MHB Distance between two points in the Cartesian plane

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The distance between two points A(x1, y1) and B(x2, y2) in the Cartesian plane is calculated using the formula AB = √((x2 - x1)² + (y2 - y1)²). The point P that divides the segment AB in the ratio λ:μ is expressed as P = (λx2 + μx1) / (λ + μ), (λy2 + μy1) / (λ + μ). This relationship is derived using the properties of similar triangles, where the ratios of the segments correspond to the ratios of the coordinates. By analyzing the horizontal and vertical legs of a right triangle formed by extending lines from point P, the equations for the coordinates of P can be established. The discussion concludes by confirming that these derived coordinates match the given formula for point P.
Poly1
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Let $AB$ be the distance between the two points $A(x_{1} ~ x_{2})$ and $B(x_{2}, ~ y_{2})$ -- e.g. $AB = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$.

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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If AP is given by $\lambda $ then

$\displaystyle AP = \frac{\lambda}{\lambda+\mu}\times AB$
 
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 $\displaystyle (x_P,y_P)$.
 
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How did you get those equations from the triangle?
 
I sort of get it, but I'm bit confused about the choice of $\lambda$ and $\mu$ in the plotting.
 
Poly said:
How did you get those equations from the triangle?

I assumed the segment AB is neither horizontal nor vertical, and arbitrarily placed a point on it and labeled it P. Then using the described triangle, from P I extended both a horizontal and a vertical line to the legs of the triangle at which the points of intersection divides the legs in the same ratio as P divides AB.
 
Hello, Poly!

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
Suppose M divides AB in the ratio \lambda\!:\!\mu \:=\:3:4

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

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

The y-coordinate is: .y_1 + \tfrac{3}{7}(y_2-y_1) \:=\:\tfrac{4}{7}y_1 + \tfrac{3}{7}y^2The coordinates of M are: .\left(\tfrac{4}{7}x_1+\tfrac{3}{7}x_2,\:\tfrac{4}{7}y_1 + \tfrac{3}{7}y_2\right)

. . =\;\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)

Compare this to the given formula.
 
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