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
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
Hello, chancey!
You could baby-talk your way through this one . . .
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.