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## Topics Covered in Distance and Section Formulae

- Distance Formula
- Distance Properties in some Geometrical Figures
- Conditions of Collinearity of Three Points
- Problems on Distance Formula
- Distance of a Point from the Origin
- Distance Formula in Geometry
- Section Formula
- Midpoint Formula
- Centroid of a Triangle
- Worksheet on Distance Formula
- Worksheet on Collinearity of Three Points
- Worksheet on Finding the Centroid of a Triangle
- Worksheet on Section Formula

### Distance Formula

The distance formula is used to calculate the distance between the points on the two-dimensional and three-dimensional coordinate plane. You can find the distance between the points by substituting the distance formula.

d = √(x2 – x1)² + (y2 – y1)²

where,

d = distance between the points

(x1, y1) and (x2, y2) are the ordered pairs of the two coordinate planes.

**Distance between three points formula:**

The distance formula for two points is used to find the distance between two points A(x1, y1, z1) and B(x2, y2, z3).

d = √(x2 – x1)² + (y2 – y1)² + (z2 – z1)²

where,

d = distance between the points

(x1, y1, z1) and (x2, y2, z2) are the ordered pairs of the two coordinate planes.

### Section Formula

The Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. This formula is used to find the centroid, incenter, and excenters of a triangle.

Do Check: Co-ordinate Geometry

### Distance and Section Formulae Solved Problems

**Example 1.**

Given a triangle ABC in which A = (4,-4) B = (0,5) and C(5,6) A point P lies on BC such that BP : PC = 3:2 find the length of line segment AP.

**Solution:**

Given BP : PC = 3 : 2

Using the section formula

The coordinates of the point p(x,y) divide the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio m1: m2.

[m1x2 + m2x1/m2 + m1 , m1y2 + m2y1/m2 + m1]

m1 = 3

m2 = 2

x1 = 0

x2 = 5

y1 = 5

y2 = 10

[3×5 + 2×0/3 + 2, 3×10 + 2×5/3 + 2]

[15/2 , 40/5]

(3,8)

Using distance formula we have

AP = √(3 – 4)² + (8 + 4)²

√1 + 144

√145

= 12.04

**Example 2.**

A(10,0) and B(5,15) are two fixed points. Find the coordinates of the point P is AB such that 5PB = AB also find the coordinates of some other points Q in AB such that AB = 6AQ.

**Solution:**

Given that,

5PB = AB

AB/PB = 5/1

AB – PB/PB = 5 – 1/1

AB/PB = 4/1

Using section formula coordinates of P are

P(x,y) = P(4×10 + 1×5/4 + 1 , 4×15 + 1 × 0/4 + 1

(40 + 5/5 , 60/5)

(9,12)

Given AB = 6AQ

AQ/AB = 1/6

AQ/AB – AQ = 1/6 – 1

AQ/QB = 1/5

Using section formula coordinates of Q are

Q(x,y) = Q(1×10 + 5×5/1 + 5, 1×15 + 5×0/1 + 5

Q(x,y) = Q(10 + 25/6, 15/6)

Q(35/6, 15/6)

**Example 3.**

Coordinates of the points which Divide the join of (1,7) and (4,3) in the ratio of 2:3 is.

**Solution:**

We know that the coordinates of the point dividing (x1, y1) and (x2,y2) in the ratio m:n is given by

(m1x1 + m2xx1/m1 + m2 , m1y2 + m2y1/m1 + m2)

Given that

(x1,y1) = (1,7)

(x2,y2) = (4,3)

(m1, m2) = 2 : 3

Coordinates of point of intersection of line segment is

(2×4 + 3×1/2 + 3, 2×3 + 3×7/2 + 3)

(8 + 3/5, 6 + 21/5)

(11/5, 27/5)

**Example 4.**

Using the section formula, the points A(7,5) B(9,3) and C(13,1) are collinear.

**Solution:**

If three points are collinear, then one of the points divides the line segment joining the order points in the ratio r : 1.

If the P is between the line A and B and AP/PB = r.

Distance AB = √(x1 – x2)² + (y2 – y1)²

= √(9 – 7)² + (3 – 5)²

= √(7)² + (-2)²

= √49 + 4

= √53

= 7.28

Distance BC = √(13 – 9)² + (1 – 3)²

= √(4)² + (-2)²

= √16 + 4

= √20

To find r

r = AB/BC = 7.28/4.47

The line divides in the ratio of 7.28 : 4.47

A line divides internally in the ratio m : n the point P = (mx1 + nx1/m + n, my2 + ny2/m + n)

m = 7.28, n = 4.47

x1 = 7

x2 = 13

y1 = 5

y2 = 1

By point B = 7.28×1 + 4.47×7/7.28 + 4.47

(94.64 + 31.29/11.75, 7.28 + 22.35/11.75)

The three points AB and C are collinear

(125.93/11.75, 29.63/11.27)

(10.7, 2.6)

**Example 5.**

Find the values of a such that PQ = QR, where P, Q, and R are the points whose coordinates are (1, – 1), (1, 3), and (a, 8) respectively.

**Solution:**

We know that

Distance AB = √(x1 – x2)² + (y2 – y1)²

PQ = √(1−1)²+(−1−3)²

=√0+(−4)²

=√ 0+16

= √16

QR =√ (1−a)²+(3−8)²

= √(1−a)²+(−5)²

= √(1−a)²+25

Therefore, PQ = QR

√16 =√ (1−a)²+25

16 = (1 – a)² + 25

(1 – a)² = 16 – 25

(1 – a)² = -9

1 – a = ±3

a = 1 ±3

a = (4, -2)

### FAQs on Distance and Section Formulae

**1. Which is section formula?**

Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter, and excenters of a triangle.

**2. What are the distance formula section formula and midpoint formula?**

The distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2+b2=c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse.

**3. Is the section formula and distance formula the same?**

The distance formula is used to find the distance between two defined points on a graph. The section formula gives the coordinates of a point that divides the line joining two points in a ratio, internally or externally.