दूरी-सूत्र

The distance formula, in coordinate geometry is used to find the distance between the two points in an $XY$ plane. The distance of a point from the $y-$axis is called its $x-$coordinate, or abscissa. The distance of a point from the $x-$axis is called its $y-$coordinate, or ordinate. The coordinates of a point on the $x-$axis are of the form $(x,0)$, and of a point on the $y-$axis are of the form $(0,y)$. To find the distance between any two points in a plane, we will use the Pythagoras theorem.

What is the distance formula?

The distance formula is the formula, which is used to find the distance between any two points, only if the coordinates are known to us. These coordinates could lie on $x-$axis or $y-$axis or both. Suppose, there are two points, say $A$ and $B$ in an $XY$ plane (see Fig. 1) The coordinates of point $A$ are $(x_{1},y_{1})$ and of $B$ are $(x_{2},y_{2})$.

Fig 1 - Distance Formula

Then the formula to find the distance between two points $AB$ is given by

$AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

Distance Formula Derivation

Let us find the distance between two points $A(x_{1},y_{1})$and $B(x_{2},y_{2})$ shown Fig.1

Draw $AD$ and $BE$ perpendicular to the $x-$axis. A perpendicular from the point $A$ on $BE$ is drawn to meet it at the point $C$.

Then, $OD=x_{1}$ , $OE=x_{2}$. So, $DE=x_{2}-x_{1}=AC$. Also, $EB=y_{2}$ , $EC=AD=y_{1}$. Hence $BC=y_{2}-y_{1}$

Now, applying the Pythagoras theorem in $\bigtriangleup ACB$ , we get

$AB^{2}=AC^{2}+BC^{2}$

$AB^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

$AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$ is the distance formula.

Example

Find the distance between the two points $A(1,2)$ and $B(3,4)$

Solution:

Let $A(1,2)=(x_{1},y_{1})$

$B(3,4)=(x_{2},y_{2})$

Distance between the two points $A$ and $B$

$AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

$AB={\sqrt {(3-1)^{2}+(4-2)^{2}}}$

$AB={\sqrt {4+4}}={\sqrt {8}}=2{\sqrt {2}}$