This article provided formulas with examples for distance formula in coordinate geometry. i.e. Distance between two points on coordinate plane formula with derivation also given different types of examples.

### What is the distance formula in coordinate geometry?

The* distance formula* is used to measure the distance (say *d)* between two points in the coordinate plane. If we consider the coordinates of the two points are A(*x*_{1}, *y*_{1}) and B(*x*_{2}, *y*_{2}) then the distance between two points ‘d’ is equal to the length of the straight line connected between these two coordinates in the plane.

#### a)The distance between points that are lying on the X-axis or Y-axis

In this case, we can calculate distance easily. It is equal to the difference between the x-coordinates.

Let us consider two points P(x_{1}, 0), Q(x_{2}, 0)

These two points lie on the X-axis because the values of y-coordinates are zero in both points.

Distance between P and Q is |x_{2} − x_{1}| units

Let us consider two points A(0, y_{1}), B(0, y_{2})

These two points lie on the Y-axis because the values of the X-coordinates are zero in both points.

Distance between A and B is |y_{2} − y_{1}| units

Note: Generally, we never say the distance in negative values.

**Example-1:**

*What is the distance between P(4, 0) and Q(6, 0)?*

These two points lie on the X-axis due to the values of y-coordinates are zero.

The distance between P and Q is the difference in the value of x-coordinates.

6 -4 = 2 units

**Example-2:**

*What is the distance between R(-4, 0) and S(-6, 0)?*

The distance between R and S is the difference in the value of x-coordinates.

-6 -(-4) =| – 2| = 2 units

**Example-3:**

*What is the distance between A(0, 2) and B(0, 9)?*

These two points lie on the Y-axis because the values of y-coordinates are zero.

The distance between A and B is the difference in the value of y-coordinates.

9 – 2 =| 7| = 7 units

**Example-4:**

*What is the distance between C(0, -9) and D(0, -5)*?

The distance between C and B is the difference in the value of y-coordinates.

– 9 – (-5) =| -4| = 4 units

Distance between two points on a line parallel to the coordinate axis

#### b)The distance between points which are parallel to X-axis or Y-axis

Let Considered two points of P(x_{1}, y_{1}) and Q(x_{2}, y_{1}).

Here the y-coordinates are equal in both points. Hence these points lie on a line and also this line is parallel to the X-axis.

Distance between two points P & Q is equal to the difference between x-coordinates.

i.e PQ = |x_{2} − x_{1}|

**Note**: Generally, we never say the distance in negative values.

Let Considered two points of R(x_{1}, y_{1}) and S(x_{1}, y_{2}).

Here the x-coordinates are equal in both points. Hence these points lie on a line and also this line is parallel to the Y-axis.

Distance between two points R & S is equal to the difference between y-coordinates.

i.e RS = |y_{2} − y_{1}|

**Example-5:**

*P and Q are two points given by (6, 4), and (−2, 4) respectively. Find the distance between P & Q.*

Here both y-coordinates are the same. Hence the distance between P and Q is the difference in the value of x-coordinates.

i.e -2 – 6 =| -8| = 8 units

**Example-6:**

R and S are two points given by (3, -4), and (3, 4) respectively. Find the distance between R & S.

Here both x-coordinates are the same. Hence the distance between R and S is the difference in the value of y-coordinates.

i.e 4 -(-4) =| 8 | = 8 units

#### c) Distance between two points on a line in the X-Y plane ( Distance Formula)

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be any two points in a plane.

As shown in the below figure.

By using P & Q points we can from ΔPRQ (right angle triangle)

According to Pythagoras’ theorem

PQ^{2} = PR^{2} + QR^{2}

PR = x_{2}-x_{1}

QR = y_{2} – y_{1}

PQ^{2} = ( x_{2}-x_{1})^{2} + (y_{2} – y_{1})^{2}

Hence, the distance between the points P & Q is

This formula is also called a distance formula

**Example -7**

*Find the distance between two points P(2, 6) and Q(6, 9)*

Here we compare these points with P(x_{1}, y_{1}) and Q(x_{2}, y_{2})

x_{1} = 2, x_{2} = 6,

y_{1} = 6, y_{2} = 9

Now using the distance formula

PQ = √25 = 5 units

#### d) Collinear points in the coordinate plane

Let P(x_{1}, y_{1}) , Q(x_{2}, y_{2}) & R(x_{3}, y_{3}) are be any three points in a plane.

The distances of their points are PQ, QR & RP

If the sum of any two distances is equal to another distance then their points lie on a straight line. Hence these points are called collinear points

i.e PQ + QR = RP (or) QR + RP = PQ (or) PQ + RP = QR

**Example -8**

*Verify that the points P(1, 5), Q(2, 3) and R(−2, −1) are collinear or not.*

Here find the distance between the points PQ, QR & RP

PQ = √ (1)^{2} + (-2)^{2} = √5

QR = √ (-4)^{2} + (-4)^{2} = √(16 x 2) = 4 √2

RP = √ (-3)^{2} + (-6)^{2} = √45 = 3 √5

Here the sum of any two distances is not equal to another distance. So these points do not lie in a straight line.

Hence these points are not collinear points.

**Example -9**

*Verify that the points P(2,7), Q(5,6) and R(8,5) are collinear or not.*

Here find the distance between the points PQ, QR & RP

PQ = √ (3)^{2} + (-1)^{2} = √10

QR = √ (3)^{2} + (-1)^{2} = √10

RP = √ (6)^{2} + (-2)^{2} = √40 = 2 √10

Here PQ + QR = RP

The sum of PQ & QR is equal to RP. So points P, Q & R lies in a straight line. Hence these points are called collinear points.

**Example – 10**

*Are the points A(2,5), B(3,8) & C(4,5) form a triangle?*

Here find the distance between the points AB, BC & CA

AB = √ (1)^{2} + (3)^{2} = √10

BC = √ (1)^{2} + (-3)^{2} = √10

CA = √ (2)^{2} + (0)^{2} = √ 4

The sum of any two of these distances is greater than the third.

Hence points A, B & C form a triangle. Also, two sides of a triangle are equal. These three points form *an isosceles triangle.*

**Example -11**

A*re the points P(5, −2), Q(6, 4) and R(7, −2) form an isosceles triangle?*

Here find the distance between the points PQ, QR & RP

PQ = √ (1)^{2} + (6)^{2} = √37

BC = √ (1)^{2} + (-6)^{2} = √37

CA = √ (2)^{2} + (0)^{2} = √ 4

The points P, Q & R form a triangle. Also, two sides of a triangle are equal. These three points form *an isosceles triangle.*

**Example -12**

*Are the points P(b, 0), Q(−b, 0), R(0, b √3 ) form a equilateral triangle.*

Here find the distance between the points PQ, QR & RP

PQ = √ (-2b)^{2} + (0)^{2} = 2b

QR = √ (b)^{2} + (b√3)^{2} = √4b^{2} = 2b

RP = √ (b)^{2} + (-b√3)^{2} = √4b^{2} = 2b

The points P, Q & R form a triangle. Also, the three sides of the triangle are equal. These three points form* an equilateral triangle.*

#### Distance formula geometry practice problems:

1) What is the distance between P(2, 3) and Q(4, 5)?

A. 1 B. 2 C. 3 D. 4

2) Find the distance between two points C(-1, -2) and D(3, -4).

A. 3 B. 4 C. 5 D. 6

3) Verify that the points E(1, -2), F(2, -5) and G(-1, -8) are collinear or not.

A. Collinear B. Not collinear

4) Are the points H(-5, -6), I(-7, -8) and J(-9, -10) form an isosceles triangle?

A. Yes B. No

5) What is the distance between K(0, 0) and L(0, 5)?

A. 1 B. 2 C. 3 D. 4

6) Find the distance between two points M(-2, -3) and N(1, -6).

A. 3 B. 4 C. 5 D. 6

7) Verify that the points O(0, 0), P(1, √3) and Q(2√3, √3 + 1) are collinear or not.

A. Collinear B. Not collinear

8) Are the points R(-1, -2), S(0, -4) and T(1, -6) form an isosceles triangle?

A. Yes B. No

9) What is the distance between U(-4, -3) and V(-6, -7)?

A. √20 B. √24 C. √28 D. √32

10) Find the distance between two points W(0, -2) and X(-4, -6).

A. √20 B. √24 C. √28 D. √32

11) Verify that the points Y(-2,-1), Z(0,-2), and A(2,-3) are collinear or not.

A.Collinear B.Not collinear

12) Are the points B(-5,-6), C(-7,-8), and D(-9,-10) form an isosceles triangle?

A.Yes B.No

**Here are the correct answers to the above questions:**

- B
- C
- Not collinear
- No
- B
- A
- Not collinear
- Yes
- √20
- √20
- Collinear
- No

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