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Section Formula
The Section formula is used to find the coordinates of the point that divides a line segment (externally or internally) into some ratio. This formula is commonly used in Mathematics and Physics. In Mathematics, it is used to find the centroid, incenters, or excenters of a triangle, while in Physics, it is used to find the center of mass, equilibrium points, etc. The section formula is widely used to find the midpoint of a line segment. Let us learn more about the section formula and its derivation on this page.
1.  What is Section Formula? 
2.  Internal Section Formula 
3.  External Section Formula 
4.  Section Formula Derivation 
5.  Section Formula for Midpoint 
6.  FAQs on Section Formula 
What is Section Formula?
When a point divides a line segment externally or internally in some ratio, we use the section formula to find the coordinates of that point. It is a handy tool used to find the coordinates of the point dividing the line segment in some ratio. This section formula can also be used to find the midpoint of a line segment and for the derivation of the midpoint formula as well.
Definition of Section Formula in Coordinate Geometry
When a point on a line segment divides it into two segments, the formula used to determine the coordinates of that point is known as the section formula. Let us say, we have a point P(x,y) that divides the line segment with marked points as A (x_{1},y_{1}) and B(x_{2},y_{2}). To find the coordinates, we use the section formula, which is mathematically expressed as:
P(x, y) = \( \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) \)
Note: Here, m and n define the division of the line segment into a ratio m:n
Section formula in coordinate geometry is mainly divided into subformulas, which are:
 Internal section formula
 External section formula
Let us learn about them one by one in the following section.
Internal Section Formula
The internal section formula is used when the line segment is divided internally by a point in a particular ratio. This formula is used to find the coordinates of a point, which falls between the two points and on the line joining these two points, in the given ratio. Observe the figure given below which shows that a line segment AB is divided by a point P(x, y) internally in a ratio such that AP: PB = m:n
Now, the coordinates of the point P in terms of the given ratio and for the given coordinates of the points A(x_{1}, y_{1}), B(x_{2}, y_{2}) is expressed as:
P(x, y) = \( \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) \)
where,
 x and y are the coordinates of point P.
 (x_{1}, y_{1}) are the coordinates of point A.
 (x_{2}, y_{2}) are the coordinates of the point B.
 m and n are the ratio values in which P divides the line internally.
External Section Formula
The external section formula is used when the line segment is divided externally by the point in the given ratio. This formula is used to find the coordinates of the point on the line segment joining the two points and falling beyond the two points, in the given ratio. Observe the following figure which shows that the line segment AB is divided by a point P(x, y) externally in a ratio such that AP: PB = m:n
Now, the coordinates of the point P in terms of the given ratio and for the given coordinates of the points A(x_{1}, y_{1}), B(x_{2}, y_{2}) is expressed as:
P(x, y) = \(\left( {\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}} \right)\)
where,
 x, and y are the coordinates of point P.
 (x_{1}, y_{1}) are the coordinates of point A.
 (x_{2}, y_{2}) are the coordinates of the point B.
 m and n are the ratio values in which P divides the line externally.
Section Formula Derivation
This formula can be proved by using two similar rightangled triangles. The hypotenuse of these two right angles is in the given ratio m: n. Let us prove the section formula using simple construction. Consider the point P (x, y) on the coordinate plane, which divides the line segment AB internally. Extend the horizontal lines and vertical lines from the three given points to form two rightangled triangles AQP and PRB, as shown in the following figure.
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:
AP/PB = AQ/PR = PQ/BR = m/n ...(1)
Now using coordinates,
AQ = x  x_{1} ...(2)
PR = x_{2}  x ...(3)
From equation (1), (2), and (3),
[x  x_{1}] / [x_{2}  x] = m / n
On solving for x, we get
x = \(\dfrac{{m{x_2} + n{x_1}}}{{m + n}}\) ..(A)
Similarly for y, we have,
PQ = y  y_{1} ...(4)
BR = y_{2}  y ...(5)
From equation (1), (4), and (5),
y = \(\dfrac{{m{y_2} + n{y_1}}}{{m + n}}\) ...(B)
Hence, from equation (A), and (B), we get,
P(x, y) = \( \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) \)
We can do the same process for the point dividing the line segment externally and we will get the following result:
P(x, y) = \( \left( {\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}} \right)\)
Section Formula for Midpoint
The midpoint formula is used to find the coordinates of the midpoint of a line segment. Here, the ratio between the two parts is 1:1. A midpoint refers to a point that is exactly in the middle of the line segment. From the following figure, we can observe that point M(x, y) is the midpoint of the line segment JK.
We can derive the midpoint formula using the section formula or find the midpoint of any line segment using the following section formula:
M(x, y) = \(\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
Substituting m = 1, n = 1, we get,
M(x, y) = \(\left( \dfrac{x_2 + x_1}2,\dfrac{y_2 + y_1}{2} \right)\)
Where
 (x, y) are the coordinates of the point M
 (x_{1}, y_{1}) are the coordinates of point J
 (x_{2}, y_{2}) are the coordinates of the point K
Important Notes on Section Formula:
 Section formula for internal division is: P(x, y) = \(\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
 Section formula for external division is: P(x, y) = \(\left( {\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}} \right)\)
 Midpoint formula is: M(x, y) = \(\left( \dfrac{x_2 + x_1}2,\dfrac{y_2 + y_1}{2} \right)\)
☛Related Articles
Examples Using Section Formula

Example 1: By using the section formula, find out the coordinates of the points that divide the line segment joining the points (8, 3) and (16, 5) in the internal ratio 3:1.
Solution:
Given: Coordinates of the endpoints of the line segment: (8, 3) and (16, 5)
Internal ratio = 3:1Let P(x,y) be the required point that divides the line segment in the given ratio. After substituting the respective values in the section formula, we get
x = \(\dfrac{3(16)+1(8)}{(3+1)}\), y = \(\dfrac{3(5)+1(3)}{(3+1)}\)
x = 14, y = 3
P = (14,3)Therefore, (14, 3) is the required point.

Example 2: Find the ratio in which a point P lying on the yaxis divides the line joining points (6, 6) and (2, 4).
Solution:
Let the ratio be m : n.
Since the point P lies on the yaxis, the abscissa will be 0 and let us assume the ordinate to be 'a'.
So, according to the formula let us take (6, 6) = (x_{1}, y_{1});
(2, 4) = (x_{2}, y_{2})
and (0, a) = (x, y)
x = \(\dfrac{{m{x_2} + n{x_1}}}{{m + n}}\)
0 = \(\dfrac{{m({2}) + n({6})}}{{m + n}}\)
0 = 2m + 6n
2m = 6n
m/n = 3/1
m : n = 3 : 1

Example 3: Find the coordinates of the point C which divides the points A (3, 1), B (1, 0) in a ratio of 2:3 externally, with the help of section formula.
Solution:
Given:
(\(x_1, y_1\)) is (3, −1)
(\(x_2, y_2\)) is (1, 0)
m:n = 2:3
To find : The coordinates of the point C(x, y)
Using the section formula for external division,
C(x, y) = \(\left( {\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}} \right)\)
C(x, y) = ((2(1)  3(3))/(2  3), (2(0)  3(1))/2  3)
C(x, y) = ((2 + 9)/(1), 3/(1))
C(x,y) = (7, 3)
Answer: The coordinates of point C are (7, 3)
FAQs on Section Formula
What is the Section Formula in Coordinate Geometry?
Section formula is used to find the coordinates of a point that divides a line segment externally or internally in some ratio. This formula can also be used to find the midpoint of a line segment. For example, if a line segment AB is divided internally by point P, then the section formula for internal division is:
P(x, y) = \(\left({\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}}\right)\)
where
 x, and y are the coordinates of point P
 (\(x_1, y_1\)) are the coordinates of point A.
 (\(x_2, y_2\)) are the coordinates of the point B.
 m and n are the ratio values in which P divides the line internally.
What are the Applications of Section Formula?
Section formula is used in various places in Mathematics and Physics. In Mathematics, we can use the section formula to find the centroid, incenters, or excenters of a triangle, etc whereas in Physics it is used to find the center of mass, equilibrium points, etc. The section formula is also widely used to find the midpoint of a line segment.
What is the Section Formula for Internal Division?
If we have a line segment AB that is divided by a point P(x, y) internally in a ratio such that AP : PB = m : n, then the section formula for internal division is:
P(x, y) = \(\left({\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}}\right)\)
where,
 x and y are the coordinates of point P
 (x_{1}, y_{1}) are the cartesian coordinates of point A
 (x_{2}, y_{2}) are the coordinates of the point B
 m and n are the ratio values in which P divides the line internally
What is the Section Formula for External Division?
When a line segment JK is divided by a point P(x, y) externally in a ratio such that JP : KP = m : n, then the section formula for the external division to find the coordinates of the point P is:
P(x, y) = \(\left({\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}}\right)\)
where,
 x, and y are the coordinates of point P.
 (x_{1}, y_{1}) are the coordinates of point J.
 (x_{2}, y_{2}) are the coordinates of the point K.
 m and n are the ratio values in which P divides the line externally.
How to Derive the Section Formula?
Section formula can be derived by constructing two right triangles and by using AA similarity. We find the ratio of the length of the sides of the triangle in terms of the given ratios, and then, on solving for x, and y, we can find the coordinates of the point that divides the line segment.
What is the Section Formula of Class 10?
There are three major section formulas that we commonly read in class 10. The formulas are listed below:
Section formula (internal division): P(x, y) = \(\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
Section formula (external division): P(x, y) = \(\left( {\dfrac{{m{x_2}  n{x_1}}}{{m  n}},\dfrac{{m{y_2}  n{y_1}}}{{m  n}}} \right)\)
Midpoint section formula: M(x, y) = \(\left( \dfrac{x_2 + x_1}2,\dfrac{y_2 + y_1}{2} \right)\)
What is the Difference between Distance Formula and Section Formula?
When a point divides a line segment externally or internally in some ratio, we use the section formula to find the coordinates of the point. The section formula is expressed as:
P(x, y) = \(\left({\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}}\right)\)
Whereas, the distance formula in coordinate geometry is used to determine the distance between the given points. The distance formula which is used to calculate the distance between the respective points \((x_1, y_1)\), and \((x_2, y_2)\) is expressed as, \(D = \sqrt{(x_2 x_1)^2 + (y_2y_1)^2}\).
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