Applications of Section Formula

As already mentioned, the section formula is useful in finding the coordinates of the point that divides any line segment, either externally or internally. The formula can also help find the midpoint of the line segment. We use the formula to divide any line segment in the desired ratio. It is also used in the derivation of the midpoint formula. Therefore, the formula has great importance in both coordinate geometry and physics. The application of the section formula in math is in finding the incenters, or excenters and centroid of a triangle. On the other hand, it helps find the center of mass, equilibrium points, etc. 

Types of Section Formula in Coordinate Geometry

So, the section formula in math helps us find the coordinates in two different ways. These two categories of the formula in the coordinate geometry are as follows:

1. Internal Section Formula
2. External Section Formula

The Internal Section Formula

The internal section formula involves finding the coordinates of a point when the line segment divides internally by a point in the desired ratio. 

Let AB be the line. And the point P divides it in the ratio m:n.

A=(x1,y1)  B=(x2,y2)

Then, P(x, y)

x=( n.x1+m.x2)/m+n

y=( n.y1+m.y2)/m+n

Since we can write m: n also as m/n:1, or simply k:1.

k=m/n

Thus, the formula shall be as follows:

P(x, y)

x= ( k.x1+x2)/k+1

y= ( k.y1+y2)/k+1

External Section Formula

The external section formula is the reverse of the internal section formula. The formula is useful when the line segment is divided externally. The external section formula helps find the coordinates of an external point on a line segment in a given ratio.

Suppose AB is the line segment. And P (x, y) divides the line segment in the ratio m: n.

P(x, y) 

x= ( n.x1-m.x2)/m-n

y=( n.y1-m.y2)/m-n

Midpoint Formula

The midpoint formula is a special case when a point (suppose P) divides the line segment into two halves. This midpoint divides the line internally in a 1:1 ratio.

Thus, applying the formula, we get:

P(x, y) 

x=(x1+x2)/2

y=(y1+y2)/2

Applications of Section Formula

The application of section formula in different mathematical concepts is as follows:

Ø  Calculation of Midpoint of a Line Segment

Let’s understand the calculation of the midpoint of a line segment through a question:

Question: R is the midpoint on a line segment P and Q. The coordinates of the mid-point R are (4, 6). The coordinates of the point Q of the line segment PQ are (8, 10). Find the coordinates of the other point, i.e. P?

Solution:

According to the question:

Q = (8, 10)

R = (4, 6)

Assume the coordinates of P = (m, n)

Applying the Midpoint section form, we get

R = ((m + 8)/2, (n + 10)/2) = (4, 6)

Solving for m,

(m + 8)/2 = 4

m + 8 = 8

m = 0

Solving for n,

(n + 10)/2 = 6

n + 10 = 12

n = 2

Answer: Thus, the coordinates of P shall be (0, 2).

Ø  Calculating the Centroid of a Triangle

Let’s understand the calculation of the centroid of a Triangle through a question:

Question: A triangle has the following vertices (4,3) (6,5), and (5,4). Calculate the centroid of this triangle. 

Solution: 

Given: (x1,y1)= (4,3)

(x2, y2)= (6,5)

(x3, y3)= ( (5,4)

Applying the centroid formula, we get

= (4+6+5/3) , (3+5+4/3)

= 15/3 and 12/3

Or (5, 3)

Answer= The coordinates of the centroids are (5, 3)

Ø  Calculation Using External Section Formula

Let’s understand the calculation using external section formula through a question:

Question: If Z (4, 5) and X (7, – 1) are two given points. Point Y divides the ZX line segment externally in the ratio 4:3. Find the coordinates of Y.

Answer: According to the question,

Z (x1,y1) =(4,5)

X(x2,y2)= (7,-1)

The given ratio of division is 4:3, 

Thus, m=4, n=3

As the question states, we shall use the external section formula for calculating the coordinates.

P(x, y) =,)

Substituting the given values in the formula,

 =4(7)-3(4))/(4-3)],[(4(-1)-3(5)/(4-3)]}

= {(28-12)/1,(-4-15)/1} ={16,-19}

Answer: Thus, the coordinates for the point Y are (16,-19)

Miscellaneous Question Solving using Section Formula

Question 1- A line segment AB is divided by the point P (externally) in the ratio 2:3. The coordinates of A and B (−3, −1) and (−1, 0), respectively. Find the coordinates of point P.

Solution: 

According to the question,

 (x₁, y₁) =(-3, −1)

 (x₂, y₂) = (-1, 0) 

m : n = 2:3

P=(x, y)

As the question states, we shall use the external section formula.

P(x, y) = {(m ×x₂ + n ×x₁)/ (m +n), (m ×y₂ + n ×y₁)/ (m +n)} 

P(x) = (2×(-1) – 3×(-3))/(2-3)

P(y) = (2× (0) – 3×(-1))/2-3)

=> P(x, y) = ((-2+9)/ (-1), 3/ (-1)

Answer= The coordinates of P(x, y) = (-7, -3)

Question 2: The two points on line segment AB has coordinates as (– 3, 3) and (12, – 7), respectively. A point P intersects the line segment AB in the ratio 2 : 3. Find the coordinates of P.

Solution: According to the question,

A (x1, y1) = (-3, 3)

B(x₂, y₂) = (12, -7)

The co-ordinates of P will be:

x=( n.x1+m.x2)/m+n

=(3.-3+2.12)/2+3

=15/5

=3

y=( n.y1+m.y2)/m+n

=(3.3+2.-7)/2+3

=-5/5

=-1

P(x, y) = (3, -1).

Conclusion 

The section formula is useful in finding the coordinates of the point that divides any line segment, either externally or internally. The formula can also help find the midpoint of the line segment. The application of section formula in maths is finding the incenters or excenters and centroid of a triangle. On the other hand, physics also helps find the centre of mass, equilibrium points, etc. There are two categories: Internal Section formula and External Section Formula.