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Thread: 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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    $\displaystyle P \in AB, AP:PB=3:1$
    p is in AB, The ratio of AP to PB is 3 to 1.... (right?)
    $\displaystyle a=(-4,6) b=(6,-7)$
    I.
    $\displaystyle y=-1.3x+0.8$
    AB:
    $\displaystyle \Delta y=13$
    $\displaystyle \Delta x=10$

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

    AP:

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


    PB:

    $\displaystyle \frac{\sqrt{269}}{4}$

    and just use $\displaystyle \Delta x$ and $\displaystyle \Delta y$

    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 $\displaystyle A\;(x_1,y_1)$ to $\displaystyle B\;(x_2,y_2)$. It is:
    $\displaystyle \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
    Well, there's a more general formula for the coordinates of the point $\displaystyle P$ that divides $\displaystyle AB$ in the ratio $\displaystyle m:n$. It is this:
    $\displaystyle \left(\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n}\ right)$
    (If you understand vector notation, this is equivalent to $\displaystyle \textbf{p}=\frac{n\textbf{a}+m\textbf{b}}{m+n}$, where $\displaystyle \textbf{a},\textbf{b},\textbf{p}$ are the position vectors of $\displaystyle A, B$ and $\displaystyle P$.)

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

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

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

    Can you complete it now?

    Grandad
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