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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)

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  1. freckles
    • one year ago
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    one sec let me fix

  2. ganeshie8
    • one year ago
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    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
    • one year ago
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    Guys I have a method for this.

  4. AaronAndyson
    • one year ago
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    A simple one...

  5. freckles
    • one year ago
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    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
    • one year ago
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    For simplicity, lets say A is origin and C is any other point : |dw:1437897811906:dw|

  7. ganeshie8
    • one year ago
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    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
    • one year ago
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    \[x = \frac{ rc + sa }{ s+r }\]

  9. ganeshie8
    • one year ago
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    we can use the same trick(component by component) to partition the segment in any ratio

  10. AaronAndyson
    • one year ago
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    \[y = \frac{ rd + sb }{ r + s }\]

  11. freckles
    • one year ago
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    Ok yeah @ganeshie8 I feel kind of dumb.. I have been doing that a very very long way.

  12. AaronAndyson
    • one year ago
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    What about my method guys?

  13. freckles
    • one year ago
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    what I wanted a method not really a formula like the way in which you get the formula

  14. AaronAndyson
    • one year ago
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    >.< Okay >.<

  15. ganeshie8
    • one year ago
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    |dw:1437898127317:dw|

  16. freckles
    • one year ago
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    but thank you

  17. AaronAndyson
    • one year ago
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    Okay @freckles

  18. ganeshie8
    • one year ago
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    |dw:1437898290504:dw|