Distance between two points in the Cartesian plane

In summary, the point $P$ which divides $AB$ in the ratio $\lambda:\mu$ is given by $\left(\dfrac{\lambda x_{2}+\mu x_{1}}{\lambda+\mu},\,\dfrac{\lambda y_{2}+\mu y_{1}}{\lambda+\mu}\right)$ if and only if $\left(\frac{\lambda}{\lambda+\mu}\right)$ is a multiple of $\mu$.
  • #1
Poly1
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0
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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  • #2
If AP is given by $\lambda $ then

$\displaystyle AP = \frac{\lambda}{\lambda+\mu}\times AB$
 
  • #3
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 $\displaystyle (x_P,y_P)$.
 
Last edited:
  • #4
How did you get those equations from the triangle?
 
  • #5
I sort of get it, but I'm bit confused about the choice of $\lambda$ and $\mu$ in the plotting.
 
  • #6
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.
 
  • #7
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 [tex]M[/tex] divides [tex]AB[/tex] in the ratio [tex]\lambda\!:\!\mu \:=\:3:4[/tex]

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.
 

FAQ: Distance between two points in the Cartesian plane

What is the formula for finding the distance between two points in the Cartesian plane?

The distance between two points (x1,y1) and (x2,y2) in the Cartesian plane can be found using the formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

How do you find the distance between two points if the coordinates are given?

To find the distance between two points with given coordinates, you can use the distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2). Simply plug in the values for x1, y1, x2, and y2 into the formula and solve for d.

Can the distance between two points on the Cartesian plane be negative?

No, the distance between two points on the Cartesian plane cannot be negative. Distance is always a positive value, representing the length of the shortest path between two points.

Is there a difference between finding the distance between two points in 2D and 3D space?

Yes, there is a difference between finding the distance between two points in 2D and 3D space. In 2D space, the distance formula is d = √((x2 - x1)^2 + (y2 - y1)^2), while in 3D space, the distance formula is d = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2).

Can the distance between two points be calculated if one of the points is at the origin?

Yes, the distance between two points can still be calculated if one of the points is at the origin (0,0). The formula for finding the distance between two points (x1,y1) and (x2,y2) is: d = √(x2^2 + y2^2). Simply substitute 0 for x1 and y1 in the formula and solve for d.

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