freckles
  • freckles
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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freckles
  • freckles
one sec let me fix
ganeshie8
  • 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)
AaronAndyson
  • AaronAndyson
Guys I have a method for this.

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AaronAndyson
  • AaronAndyson
A simple one...
freckles
  • 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….
ganeshie8
  • ganeshie8
For simplicity, lets say A is origin and C is any other point : |dw:1437897811906:dw|
ganeshie8
  • 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
AaronAndyson
  • AaronAndyson
\[x = \frac{ rc + sa }{ s+r }\]
ganeshie8
  • ganeshie8
we can use the same trick(component by component) to partition the segment in any ratio
AaronAndyson
  • AaronAndyson
\[y = \frac{ rd + sb }{ r + s }\]
freckles
  • freckles
Ok yeah @ganeshie8 I feel kind of dumb.. I have been doing that a very very long way.
AaronAndyson
  • AaronAndyson
What about my method guys?
freckles
  • freckles
what I wanted a method not really a formula like the way in which you get the formula
AaronAndyson
  • AaronAndyson
>.< Okay >.<
ganeshie8
  • ganeshie8
|dw:1437898127317:dw|
freckles
  • freckles
but thank you
AaronAndyson
  • AaronAndyson
Okay @freckles
ganeshie8
  • ganeshie8
|dw:1437898290504:dw|
ganeshie8
  • ganeshie8
so the x coordinate of point B is \(5\) similarly we can work the y coordinate also
ganeshie8
  • ganeshie8
It seems you have derived correct formulas for both internal and external partitioning
freckles
  • freckles
\[\frac{y}{6-y}=\frac{5}{2} \\ 2y=30-5y \\ 7y=30 \\ y=\frac{30}{7} \\ \text{ and yeah } \\ \frac{x}{7-x}=\frac{5}{2} \\ 2x=35-5x \\ 7x=35 \\ x=5 \\ (5,\frac{30}{7})\]
freckles
  • freckles
|dw:1437898739186:dw|