Section Formula

Section Formula is a handy tool in both Math and Physics for pinpointing the coordinates of a point that splits a line segment, whether it's an internal or external division. In Math, it's applied to discover essential points like the centroid, incenters, or excenters of a triangle. Meanwhile, in Physics, it comes in handy for determining the center of mass and equilibrium points. This formula is also frequently used to identify the midpoint of a line segment. This article delve into the section formula and understand how it's derived.

What is Section Formula?

A line segment can be divided into two parts by placing a point on the line in between the two extreme points of the line segment or maybe somewhere externally. In coordinate geometry, section formulas may be used to find a relation between the coordinates of the point that divides the line segment into two parts, and the ratio in which the line segment is divided. If we are provided with the coordinates of the point that divides the line segment, we can use the section formula in Mathematics to determine the ratio in which the line segment is divided by the point.

Alternatively, we can figure out the coordinate of the point that divides the line segment in a particular ratio. The section formula may be used to determine the incentre, excentre, and centroid of a triangle. Even in Physics, it is sometimes used to find the centre of mass (COM) of a rigid system of particles.

The section formula is of two types depending upon the location of the point dividing them into two parts:

1. Section formula for internal division (used when the point divides the line segment internally).

2. Section formula for external division (used when the point divides the line segment externally). 

Definition of Section Formula in Coordinate Geometry

When you have a line segment with two points, A and B, and there's another point, P, on that line segment, the formula to figure out the coordinates of P is called the section formula. Let's say the coordinates of A are $(x_1, y_1)$ and B are $(x_2, y_2)$, and the coordinates of P are (x, y). We can use the section formula to find the coordinates of P, and it's expressed mathematically as:

$P(x, y) = \left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$

In coordinate geometry, we use the section formula to divide a line segment. This division is defined by two numbers, m and n, representing the ratio m:n.

Internal Section Formula

The internal section formula is used when the point divides the line segment internally, that is, the point lies between the two extreme points of the line segment.

Let AB be a line segment where A(x₁, y₁) and B(x₂, y₂).

Let P(x, y) be the point that divides the line segment in the m:n ratio internally.

By using section formula internally,

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\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m+n}, \frac{my_{2} + ny_{1}}{m+n})\]

\[x = (\frac{mx_{2} + nx_{1}}{m+n})\]

\[y = (\frac{my_{2} + ny_{1}}{m+n})\]

Hence, we get the coordinates as P(x, y) which divide the line in the  m:n ratio.

Sometimes, m:n is taken to be k:1. In such a case, the formula stands as:

\[P(x, y) = P(\frac{kx_{2} + x_{1}}{k+1}, \frac{ky_{2} + y_{1}}{k+1})\]

\[x = (\frac{kx_{2} + x_{1}}{k+1})\]

\[y = (\frac{ky_{2} + y_{1}}{k + 1})\]

External Section Formula

The External section formula is used when the point divides the line segment externally, that is, the point lies outside the two extreme points of the line segment.

Let AB be a line segment where A(x₁, y₁) and B(x₂, y₂).

Let P(x, y) be the point which divides the line segment in the m:n ratio externally.

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\[P(x, y) = P(\frac{mx_{2} - nx_{1}}{m - n} , \frac{my_{2} - ny_{1}}{m - n})\]

\[x = (\frac{m_{2} - nx_{1}}{m - n})\]

\[y = (\frac{my_{2} - ny_{1}}{m - n})\]

The section formula is not only valid in two dimensions but also for coordinate geometry in three dimensions. 

Let AB be a line segment with A(x₁, y₁, z₁) and B(x₂, y₂, z₂).

Let P(x, y, z) be the point that divides the line segment in the m:n ratio internally.

Hence section formula externally: \[P(x, y, z) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n}, \frac{mz_{2} + nz_{1}}{m + n})\]

\[x = (\frac{mx_{2} + nx_{1}}{m + n})\]

\[y = (\frac{my_{2} + ny_{1}}{m + n})\]

\[z = (\frac{mz_{2} + nz_{1}}{m + n})\]

Section Formula Derivation

To understand this formula, we'll use two right-angled triangles. The hypotenuses of these triangles are in a specific ratio, m:n. Let's prove the section formula with a simple method. Imagine a point P (x, y) on a coordinate plane that divides the line AB internally. Extend horizontal and vertical lines from three points to create two right-angled triangles, AQP and PRB, as shown in the figure.

Section formula proof:

In the two right triangles AQP and PRB,

∠PAQ = ∠BPR (corresponding angles)

∠PQA = ∠ BRP = 90°

we can see that both triangles are similar by AA similarity.

So we can write:

$\dfrac{AP}{PB} = \dfrac{AQ}{PR} = \dfrac{PQ}{BR} = \dfrac{m}{n}$ ...(1)

Now using coordinates,

$AQ = x - x_1$ ...(2)

$PR = x_2 - x$ ...(3)

From equation (1), (2), and (3),

$\dfrac{[x - x_1]}{[x_2 - x]} = \dfrac{m}{n}$

On solving for x, we get

$x = \left(\dfrac{mx_2+nx_1}{m+n}\right)$ ..(A)

Similarly for y, we have,

$PQ = y - y_1$ ...(4)

$BR = y_2 - y$ ...(5)

From equation (1), (4), and (5),

$y = \left(\dfrac{my_2+ny_1}{m+n}\right)$...(B)

Hence, from equation (A), and (B), we get,

$P(x, y) = \left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$ which is internal section formula

For the external section formula we can write the above equation as:

$P(x, y) = \left(\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}\right)$

Special Case: Midpoint Formula

When the point P(x, y) divides the line segment into two halves, we may say that P(x, y) is the midpoint of the line segment. By the use of section and midpoint formula:

m : n = 1 : 1, since the line is divided into equal parts, we have.

\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[P(x, y) = P(\frac{(1)x_{2} + (1)x_{1}}{1 + 1}, \frac{(1)y_{2} + (1)y_{1}}{1 + 1})\]

\[P(x, y) = P(\frac{x_{2} + x_{1}}{2}, \frac{y_{2} + y_{1}}{2})\]

∴ \[x = (\frac{x_{2} + x_{1}}{2}), y = (\frac{y_{2} + y_{1}}{2})\]

Solved Examples

Q1. Find the coordinates of a point M which divides a line segment PQ in the ratio 2:3. PQ line segment joins the points P (2,1) and Q (-3,6). Does the point M lie on the line 5y - x = 15?

Sol. Given,

P (2, 1), Q (-3, 6), 

Let the x coordinate of M be x and y coordinate be y: M (x,y)

m : n = 2 : 3

By section formula,

\[M(x, y) = M(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[M(x, y) = M(\frac{2(-3) + 3(2)}{2 + 3}, \frac{2(6) + 3(1)}{2 + 3})\]

\[M(x, y) = M(\frac{-6+6}{5}, \frac{15}{5})\]

M(x, y) = M(0, 3)

x = 0, y = 3

Hence the point M is (0,3).

Putting the values of x and y in 5y - x = 15,

5(3) - (0) - 15 = 0

Hence, point M lies on the line 5y - x = 15.

Q2. Find the midpoint of the line segment AB which joins A (4,8) and (2,4).

Sol. Given,

A (4, 8), B (2, 4)

Let the x coordinate of midpoint be x and y coordinate y: P (x,y).

m : n = 1 : 1

By the use of section and midpoint formula,

\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[P(x, y) = P(\frac{(1)x_{2} + (1)x_{1}}{1 + 1}, \frac{(1)y_{2} + (1)y_{1}}{1 + 1})\]

\[P(x, y) = P(\frac{x_{1} + x_{2}}{2}, \frac{y_{2} + y_{1}}{2})\]

\[P(x, y) = P(\frac{4 + 2}{2}, \frac{8 + 4}{2})\]

P(x, y) = P(3, 6)

∴ x = 3, y = 6

Hence, the midpoint of AB is P (3,6).

So now we have learned about the Section Formula and its applications through solved examples. To learn more about section formulas, you must solve more problems related to them. 

Practice Questions:

1. Find the coordinates of a point C which will divide the line segment AB internally in the ratio of 1: 2, where the coordinates of point A and B are (−3,1), and (3,-6) respectively.

Ans: (−1, −4/3​)

2. The Section formula is used to find the coordinates of the point that divide a line segment (externally or internally) into some ratio.

Ans: True

Conclusion

The Section Formula is a mathematical tool used to find the coordinates of a point that divides a line segment into two parts. Vedantu explains in detail about two variations: Section formula for external division and internal division. The external section formula helps find the coordinates when the point lies outside the line segment, while the internal section formula deals with points inside the segment. The Section formula derivation and proof are crucial for understanding its application. Whether externally or internally, this formula serves as a valuable tool for calculating coordinates and understanding the division of line segments in geometry.