Section Formula
The Section Formula is a fundamental result in Coordinate Geometry that enables us to find the coordinates of a point that divides a line segment joining two given points in a specified ratio. It is one of the most important formulas in the CBSE Class 10 Mathematics curriculum (Chapter 7: Coordinate Geometry) and is extensively tested in board examinations. The formula comes in two variants: one for internal division (where the dividing point lies between the two endpoints) and one for external division (where the dividing point lies outside the segment, on its extension). The internal division formula is the primary focus of Class 10 and is derived using the properties of similar triangles. Understanding the Section Formula unlocks the ability to solve a wide range of geometric problems: finding the coordinates of centroids, incentres, and other special points of triangles; determining the ratio in which a given point divides a line segment; finding the point that divides a median or diagonal at a specific ratio; and solving problems involving trisection of line segments. The midpoint formula, one of the most frequently used results in coordinate geometry, is a special case of the Section Formula where the ratio is 1:1. Beyond academic mathematics, the Section Formula has applications in computer graphics (interpolation between points), navigation (finding intermediate waypoints), and physics (centre of mass calculations). This comprehensive guide covers the derivation, both forms of the formula, and an extensive set of solved examples.
What is Section Formula - Internal Division, External Division, Derivation & Examples?
Section Formula (Internal Division): If a point P(x, y) divides the line segment joining A(x_1, y_1) and B(x_2, y_2) internally in the ratio m:n, then:
x = (mx_2 + nx_1) / (m + n)
y = (my_2 + ny_1) / (m + n)
Key Understanding: The point P divides AB internally in the ratio m:n means that AP:PB = m:n, and P lies between A and B on the segment AB.
Section Formula (External Division): If P divides AB externally in the ratio m:n (P lies on the extension of AB beyond B), then:
x = (mx_2 - nx_1) / (m - n)
y = (my_2 - ny_1) / (m - n) [where m is not equal to n]
Midpoint Formula (Special Case): When m:n = 1:1, the point P is the midpoint of AB:
x = (x_1 + x_2) / 2, y = (y_1 + y_2) / 2
Important Observations:
- In the internal division formula, the coordinates of P are a weighted average of the coordinates of A and B, with weights n and m respectively (note: weight m is applied to B's coordinates, and weight n to A's).
- The formula can be remembered as: "the ratio arm towards B multiplied by B's coordinates, plus the ratio arm towards A multiplied by A's coordinates, all divided by the sum of the ratio arms."
- If P divides AB in the ratio k:1, the formula simplifies to: x = (kx_2 + x_1)/(k + 1), y = (ky_2 + y_1)/(k + 1).
- The external division formula is obtained by replacing n with -n in the internal division formula.
Section Formula Formula
Section Formula:
Internal Division (m:n)
P = ((mx_2 + nx_1)/(m+n), (my_2 + ny_1)/(m+n))
External Division (m:n)
P = ((mx_2 - nx_1)/(m-n), (my_2 - ny_1)/(m-n))
Midpoint Formula (m:n = 1:1)
M = ((x_1 + x_2)/2, (y_1 + y_2)/2)
Finding the Ratio of Division: If a point P(x, y) lies on the segment AB and we need to find the ratio AP:PB, let it be k:1. Then:
x = (kx_2 + x_1)/(k + 1) implies k = (x - x_1)/(x_2 - x) [if x_2 is not equal to x]
Centroid of a Triangle:
The centroid G of a triangle with vertices A(x_1, y_1), B(x_2, y_2), and C(x_3, y_3) is:
G = ((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)
The centroid divides each median in the ratio 2:1 from the vertex.
Derivation and Proof
Derivation of the Section Formula (Internal Division):
Given: Points A(x_1, y_1) and B(x_2, y_2). Point P(x, y) divides AB internally in the ratio m:n.
To Find: The coordinates (x, y) of P in terms of x_1, y_1, x_2, y_2, m, n.
Construction: Draw perpendiculars AL, PM, and BN from A, P, and B respectively to the x-axis. Draw AK perpendicular to PM and PJ perpendicular to BN.
Step 1: From the construction: AK is a horizontal segment from A to the vertical through P. So AK = x - x_1 (horizontal distance). Similarly, PK = y - y_1 (vertical distance from A level to P).
Step 2: Also: PJ = x_2 - x (horizontal from P to B's vertical). BJ = y_2 - y.
Step 3: Triangles APK and PBJ are similar (AA criterion: angle AKP = angle PJB = 90 degrees, and angle APK = angle PBJ as corresponding angles since AK || PJ).
Step 4: From the similarity: AP/PB = AK/PJ = PK/BJ.
Since AP/PB = m/n (given):
m/n = (x - x_1)/(x_2 - x) ... (i)
Step 5: From (i): m(x_2 - x) = n(x - x_1).
mx_2 - mx = nx - nx_1
mx_2 + nx_1 = mx + nx = x(m + n)
x = (mx_2 + nx_1)/(m + n)
Step 6: Similarly, from m/n = PK/BJ = (y - y_1)/(y_2 - y):
m(y_2 - y) = n(y - y_1)
my_2 + ny_1 = y(m + n)
y = (my_2 + ny_1)/(m + n)
Conclusion: The coordinates of P are ((mx_2 + nx_1)/(m+n), (my_2 + ny_1)/(m+n)). QED.
Derivation of the Midpoint Formula:
Setting m = n = 1 in the Section Formula: x = (1.x_2 + 1.x_1)/(1+1) = (x_1 + x_2)/2. Similarly, y = (y_1 + y_2)/2. This is the midpoint formula.
Types and Properties
Problems on the Section Formula fall into several categories:
Type 1: Finding the Point of Division
Given two endpoints and a ratio, find the coordinates of the dividing point. Direct application of the formula.
Type 2: Finding the Ratio of Division
Given two endpoints and a point on the line segment, find the ratio in which the point divides the segment. Set up the formula with ratio k:1 and solve for k.
Type 3: Trisection Points
Find the two points that trisect a line segment (divide it into three equal parts). The ratios are 1:2 and 2:1.
Type 4: Finding the Centroid
Given the vertices of a triangle, find the centroid using the formula G = ((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3).
Type 5: Point on an Axis Dividing a Segment
Find the ratio in which the x-axis (y = 0) or y-axis (x = 0) divides a given line segment. Set the appropriate coordinate to 0 and solve.
Type 6: Finding Unknown Coordinates
Given the dividing point, the ratio, and one endpoint, find the other endpoint.
Type 7: Combined Problems
Problems that combine the Section Formula with the Distance Formula, collinearity conditions, or area formulas.
Methods
Method 1: Direct Application
Given A(x_1, y_1), B(x_2, y_2), and ratio m:n, substitute directly.
Example: A(2, 3), B(8, 9), ratio 2:1. P = ((2x8 + 1x2)/(2+1), (2x9 + 1x3)/(2+1)) = (18/3, 21/3) = (6, 7).
Method 2: Finding Ratio (k:1 method)
Let the ratio be k:1. Set up x = (kx_2 + x_1)/(k+1) or y = (ky_2 + y_1)/(k+1). Substitute the known coordinates of the dividing point and solve for k.
Example: A(1, 2), B(4, 5), P(3, 4). 3 = (4k + 1)/(k+1). 3k + 3 = 4k + 1. k = 2. Ratio is 2:1.
Method 3: Trisection
For trisection of AB, the two points are: P1 (ratio 1:2 from A) and P2 (ratio 2:1 from A).
P1 = ((1.x_2 + 2.x_1)/3, (1.y_2 + 2.y_1)/3). P2 = ((2.x_2 + 1.x_1)/3, (2.y_2 + 1.y_1)/3).
Method 4: Axis Division
To find where the x-axis divides AB: set y = 0 in the formula. (my_2 + ny_1)/(m+n) = 0, so my_2 + ny_1 = 0, giving m/n = -y_1/y_2.
Example: A(2, -3), B(5, 6). x-axis divides AB in ratio m:n where m/n = -(-3)/6 = 1/2. So ratio = 1:2.
Method 5: Finding an Unknown Endpoint
If M is the midpoint of AB, and A and M are known: B = (2x_M - x_A, 2y_M - y_A).
If P divides AB in ratio m:n, and A and P are known: use the formula to express B in terms of the other quantities.
Tips:
- In the formula, m is the ratio arm closer to B, and n is closer to A. Think: "m times B's coordinate + n times A's coordinate."
- Always verify by checking that P lies between A and B (for internal division) or beyond (for external).
- For negative ratio values, the division is external.
- The centroid formula is just the average of all three vertices' coordinates.
Solved Examples
Example 1: Finding a Point Dividing a Segment Internally
Problem: Find the coordinates of the point that divides the line segment joining A(1, -2) and B(4, 7) in the ratio 2:1 internally.
Solution:
Step 1: Using the Section Formula with m = 2, n = 1: x = (2 x 4 + 1 x 1)/(2 + 1) = (8 + 1)/3 = 9/3 = 3.
Step 2: y = (2 x 7 + 1 x (-2))/(2 + 1) = (14 - 2)/3 = 12/3 = 4.
Answer: The point is (3, 4).
Example 2: Finding the Ratio of Division
Problem: Find the ratio in which the point (3, 5) divides the line segment joining A(1, 3) and B(7, 9).
Solution:
Step 1: Let the ratio be k:1. Using the x-coordinate: 3 = (7k + 1)/(k + 1).
Step 2: 3(k + 1) = 7k + 1. 3k + 3 = 7k + 1. 2 = 4k. k = 1/2.
Step 3: The ratio is (1/2):1 = 1:2.
Verification: Using y: y = (9 x (1/2) + 3)/(1/2 + 1) = (4.5 + 3)/1.5 = 7.5/1.5 = 5. Correct!
Answer: The ratio is 1:2.
Example 3: Trisection Points
Problem: Find the coordinates of the points of trisection of the line segment joining A(-2, -3) and B(4, 9).
Solution:
Step 1: Point P1 divides AB in ratio 1:2 (one-third from A). x = (1 x 4 + 2 x (-2))/(1+2) = (4 - 4)/3 = 0. y = (1 x 9 + 2 x (-3))/(1+2) = (9 - 6)/3 = 1. P1 = (0, 1).
Step 2: Point P2 divides AB in ratio 2:1 (two-thirds from A). x = (2 x 4 + 1 x (-2))/(2+1) = (8 - 2)/3 = 2. y = (2 x 9 + 1 x (-3))/(2+1) = (18 - 3)/3 = 5. P2 = (2, 5).
Answer: The trisection points are (0, 1) and (2, 5).
Example 4: Finding the Centroid
Problem: Find the centroid of the triangle with vertices A(1, 1), B(5, 3), and C(3, 7).
Solution:
Step 1: G = ((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3) = ((1 + 5 + 3)/3, (1 + 3 + 7)/3) = (9/3, 11/3) = (3, 11/3).
Answer: The centroid is (3, 11/3).
Example 5: Ratio of Division by the X-Axis
Problem: In what ratio does the x-axis divide the line segment joining A(2, -3) and B(5, 6)?
Solution:
Step 1: On the x-axis, y = 0. Let the ratio be k:1.
Step 2: y = (k x 6 + 1 x (-3))/(k + 1) = 0.
Step 3: 6k - 3 = 0. k = 1/2.
Step 4: The ratio is (1/2):1 = 1:2.
Verification: x = ((1/2)(5) + 2)/((1/2) + 1) = (2.5 + 2)/1.5 = 4.5/1.5 = 3. The point is (3, 0), which lies on the x-axis. Correct.
Answer: The x-axis divides AB in the ratio 1:2.
Example 6: Finding an Unknown Vertex of a Triangle
Problem: The centroid of a triangle is G(2, 3). Two vertices are A(1, 1) and B(3, 5). Find the third vertex C.
Solution:
Step 1: Let C = (x, y). G = ((1 + 3 + x)/3, (1 + 5 + y)/3) = (2, 3).
Step 2: (4 + x)/3 = 2, so 4 + x = 6, x = 2.
Step 3: (6 + y)/3 = 3, so 6 + y = 9, y = 3.
Answer: The third vertex is C(2, 3). (Interestingly, it coincides with the centroid, which happens when A, B, C form a specific configuration.)
Example 7: Proving a Point Lies on a Line Segment
Problem: Show that the point P(5, -2) lies on the line segment joining A(1, 2) and B(7, -4), and find the ratio AP:PB.
Solution:
Step 1: Let P divide AB in ratio k:1. x-coordinate: 5 = (7k + 1)/(k + 1). 5k + 5 = 7k + 1. 4 = 2k. k = 2.
Step 2: Verify with y: y = ((-4)(2) + 2)/(2 + 1) = (-8 + 2)/3 = -6/3 = -2. Correct!
Step 3: Since k = 2 > 0, P divides AB internally. The ratio AP:PB = 2:1.
Answer: P lies on AB and divides it in the ratio 2:1.
Example 8: Ratio of Division by the Y-Axis
Problem: Find the ratio in which the y-axis divides the segment joining A(-6, 4) and B(8, -2).
Solution:
Step 1: On the y-axis, x = 0. Let the ratio be k:1.
Step 2: 0 = (8k + (-6))/(k + 1). So 8k - 6 = 0. k = 3/4.
Step 3: Ratio = 3/4 : 1 = 3:4.
Verification: y = ((-2)(3/4) + 4)/(3/4 + 1) = (-1.5 + 4)/1.75 = 2.5/1.75 = 10/7. The point of intersection is (0, 10/7).
Answer: The y-axis divides AB in the ratio 3:4.
Example 9: Section Formula with Negative Coordinates
Problem: Find the point that divides the segment from A(-4, -5) to B(6, 10) in the ratio 3:2 internally.
Solution:
Step 1: x = (3 x 6 + 2 x (-4))/(3 + 2) = (18 - 8)/5 = 10/5 = 2.
Step 2: y = (3 x 10 + 2 x (-5))/(3 + 2) = (30 - 10)/5 = 20/5 = 4.
Answer: The point is (2, 4).
Example 10: Application: Median Length
Problem: Find the length of the median from vertex A(2, 5) to side BC of the triangle with B(-1, 3) and C(5, 1).
Solution:
Step 1: The median from A goes to the midpoint M of BC. M = ((-1 + 5)/2, (3 + 1)/2) = (2, 2).
Step 2: Length of median AM = sqrt[(2 - 2)^2 + (5 - 2)^2] = sqrt[0 + 9] = 3 units.
Answer: The length of the median is 3 units.
Real-World Applications
The Section Formula has significant applications across mathematics, science, and technology.
Centre of Mass in Physics: The Section Formula is essentially the same as the centre of mass formula for two particles. If two particles of masses m and n are placed at points A and B, their centre of mass divides AB in the ratio m:n from A towards B. The Section Formula gives the exact position of this centre of mass. For three or more particles, the centroid formula (which is derived from the Section Formula) gives the centre of mass.
Computer Graphics: Linear interpolation between two points is a direct application of the Section Formula. When animating objects, rendering gradients, or blending colours, the coordinates (or colour values) are interpolated using the Section Formula with varying ratios. Bezier curves, used extensively in vector graphics and font design, are built from repeated application of the Section Formula.
Navigation and Waypoints: Pilots and sailors calculate intermediate waypoints along a route using the Section Formula. If the starting and ending coordinates are known, waypoints at specific fractions of the journey are found by dividing the segment in the appropriate ratio. GPS navigation uses this principle for estimated position between known reference points.
Surveying and Land Division: When a plot of land needs to be divided in a specific ratio along a boundary, the Section Formula provides the exact coordinates of the division point. This is essential for fair land division, property demarcation, and urban planning.
Data Visualisation: In data plotting and chart rendering, the Section Formula is used to position tick marks at regular intervals along axes, to place labels at specific fractions of a scale, and to compute interpolated data points for smooth curve rendering.
Structural Engineering: In truss analysis and beam calculations, the positions of load-bearing points that divide structural members in specific ratios are computed using the Section Formula. This ensures accurate load distribution and structural integrity.
Key Points to Remember
- The Section Formula gives the coordinates of the point dividing a segment joining A(x_1,y_1) and B(x_2,y_2) in ratio m:n internally: P = ((mx_2+nx_1)/(m+n), (my_2+ny_1)/(m+n)).
- The midpoint formula is a special case with m:n = 1:1: M = ((x_1+x_2)/2, (y_1+y_2)/2).
- For external division, replace n with -n: P = ((mx_2-nx_1)/(m-n), (my_2-ny_1)/(m-n)).
- To find the ratio in which a point divides a segment, use the k:1 method and solve for k.
- Trisection points divide a segment in ratios 1:2 and 2:1 from one endpoint.
- The centroid of a triangle = ((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3) and divides each median in ratio 2:1 from the vertex.
- To find where an axis divides a segment: set x=0 (for y-axis) or y=0 (for x-axis) and solve for the ratio.
- A negative value of k in the k:1 method indicates external division.
- In the formula, m multiplies B's coordinates and n multiplies A's coordinates. Remember: "m for the far point, n for the near point" (from A's perspective).
- The Section Formula is widely tested in CBSE board exams (2-4 marks) in problems involving dividing points, centroids, axis intersections, and finding unknown vertices.
Practice Problems
- Find the coordinates of the point dividing the segment from A(-3, 4) to B(7, -6) in the ratio 3:2.
- In what ratio does the point (5, 6) divide the segment joining (3, 2) and (9, 14)?
- Find the trisection points of the segment joining (-1, 4) and (8, -2).
- Find the centroid of the triangle with vertices (-2, 3), (4, -1), and (7, 5).
- In what ratio does the y-axis divide the segment joining (-5, 3) and (10, -6)?
- The midpoint of AB is (4, 5). If A is (2, 3), find B.
- If the centroid of a triangle with vertices (a, 1), (2, b), and (3, 4) is (2, 3), find a and b.
- Prove that the medians of a triangle with vertices (0, 0), (6, 0), and (0, 8) are concurrent and find their point of intersection.
Frequently Asked Questions
Q1. What is the Section Formula?
The Section Formula gives the coordinates of a point that divides a line segment joining two given points in a specified ratio. If P divides the segment from A(x_1,y_1) to B(x_2,y_2) internally in ratio m:n, then P = ((mx_2+nx_1)/(m+n), (my_2+ny_1)/(m+n)).
Q2. How is the midpoint formula related to the Section Formula?
The midpoint formula is the Section Formula with ratio 1:1. Setting m=1 and n=1 gives P = ((x_2+x_1)/2, (y_2+y_1)/2), which is the midpoint of the segment.
Q3. What does internal vs external division mean?
Internal division means the point P lies between A and B on the segment. External division means P lies on the extension of the segment beyond one of the endpoints. For external division, the formula uses subtraction instead of addition in both numerator and denominator.
Q4. How do you find the ratio in which a given point divides a segment?
Use the k:1 method: let the ratio be k:1, substitute the known point's coordinates into the Section Formula, and solve for k. If k is positive, the division is internal. If k is negative, the division is external.
Q5. What is the centroid formula and how is it derived?
The centroid G of a triangle with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) is G = ((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3). It is derived by first finding the midpoint of one side, then using the Section Formula to find the point that divides the median from the opposite vertex to this midpoint in ratio 2:1.
Q6. In the formula, which coordinate gets multiplied by m and which by n?
In the internal division formula for ratio m:n from A to B: the coordinate of B (the point further from A in the ratio) gets multiplied by m, and the coordinate of A gets multiplied by n. Think of it as: 'm times far + n times near, divided by m+n.'
Q7. How do you find where an axis divides a segment?
For the x-axis: set y=0 in the formula and solve for the ratio. For the y-axis: set x=0 and solve. The y-coordinate of a point on the x-axis is 0, and the x-coordinate of a point on the y-axis is 0.
Q8. Can the Section Formula be used to prove collinearity?
Yes. If three points A, B, C are such that B divides AC in some ratio k:1 (where k is a real number), then A, B, C are collinear. The Section Formula provides the coordinates, and if they match the given point B, collinearity is proved.
Related Topics
- Midpoint Formula
- Distance Formula
- Area of Triangle Using Coordinates
- External Division (Section Formula)
- Introduction to Coordinate Geometry
- Cartesian Plane
- Plotting Points in Four Quadrants
- Abscissa and Ordinate
- Centroid of a Triangle
- Collinearity Using Distance Formula
- Coordinate Geometry Word Problems
- Equation of a Line (Introduction)
- Quadrilateral Using Coordinate Geometry
