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Why is the point $P$ which divides $AB$ in the ratio $\lambda:\mu$ given by $\displaystyle ~~ \bigg(\frac{\lambda x_{2}+\mu x_{1}}{\lambda+\mu}, ~ \frac{\lambda y_{2}+\mu y_{1}}{\lambda+\mu}\bigg)$? How do you show that?

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- Thread starter
- #1

Why is the point $P$ which divides $AB$ in the ratio $\lambda:\mu$ given by $\displaystyle ~~ \bigg(\frac{\lambda x_{2}+\mu x_{1}}{\lambda+\mu}, ~ \frac{\lambda y_{2}+\mu y_{1}}{\lambda+\mu}\bigg)$? How do you show that?

- Feb 1, 2012

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If AP is given by $\lambda $ then

$\displaystyle AP = \frac{\lambda}{\lambda+\mu}\times AB$

$\displaystyle AP = \frac{\lambda}{\lambda+\mu}\times AB$

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Let point *P* be $\displaystyle (x_P,y_P)$.

Using a right triangle where*AB* is the hypotenuse, and working with the legs, you could set up the following equations:

Horizontal leg:

$\displaystyle \frac{\lambda}{\lambda+\mu}=\frac{x_P-x_1}{x_2-x_1}$

Vertical leg:

$\displaystyle \frac{\lambda}{\lambda+\mu}=\frac{y_P-y_1}{y_2-y_1}$

Now solve for $\displaystyle (x_P,y_P)$.

Using a right triangle where

Horizontal leg:

$\displaystyle \frac{\lambda}{\lambda+\mu}=\frac{x_P-x_1}{x_2-x_1}$

Vertical leg:

$\displaystyle \frac{\lambda}{\lambda+\mu}=\frac{y_P-y_1}{y_2-y_1}$

Now solve for $\displaystyle (x_P,y_P)$.

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I assumed the segmentHow did you get those equations from the triangle?

Let $AB$ be the distance between the two points $A(x_1,\,y_1)$ and $B(x_2,\,y_2)$

. . $AB \:=\: \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . We don't need this.

Why is the point $P$ which divides $AB$ in the ratio $\lambda\!:\!\mu$ given by $\left(\dfrac{\lambda x_{2}+\mu x_{1}}{\lambda+\mu},\,\dfrac{\lambda y_{2}+\mu y_{1}}{\lambda+\mu}\right)$?

How do you show that?

Code:

```
B
o(x2,y2)
* | :
* | :
M * | y2-y1
o | :
* | :
(x1,y1) o-----------------------+ -
A - - - - x2-x1 - - - - C
```

Then [tex]M[/tex] is [tex]\tfrac{3}{7}[/tex] of the way from [tex]A[/tex] to [tex]B.[/tex]

The x-coordinate is: .[tex]x_1 + \tfrac{3}{7}(x_1-x_1) \:=\:\tfrac{4}{7}x_1 + \tfrac{3}{7}x_2[/tex]

The y-coordinate is: .[tex]y_1 + \tfrac{3}{7}(y_2-y_1) \:=\:\tfrac{4}{7}y_1 + \tfrac{3}{7}y^2[/tex]

The coordinates of [tex]M[/tex] are: .[tex]\left(\tfrac{4}{7}x_1+\tfrac{3}{7}x_2,\:\tfrac{4}{7}y_1 + \tfrac{3}{7}y_2\right) [/tex]

. . [tex]=\;\left(\frac{4x_1 + 3x_2}{7},\:\frac{4y_1+3y_2}{7}\right) \;=\; \left(\frac{3x_2+4x_1}{3+4},\:\frac{3y_2+4y_1}{3+4}\right)[/tex]

Compare this to the given formula.