Question

A segment with endpoints A (a,b) and C (c,d) is partitioned by a point B such that AB and BC form a r:s ratio. Find B.
We will call B (x,y).
I see people in algebra get this as a question sometimes. To me it is pretty lengthy. I wonder if they actually have a formula for it they use (that they don't give :p) or if there is a simpler way. So this is what this post is about, deriving a formula for it. Like is there an easier way than the way I have chosen. I just can't imagine an algebra student doing all of these steps. (Also I typed this in Microsoft word so I will be looking for type-o)

Answer 1

one sec let me fix

Answer 2

we can simply work x coordinate and y coordinate separately, below should work :

x = a + |c-a|*r/(r+s)

y = b + |d-b|*s/(r+s)

Answer 3

Guys I have a method for this.

Answer 4

A simple one...

Answer 5

First equation: \[\frac{AB}{BC}=\frac{r}{s}\]
Second equation(s): \[\frac{b-d}{a-c}=\frac{y-d}{x-c}=\frac{y-b}{x-a}\]
So the first equation: \[\frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(x-c)^2+(x-d)^2}}=\frac{r}{s}\]
Square both sides: \[\frac{(x-a)^2+(y-b)^2}{(x-c)^2+(y-d)^2}=\frac{r^2}{s^2}\]
Now the second equation(s) gives us:
\[\frac{b-d}{a-c}(x-c)=y-d\]
And
\[\frac{b-d}{a-c}(x-a)=y-b\]
So making these subs into our first equation gives:
\[\frac{(x-a)^2+(\frac{b-d}{a-c})^2(x-a)^2}{(x-c)^2+(\frac{b-d}{a-c})^2(x-c)^2}=\frac{r^2}{s^2}\]
Adding like terms on top and bottom on left hand side gives:
\[\frac{(1+(\frac{b-d}{a-c})^2)(x-a)^2}{(1+(\frac{b-d}{a-c})^2}(x-c)^2}=\frac{r^2}{s^2}\]
So simplifying equation just a bit gives:
\[(\frac{x-a}{x-c})^2=\frac{r^2}{s^2}\]
So this means we have:
\[\frac{x-a}{x-c}=\frac{r}{s} \text{ or } \frac{x-a}{x-c}=-\frac{r}{s}\]
So multiplying (x-c) on both sides…
\[x-a=\frac{r}{s}(x-c) \text{ or } x-a=\frac{-r}{s}(x-c) \]
Distributing a bit:
\[x-a=\frac{r}{s}x-\frac{r}{s}c \text{ or } x-a=\frac{-r}{s}x+\frac{r}{s}c\]
Putting x terms on one side and non-x terms on opposing side:
\[(1-\frac{r}{s})x=-\frac{r}{s}c+a \text{ or } (1+\frac{r}{s})x=\frac{r}{s}c+a\]
Finally solving for … x:
\[x=\frac{\frac{-r}{s}c+a}{1-\frac{r}{s}} \text{ or } x=\frac{\frac{r}{s}c+a}{1+\frac{r}{s}}\]
Making things pretty by getting rid of the compound fraction…
That is we are going to multiply s on top and bottom…
\[x=\frac{-rc+sa}{s-r} \text{ or } x=\frac{rc+as}{s+r}\]
Now we can find y using one of those second equation(s):
\[\frac{b-d}{a-c}(x-c)=y-d\]
Add d on both sides…
\[y=\frac{b-d}{a-c}(x-c)+d=\frac{b-d}{a-c}(\frac{-rc+sa}{s-r}-c)+d \\ \text{ or } \\ y=\frac{b-d}{a-c}(x-c)+d=\frac{b-d}{a-c}(\frac{rc+sa}{s+r}-c)+d\]
I guess we can combine the terms for y by finding a common denominator.
\[y=\frac{b-d}{a-c}(\frac{-rc+sa}{s-r}-c)+d=\frac{(b-d)(-rc+sa)+d(a-c)(s-r)}{(a-c)(s-r)} \\ \text{ or } \\ y =\frac{b-d}{a-c}(\frac{rc+sa}{s+r}-c)+d=\frac{(b-d)(rc+sa)+d(a-c)(s+r)}{(a-c)(s+r)}\]
Anyways there will be one pair (x,y) that we are looking for. We can rule out the other point based on where the given coordinates are….

Answer 6

For simplicity, lets say A is origin and C is any other point :