## freckles one year ago 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)

1. freckles

one sec let me fix

2. ganeshie8

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)

3. AaronAndyson

Guys I have a method for this.

4. AaronAndyson

A simple one...

5. freckles

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….

6. ganeshie8

For simplicity, lets say A is origin and C is any other point : |dw:1437897811906:dw|

7. ganeshie8

Recall how we compute midpoint of AC : we work it component by component x cordinate = (0+7)/2 = 3.5 y coordinate = (6+0)/2 = 3

8. AaronAndyson

$x = \frac{ rc + sa }{ s+r }$

9. ganeshie8

we can use the same trick(component by component) to partition the segment in any ratio

10. AaronAndyson

$y = \frac{ rd + sb }{ r + s }$

11. freckles

Ok yeah @ganeshie8 I feel kind of dumb.. I have been doing that a very very long way.

12. AaronAndyson

13. freckles

what I wanted a method not really a formula like the way in which you get the formula

14. AaronAndyson

>.< Okay >.<

15. ganeshie8

|dw:1437898127317:dw|

16. freckles

but thank you

17. AaronAndyson

Okay @freckles

18. ganeshie8

|dw:1437898290504:dw|