1. 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

2. $P \in AB, AP:PB=3:1$
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,

3. Hello scubasteve94
Originally Posted by scubasteve94
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 $m = 3, n = -2$.

(i) $(\tfrac72,-\tfrac{15}{4})$; (ii) $(26,-33)$