To calculate the **distance between two points** in a plane, we have to use distance formula as per described in coordinate geometry. With the help of this formula, we can find the distance between any two points marked in an x-y coordinate. This is a significant topic explained in Class 10 Maths Chapter 7.

Suppose, City A is located 50km east and 20km north from the city B. Then how can we calculate the distance between the two cities, without actually measuring it. Here, we can make use of **Pythagoras theorem**, by forming a right-triangle in a plane using the directions given. Now let us read more to know how we can use this theorem to find the distance between to cartesian point.

**Also, read:**

## 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.

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 the x-axis.

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

## Distance Formula

Consider the following situation.

A boy walks towards north 30 meters and took a turn to the east and walked for 40 meters more. How do we calculate the shortest distance between the 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}Â = AB^{2} + BC^{2}

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

Hence, we got the distance between the start point and the endpoint.Â In the same way, the distance between two points in a coordinate plane is also calculated using the Pythagorean theorem or right-angles triangle theorem. Let us see how we can calculate.

### Distance between a point from the origin

Suppose a point P(x,y) in the xy – plane as shown in the figure below:

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 the x-axis.

By Pythagoras theorem,

OP^{2}Â = x^{2Â }+ y^{2}

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

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

Now, let us consider a more generic case. Let us take two points P(x_{1},y_{1})Â and Q(x_{2},y_{2})Â in the coordinate plane. Let us represent this in a figure.

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 the x-axis.

**Distance between points P and Q are calculated as follows:**

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

â‡’ PR = ST

Coordinates of S and T are (x_{1, }0) and Â (x_{2, }0)Â respectively.

OSÂ = Â x_{1Â }and OTÂ = x_{2}

STÂ = OT – OSÂ = x_{2}– x_{1}Â = PR

Similarly,

PS = Â RT

QR = QT – RT = QT – PS =Â y_{2}– y_{1}

By Pythagoras theorem,

PQ^{2Â }= PR^{2Â }+ QR^{2}

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

Therefore,

Distance between two points (x_{1},y_{1})Â and (x_{2},y_{2})Â is given by:

PQ = âˆš[(x_{2}– x_{1})^{2}+ (y_{2}– y_{1})^{2}]

It is known as **distance formula.**

Observe that (x_{2}– x_{1})^{2}Â is the square of the difference in x â€“ coordinates of P and Q and is always positive. The same can be said about (y_{2}– y_{1})^{2Â }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

**Example 1: Find the value of a, if the distance between the points P(3,-6) and Q(-3,a) is 10 units.**

Solution: By distance formula,

Distance between points (3,-6) and (-3,a) = âˆš[(-3-3)^{2}+(a+6)^{2}] = 10 units

Squaring both sides of the equation gives,

(-6)^{2Â }+ (a+6)^{2}Â = 100

(a+6)^{2}Â = 100 – 36Â = 64Â

Taking root on both the sides, we get;

a + 6 = Â±8

**Case I, consider +8,**

a+6= 8Â ,

a = 8-6 = 2

**Case II, consider -8**

a+6Â = -8

a = -8 – 6

a = -14

Therefore, the coordinates are either P(3,-6) and Q(-3,2) or P(3,-6) and Q(-3,-14).

**Example 2:Â Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).**

Solution:Â Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Given, AP = BP.

So, AP^{2Â }= BP2

(x â€“ 7)^{2Â }+ (y â€“ 1)^{2Â }= (x â€“ 3)^{2Â }+ (y â€“ 5)^{2Â }

x^{2Â }â€“ 14x + 49 + y^{2Â }â€“ 2y + 1 = x^{2Â }â€“ 6x + 9 + y^{2Â }â€“ 10y + 25

x â€“ y = 2

which is the required relation.

**Example 3:Â Find a point on the y-axis which is equidistant from the points A(6, 5) and B(â€“ 4, 3).**

Solution:Â We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B. Then:

(6 â€“ 0)^{2Â }+ (5 â€“ y)^{2Â }= (â€“ 4 â€“ 0)^{2Â }+ (3 â€“ y)^{2}

36 + 25 + y^{2}â€“ 10y = 16 + 9 + y^{2Â }â€“ 6y

4y = 36

y = 9

So, the required point is (0, 9).

Verification:

AP =Â âˆš[(6 â€“ 0)^{2Â }+ (5 â€“ y)^{2}]

= âˆš(36+16)

= âˆš52

BP =Â Â âˆš[(-4-0)^{2}+(3-9)^{2}]

=âˆš(16+36)

=âˆš52

Hence, we conclude that, point (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.

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