External Division (Section Formula)

Q: What is external division of a line segment?

When a point P divides line segment AB such that P lies outside the segment (on the line extended beyond A or B), with AP/PB = m/n, it is called external division in ratio m : n.

Q: What is the formula for external division?

P(x, y) = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)), where A(x₁, y₁) and B(x₂, y₂) are endpoints and m : n is the external ratio.

Q: How is external division different from internal division?

In internal division, P lies between A and B. In external division, P lies outside the segment. The formulas differ: internal uses +, external uses −.

Q: What happens when m equals n in external division?

The formula is undefined because the denominator (m − n) becomes zero. Geometrically, the external division point goes to infinity.

Q: How do you find the ratio of external division?

Assume the ratio is k : 1. Substitute the known point coordinates into the external division formula and solve for k.

Q: Can external division give negative coordinates?

Yes. Depending on the positions of A and B and the ratio, the external division point can have negative x or y coordinates.

Q: What is a harmonic conjugate?

If a point P₁ divides AB internally in ratio m : n, and P₂ divides AB externally in the same ratio m : n, then P₁ and P₂ are harmonic conjugates with respect to A and B.

Q: Is external division in CBSE Class 10 syllabus?

It is part of the coordinate geometry chapter and may appear in board exams as a higher-order question. The internal section formula is more frequently tested.

Class 10Coordinate Geometry

The section formula for external division gives the coordinates of a point that divides a line segment externally in a given ratio. Unlike internal division where the point lies between the endpoints, in external division the point lies outside the segment — on the line extended beyond one endpoint.

If a point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) externally in the ratio m : n, then P lies on the line AB produced (extended) beyond B or beyond A.

This topic extends the section formula learned earlier and appears in coordinate geometry problems in CBSE Class 10.

What is External Division?

Definition: A point P is said to divide the line segment AB externally in the ratio m : n if P lies on the line AB extended (outside the segment) such that AP/PB = m/n.

Key difference from internal division:

Internal division: P lies between A and B → AP + PB = AB.
External division: P lies outside A and B → AP − PB = AB (or PB − AP = AB).

Visual representation:

If m > n: P lies beyond B (on the side of B away from A).
If m < n: P lies beyond A (on the side of A away from B).
If m = n: External division is undefined (the point is at infinity).

External Division (Section Formula) Formula

Formula for External Division:

P(x, y) = ( (mx₂ − nx₁)/(m − n) , (my₂ − ny₁)/(m − n) )

Where:

A(x₁, y₁) and B(x₂, y₂) are the endpoints
m : n is the ratio of external division
m ≠ n (if m = n, the formula is undefined)

Comparison:

Type	Formula
Internal Division (m : n)	P = ( (mx₂ + nx₁)/(m + n) , (my₂ + ny₁)/(m + n) )
External Division (m : n)	P = ( (mx₂ − nx₁)/(m − n) , (my₂ − ny₁)/(m − n) )

Memory aid: For external division, replace all + signs in the internal formula with − signs.

Derivation and Proof

Deriving the external division formula:

Let A(x₁, y₁) and B(x₂, y₂) be the endpoints.
Let P(x, y) divide AB externally in ratio m : n, so AP/PB = m/n.
Treating the internal division formula with ratio m : (−n) (negative ratio for external division):
x = (m·x₂ + (−n)·x₁) / (m + (−n)) = (mx₂ − nx₁) / (m − n)
y = (m·y₂ + (−n)·y₁) / (m + (−n)) = (my₂ − ny₁) / (m − n)

Key insight: External division in ratio m : n is the same as internal division in ratio m : (−n). This is why the + signs become − signs.

Verification approach: After finding P, verify that:

A, B, P are collinear.
AP/PB = m/n.
P lies outside segment AB.

Types and Properties

Situations involving external division:

Type 1: Given two points and ratio — find the point of external division directly.
Type 2: Given endpoints and the external point — find the ratio of external division.
Type 3: Extending a line segment beyond one endpoint by a given length.
Type 4: Problems involving both internal and external division together.

Special cases:

If m : n = 2 : 1, P lies beyond B at distance equal to AB from B.
If m : n = 1 : 2, P lies beyond A at distance AB/2 from A.
If m = n, external division is undefined (parallel lines concept — point at infinity).

Methods

Steps to find the point of external division:

Identify the two endpoints A(x₁, y₁) and B(x₂, y₂).
Note the ratio m : n of external division.
Apply the formula: x = (mx₂ − nx₁)/(m − n), y = (my₂ − ny₁)/(m − n).
Simplify the coordinates.
Verify the result if required (check collinearity and ratio).

Finding the ratio of external division:

Let the ratio be k : 1.
Substitute into the formula: P(x, y) = ( (kx₂ − x₁)/(k − 1) , (ky₂ − y₁)/(k − 1) ).
Use one coordinate (x or y) to solve for k.
Verify with the other coordinate.
If k > 0, P divides AB externally in ratio k : 1.

Solved Examples

Example 1: Basic External Division

Problem: Find the coordinates of the point that divides the line segment joining A(2, 3) and B(6, 7) externally in the ratio 3 : 1.

Solution:

Given:

A(x₁, y₁) = (2, 3), B(x₂, y₂) = (6, 7)
Ratio m : n = 3 : 1 (external)

Using the formula:

x = (mx₂ − nx₁)/(m − n) = (3×6 − 1×2)/(3 − 1) = (18 − 2)/2 = 16/2 = 8
y = (my₂ − ny₁)/(m − n) = (3×7 − 1×3)/(3 − 1) = (21 − 3)/2 = 18/2 = 9

Answer: P = (8, 9).

Example 2: External Division with Ratio 2 : 5

Problem: Find the point dividing the join of A(−4, 6) and B(8, −3) externally in the ratio 2 : 5.

Solution:

Given:

A(−4, 6), B(8, −3), m : n = 2 : 5

Using the formula:

x = (2×8 − 5×(−4))/(2 − 5) = (16 + 20)/(−3) = 36/(−3) = −12
y = (2×(−3) − 5×6)/(2 − 5) = (−6 − 30)/(−3) = −36/(−3) = 12

Answer: P = (−12, 12).

Example 3: Finding the Ratio of External Division

Problem: In what ratio does the point P(5, 4) divide the line segment joining A(3, 2) and B(4, 3) externally?

Solution:

Given: Let the ratio be k : 1.

Using x-coordinate:

5 = (k×4 − 1×3)/(k − 1)
5(k − 1) = 4k − 3
5k − 5 = 4k − 3
k = 2

Verify with y-coordinate:

y = (2×3 − 1×2)/(2 − 1) = (6 − 2)/1 = 4 ✓

Answer: P divides AB externally in the ratio 2 : 1.

Example 4: Negative Coordinates

Problem: Find the point that divides the join of A(−1, −5) and B(4, 5) externally in the ratio 3 : 2.

Solution:

Given: A(−1, −5), B(4, 5), m : n = 3 : 2

Using the formula:

x = (3×4 − 2×(−1))/(3 − 2) = (12 + 2)/1 = 14
y = (3×5 − 2×(−5))/(3 − 2) = (15 + 10)/1 = 25

Answer: P = (14, 25).

Example 5: External Division with Equal Halves

Problem: Find the point dividing A(1, 2) and B(5, 6) externally in ratio 5 : 3.

Solution:

Given: A(1, 2), B(5, 6), m : n = 5 : 3

Using the formula:

x = (5×5 − 3×1)/(5 − 3) = (25 − 3)/2 = 22/2 = 11
y = (5×6 − 3×2)/(5 − 3) = (30 − 6)/2 = 24/2 = 12

Answer: P = (11, 12).

Example 6: External vs Internal Division Comparison

Problem: A = (2, 1), B = (8, 7). Find both the internal and external points of division in ratio 2 : 1.

Solution:

Internal division (2 : 1):

x = (2×8 + 1×2)/(2 + 1) = 18/3 = 6
y = (2×7 + 1×1)/(2 + 1) = 15/3 = 5
P_internal = (6, 5)

External division (2 : 1):

x = (2×8 − 1×2)/(2 − 1) = 14/1 = 14
y = (2×7 − 1×1)/(2 − 1) = 13/1 = 13
P_external = (14, 13)

Answer: Internal: (6, 5). External: (14, 13).

Example 7: Verifying External Division

Problem: Verify that P(11, −1) divides A(1, 4) and B(7, 1) externally in ratio 5 : 2.

Solution:

Step 1: Apply formula with m = 5, n = 2:

x = (5×7 − 2×1)/(5 − 2) = (35 − 2)/3 = 33/3 = 11 ✓
y = (5×1 − 2×4)/(5 − 2) = (5 − 8)/3 = −3/3 = −1 ✓

Step 2: Verify with distances:

AP = √((11−1)² + (−1−4)²) = √(100 + 25) = √125 = 5√5
PB = √((11−7)² + (−1−1)²) = √(16 + 4) = √20 = 2√5
AP/PB = 5√5 / 2√5 = 5/2 ✓

Answer: Verified. P(11, −1) divides AB externally in ratio 5 : 2.

Example 8: Finding a Point on Extended Line

Problem: A line segment has endpoints A(3, −2) and B(−1, 4). Find the point on the line AB produced beyond B such that it is at twice the distance of AB from A.

Solution:

Given: AP = 2 × AB → AP/AB = 2 → AP/(AP − AB) = 2/1 → AP/PB cannot be directly set.

Since AP = 2·AB and P is beyond B: AP = AB + BP → 2·AB = AB + BP → BP = AB.

So AP : PB = 2 : 1 (external division).

Using formula:

x = (2×(−1) − 1×3)/(2 − 1) = (−2 − 3)/1 = −5
y = (2×4 − 1×(−2))/(2 − 1) = (8 + 2)/1 = 10

Answer: P = (−5, 10).

Example 9: Harmonic Conjugate

Problem: Find both the internal and external points of division of A(1, 3) and B(9, 11) in ratio 3 : 5.

Solution:

Internal (3 : 5):

x = (3×9 + 5×1)/8 = (27 + 5)/8 = 32/8 = 4
y = (3×11 + 5×3)/8 = (33 + 15)/8 = 48/8 = 6
P₁ = (4, 6)

External (3 : 5):

x = (3×9 − 5×1)/(3 − 5) = (27 − 5)/(−2) = 22/(−2) = −11
y = (3×11 − 5×3)/(3 − 5) = (33 − 15)/(−2) = 18/(−2) = −9
P₂ = (−11, −9)

Answer: Internal: (4, 6). External: (−11, −9). These are called harmonic conjugates of each other with respect to A and B.

Example 10: Word Problem — External Division

Problem: Two towns A(2, 3) and B(10, 9) are connected by a road. A signpost P is placed on the road extended beyond B such that AP : PB = 7 : 3. Find the position of P.

Solution:

Given: A(2, 3), B(10, 9), external ratio 7 : 3

Using formula:

x = (7×10 − 3×2)/(7 − 3) = (70 − 6)/4 = 64/4 = 16
y = (7×9 − 3×3)/(7 − 3) = (63 − 9)/4 = 54/4 = 13.5

Answer: The signpost P is at (16, 13.5).

Real-World Applications

Uses of external division:

Geometry proofs: Finding points on lines extended beyond given endpoints.
Harmonic conjugates: In projective geometry, internal and external division in the same ratio give harmonic conjugate points.
Physics: Locating image positions in optics (lens formula involves external division when image is virtual).
Computer graphics: Extrapolation of line segments for rendering and animation.
Engineering: Extending structural lines to find intersection points.

Key Points to Remember

External division formula: P = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)).
In external division, the point P lies outside the line segment AB.
Replace + with − in the internal division formula to get the external formula.
External division in ratio m : n = internal division in ratio m : (−n).
If m > n: P is beyond B. If m < n: P is beyond A.
External division is undefined when m = n (denominator becomes zero).
Always verify by checking both coordinates and the ratio AP/PB.
The internal and external division points in the same ratio are called harmonic conjugates.
A, P_internal, B, P_external are collinear.
External division is less common in CBSE but appears in important board questions and competitive exams.

Practice Problems

Find the point dividing the join of (3, −1) and (8, 9) externally in the ratio 2 : 3.
In what ratio does the point (−3, 5) divide the segment joining (1, 3) and (−1, 4) externally?
A(0, 0) and B(6, 8). Find the point on line AB produced beyond B such that AP = 5·AB/2.
Find both the internal and external division points of (−2, 5) and (3, −5) in ratio 4 : 1.
Prove that the external division point of (1, 1) and (5, 5) in ratio 3 : 1 is (7, 7).
Find the coordinates of the point which divides externally the line joining (4, −3) and (−2, 5) in ratio 3 : 7.

Frequently Asked Questions

Q1. What is external division of a line segment?

When a point P divides line segment AB such that P lies outside the segment (on the line extended beyond A or B), with AP/PB = m/n, it is called external division in ratio m : n.

Q2. What is the formula for external division?

P(x, y) = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)), where A(x₁, y₁) and B(x₂, y₂) are endpoints and m : n is the external ratio.

Q3. How is external division different from internal division?

In internal division, P lies between A and B. In external division, P lies outside the segment. The formulas differ: internal uses +, external uses −.

Q4. What happens when m equals n in external division?

The formula is undefined because the denominator (m − n) becomes zero. Geometrically, the external division point goes to infinity.

Q5. How do you find the ratio of external division?

Assume the ratio is k : 1. Substitute the known point coordinates into the external division formula and solve for k.

Q6. Can external division give negative coordinates?

Yes. Depending on the positions of A and B and the ratio, the external division point can have negative x or y coordinates.

Q7. What is a harmonic conjugate?

If a point P₁ divides AB internally in ratio m : n, and P₂ divides AB externally in the same ratio m : n, then P₁ and P₂ are harmonic conjugates with respect to A and B.

Q8. Is external division in CBSE Class 10 syllabus?

It is part of the coordinate geometry chapter and may appear in board exams as a higher-order question. The internal section formula is more frequently tested.

External Division (Section Formula)

What is External Division?

External Division (Section Formula) Formula

Derivation and Proof

Types and Properties

Methods

Solved Examples

Example 1: Basic External Division

Example 2: External Division with Ratio 2 : 5

Example 3: Finding the Ratio of External Division

Example 4: Negative Coordinates

Example 5: External Division with Equal Halves

Example 6: External vs Internal Division Comparison

Example 7: Verifying External Division

Example 8: Finding a Point on Extended Line

Example 9: Harmonic Conjugate

Example 10: Word Problem — External Division

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What is external division of a line segment?

Q2. What is the formula for external division?

Q3. How is external division different from internal division?

Q4. What happens when m equals n in external division?

Q5. How do you find the ratio of external division?

Q6. Can external division give negative coordinates?

Q7. What is a harmonic conjugate?

Q8. Is external division in CBSE Class 10 syllabus?

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