# Distance Between Two Points

Coordinates of a point

Two-dimensional geometry deals with coordinates of points, distance between the points etc.
Coordinates of a point is a pair of numbers which exactly define the location of that point in a coordinate plane.

Figure 1-Coordinates of a point

Coordinates of the point P in the two dimensional plane is (x,y) which means, P is x units away from y-axis and y units away from x-axis.

Coordinates of a point on thex-axis is of the form (a,0), where a is the distance of the point from theorigin.

Coordinates of a point on the y-axis is of the form (0,a), where a is the distance of the point from the origin.

Distance between two points in a plane

Consider the following situation.

A boy walks towards north 30 meters and took a turn to east and walked for 40 meters more. How do we calculate the shortest distance between initial place and final place?

A pictorial representation of the above situation is shown:

The initial point is A and final point is C. The distance between the points A, B is 30 m and between points B, C is 40 m.

The shortest distance between points A and C is AC.

This is calculated using Pythagoras theorem,

$AC^2$ = $AB^2~+~BC^2$

$AC$ = $\sqrt{30^2~+~40^2}$ = 50 m

Distance between two points in a coordinate plane is also calculated using the above method.

Consider a point P(x,y) in the xy – plane.

Figure 2-Distance between a point from origin

Let us calculate the distance between the point P and the origin.P is x units away from y-axis and y units away from x-axis.

By Pythagoras theorem,

$OP^2$ = $x^2~+~y^2$

$OP$ = $\sqrt{x^2~+~y^2}$

Therefore distance between any point (x,y) in xy-plane and the origin (0,0) is $\sqrt{x^2~+~y^2}$

Now, let us consider a more generic case. Let us take two points $P(x_1,y_1)$  and $Q(x_2,y_2)$  in the coordinate plane. Let us represent this in a figure.

Figure 3-Distance between points

Note that we have taken our points P and Q in the first quadrant itself. What if the points are in other quadrants? As you shall observe in the following discussion, the final formula still remains the same, irrespective of which quadrant P and Q lie in.

PS,QT are perpendicular to x – axis and PR is parallel to x – axis.

Distance between points P and Q is calculated as follows.

S and T are the points on x – axis which are end points of two parallel line segments PS and QT respectively.

⇒ PR = ST

Coordinates of S and T are $(x_1~,~0)$ and $(x_2~,~0)$  respectively.

$OS$ = $x_1~and~OT$ = $x_2$

$ST$ = $OT~-~OS$ = $x_2~-~x_1$ = $PR$

Similarly,

PS =  RT

QR = QT – RT = QT – PS = $y_2~-~y_1$

By Pythagoras theorem,

$PQ^2$ = $PR^2~+~QR^2$

$PQ$ = $\sqrt{(x_2~-~x_1 )^2~+~(y_2~-~y_1 )^2 }$

Therefore,

Distance between two points $(x_1,y_1)$  and $(x_2,y_2)$ is $\sqrt{(x_2~-~x_1 )^2~+~(y_2~-~y_1 )^2}$  . It is known as distance formula.

Observe that $(x_2~-~x_1 )^2$  is the square of the difference in x – coordinates of P and Q and is always positive. The same can be said about $(y_2~-~y_1 )^2$  as well. Use this point and try to see for yourself why the formula remains the same for any coordinates of P and Q, in any quadrant.

Example: Find the value of a, if the distance between the points P(3,-6) and Q(-3,a) is 10 units.

By distance formula,

Distance between points (3,-6) and (-3,a) = $\sqrt{(-3~-3)^2~+~(-6~-~a)^2 }$ = 10 units

Squaring both sides of the equation gives,

$(-6)^2~+~(-6~-~a)^2$ = $100$

$(-6~-~a)^2$ = $100~-~36$ = $64$

$-6~-~a$ = $±8$

Case I,

$-6~-~a$ = $8$ ,         $a$ = $-6~-~8$ = $-14$

Case II,

$-6~-~a$ = $-8$ ,         $a$ =

$-6~+~8$ = $2$<

Therefore,

Coordinates of Q can be (-3 ,-14) or (-3 ,2).

To learn more about coordinate geometry and related topics, log onto www.byjus.com. To watch interesting videos on the topic, download Byju’s – The Learning App from Google Play Store.

#### Practise This Question

The points (2, –1), (3, 4), (–2, 3) and (–3, –2) taken in the order from the vertices of _________________