Thread: Midpoints and ratios

1. Midpoints and ratios

Points A and B are located at $\displaystyle (a,b)$ and $\displaystyle (c,d)$ respectively. Point C divides $\displaystyle AB$ into the ratio $\displaystyle p:q$. Find the coordinates of Point C.

Thanks... this is a hard one i think... maybe i need to brush up on my analysis

2. AB can be represented as (c-a,d-b)
since C divides AB in the ratio p:q, $\displaystyle AC=\frac{p}{p+q}AB$
$\displaystyle AC = (\frac{p(c-a)}{p+q},\frac{p(d-b)}{p+q})$

C = A+AC
$\displaystyle C =(a+\frac{p(c-a)}{p+q},b+\frac{p(d-b)}{p+q})$
$\displaystyle =(\frac{ap+aq+cp-ap}{p+q},\frac{bp+bq+dp-bp}{p+q})$
$\displaystyle =(\frac{aq+cp}{p+q},\frac{bq+dp}{p+q})$

3. Originally Posted by badgerigar
AB can be represented as (c-a,d-b)
since C divides AB in the ratio p:q, $\displaystyle AC=\frac{p}{p+q}AB$
$\displaystyle AC = (\frac{p(c-a)}{p+q},\frac{p(d-b)}{p+q})$

C = A+AC
$\displaystyle C =(a+\frac{p(c-a)}{p+q},b+\frac{p(d-b)}{p+q})$
$\displaystyle =(\frac{ap+aq+cp-ap}{p+q},\frac{bp+bq+dp-bp}{p+q})$
$\displaystyle =(\frac{aq+cp}{p+q},\frac{bq+dp}{p+q})$
Thanks heaps badgerigar! I notice how you switched the ordered pairs from coordinate points into length units, that was smart.

This probably isn't that important seeing as the main aim has been achieved, but for all purposes shouldn't AB be (|c-a|, |d-b|)

So the final coordinates become

$\displaystyle \left(a+\frac{p|c-a|}{p+q}, b+\frac{p|d-b|}{p+q}\right)$

Thanks again

4. This probably isn't that important seeing as the main aim has been achieved, but for all purposes shouldn't AB be (|c-a|, |d-b|)

So the final coordinates become

No. If c-a is negative that means that point B is left of point A. taking the absolute value sign will have you adding a positive value to the x value of A, since \frac{p}{p+q} is positive. This will put C to the right of A, so not between A and B.

If you want to think about an example try A=(3,0) B=(0,0) and the ratio 1:2

5. Originally Posted by badgerigar
$\displaystyle =\left(\frac{cp+aq}{p+q},\frac{dp+bq}{p+q}\right)$
Well, i say this result is quite nice and worth memorizing. Thanks!