{4}

    and just use \Delta x and \Delta y

    to figure out the coordinates.
    If this is correct just do the same for II,
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  3. #3
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    Hello scubasteve94
    Quote Originally Posted by scubasteve94 View Post
    If A = (-4, 6) and B = (6, -7) find:
    I) the coordinates of P, where P AB and AP:PB = 3:1
    ii) the coordinates of P, where P AB and AP:AB = 3:1
    I'm sure you know the formula for finding the coordinates of the mid-point of the line joining A\;(x_1,y_1) to B\;(x_2,y_2). It is:
    \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
    Well, there's a more general formula for the coordinates of the point P that divides AB in the ratio m:n. It is this:
    \left(\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n}\  right)
    (If you understand vector notation, this is equivalent to \textbf{p}=\frac{n\textbf{a}+m\textbf{b}}{m+n}, where \textbf{a},\textbf{b},\textbf{p} are the position vectors of A, B and P.)

    For part (i), AP:PB = 3:1, so m=3, n = 1. So plug these values into the formula, using x_1=-4,y_1=6,x_2=6,y_2=-7.

    For part (ii), we have AP:AB=3:1. So P lies outside the line segment AB. Draw a diagram if you're not sure. You'll then find that the ratio AP:PB = 3:-2. So you'll need to use the values m = 3, n = -2.

    Here are my answers:
    Spoiler:
    (i) (\tfrac72,-\tfrac{15}{4}); (ii) (26,-33)

    Can you complete it now?

    Grandad
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