Sales Toll Free No: 1-855-666-7446

# Distance between Two Points

Top
 Sub Topics Coordinate system uses a set of numbers to identify any point in a plane or space uniquely. There are various types of coordinate systems such as number line, Cartesian plane, n-dimensional Euclidean space, polar coordinates and so on. Each of these systems will have a center point or origin with respect to whom each point will be uniquely defined.In real life, the word distance comes in use quite frequently. If there are two points p, q on a number line, that is, one dimensional system the distance will be |p - q|. But when it comes to a Cartesian plane a formula is applied to get the distance between two points.The distance between two points is given by $d = \sqrt{dp^2 + dq^2}$.Here, ‘d’ is the distance. ‘p’ is the coordinates of x - axis. ‘q’ is the coordinates of y - axis.‘dp’ is the difference between the x - coordinates of the points.‘dq’ is the difference between the y - coordinates of the points.Suppose we have coordinates of the points, then we use the above formula for finding the distance between the coordinates of the points.

## Vertical and Horizontal Lines

Slope of line is basically the measure of the inclination of a line. Here, we will discuss about vertical and horizontal lines. In a coordinate plane, there are two axes known as x and y -axis. These axes are represented by horizontal and vertical lines respectively. When x - coordinate is constant and y - coordinate changes, then line is called as vertical line.

A line that moves straight up and down and also parallel to y - axis of the coordinate plane is known as vertical line. In case of vertical line, slope is not defined. Equation of a vertical line is x = u.

When y - coordinate remains constant and x - coordinate changes, then the line formed is horizontal.

A line that moves straight from left to right and also parallel to the x - axis of a coordinate plane is known as horizontal line. Slope of a horizontal line is zero. Equation of horizontal line is y = v. Here, value of ‘y’ shows the coordinates of any point on the line and ‘v’ shows the line which crosses x - axis.

We can also draw points in negative direction. If we plot the coordinates in reverse direction from origin or left side of origin point, then we get negative line. In case of vertical lines, points above the origin is positive and below the origin point is negative. We can plot vertical horizontal lines in negative direction also.

## Formula

Given there are two points in a Cartesian plane $(x_{1},y_{1})$ and $(x_{2},y_{2})$. Then the distance between the two points can be calculated by using the formula.
$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
This formula is actually a variation of Pythagoras theorem. How? We shall see soon.

## Derivation

For Pythagoras formula we know that for a right angled triangle where p is the base length, q is the perpendicular length and r is the hypotenuse length.

$r^{2}=p^{2}+q^{2}$

Also we know that in a one-dimensional axis, the distance between two points p and q is |p - q|. Let us take an example of two points A(4, 3) and B(2, 1).

We draw a straight line between them.

We do make a perpendicular triangle as given.

The third point C can be obtained as (4, 1) as the x-coordinate point A and y-coordinate as point B. The distance AC and BC can easily be calculated as these lines are parallel to Y-axis and X-axis respectively.
Now we have ABC as the right angled triangle with Ab as the hypotenuse, BC as base and AC as the perpendicular. Using the Pythagoras theorem we can have,
$AB^{2} = AC^{2} + BC^{2}$
We can see that AC = Difference between the y-coordinates of A and B; BC = Difference between the x-coordinates of B and C. Hence, we have
$AB^{2} = (3 - 1)^{2} + (4 - 2)^{2}$
AB = $d = \sqrt{(3-1)^{2}+(4-2)^{2}}$ = $\sqrt{4 + 4}$ = 2$\sqrt{2}$
Hence, we get our formula for the distance between the two points in a Cartesian plane. If the distance between x-coordinates is dp, and y-coordinates is dq the distance d between the points will be:
$d = \sqrt{dp^2 + dq^2}$

## Examples

$d = \sqrt{(-1-3)^{2}+(3-1)^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2\sqrt{5}$.
Hence, the distance between these two points are $2\sqrt{5}$.