The questions asks to find the point that divides the interval AB ( A(-7, 5), B(3, 1) ) into the ratio 3:1

I'm looking at the answer but I can't work out why they changed 3:1 to 3:-1 and therefore get a different answer to me

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- Oct 22nd 2006, 09:59 PMchanceyMidpoint formula question
The questions asks to find the point that divides the interval AB ( A(-7, 5), B(3, 1) ) into the ratio 3:1

I'm looking at the answer but I can't work out why they changed 3:1 to 3:-1 and therefore get a different answer to me - Oct 23rd 2006, 12:36 AMearboth
Hi,

you are looking for a unknown point U so that the distances

$\displaystyle \frac{AU}{UB}=\frac{3}{1}$

$\displaystyle AU=3\cdot UB$

UB = AB - AU . So you get:

$\displaystyle AU=3\cdot (AB-AU)$. Thus:

$\displaystyle AU=3AB-3AU$

$\displaystyle AU=\frac{3}{4}\cdot AB$

Because $\displaystyle B(x_A+10,y_A+(-4))$ your unknown point is now well-known:

$\displaystyle U(x_A+\frac{3}{4}\cdot 10,y_A+\frac{3}{4}\cdot (-4))$. That means: $\displaystyle U(0.5,2)$

I've attached a diagram to show you the situation. In the textbox you find the ratio.

EB - Oct 23rd 2006, 06:08 AMSoroban
Hello, chancey!

You could baby-talk your way through this one . . .

Quote:

Find the point that divides the line segment $\displaystyle AB$ into the ratio $\displaystyle 3:1$

given: $\displaystyle A(-7, 5),\;B(3, 1)$

We want a point $\displaystyle P$ which is $\displaystyle \frac{3}{4}$ the distance from $\displaystyle A$ to $\displaystyle B.$

. . $\displaystyle A$ . . . . . . . . . . . $\displaystyle P$ . . .$\displaystyle B$

. . $\displaystyle * - - + - - + - - \bullet - - *$

To go from $\displaystyle A(-7,5)$ to $\displaystyle B(3,1)$, we'd move: $\displaystyle \begin{Bmatrix}\text{10 units right} \\ \text{4 units down}\end{Bmatrix}$

To go three-fourths the distance, we'd move: $\displaystyle \begin{Bmatrix}\frac{3}{4}\cdot10 = \frac{15}{2}\text{ units right} \\ \frac{3}{4}\cdot4 = \text{3 unit down}\end{Bmatrix}$

Hence, $\displaystyle P$ is: .$\displaystyle \begin{Bmatrix}\text{-}7 + \frac{15}{2} = \frac{1}{2} \\ 5 + (\text{-}3) = 2\end{Bmatrix} \quad \Rightarrow\quad P\left(\frac{1}{2},\,2\right)$

Of course, this is basically Earboth's solution.