Division of Line Segment

Here we will discuss about internal and external division of line segment.

To find the co-ordinates of the point dividing the line segment joining two given points in a given ratio:

(i) Internal Division of line segment:

Let (x₁, y₁) and (x₂, y₂) be the cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ internally in a given ratio m : n (say), i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

Let, (x, y) be the required co-ordinate of R . From P, Q and R, draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM at T.

Then, 

PS = LN = ON - OL = x – x₁;

PT = LM = OM – OL = x₂ - x₁;

RS = RN – SN = RN – PL = y - y₁;

and QT = QM – TM = QM – PL = y₂ – y₁

Again, PR/RQ = m/n

or, RQ/PR = n/m

or, RQ/PR + 1 = n/m + 1

or, (RQ + PR/PR) = (m + n)/m

o, PQ/PR = (m + n)/m

Now, by construction, the triangles PRS and PQT are similar; hence,

PS/PT = RS/QT = PR/PQ

Taking, PS/PT = PR/PQ we get,

(x - x₁)/(x₂ - x₁) = m/(m + n)

or, x (m + n) – x₁ (m + n) = mx₂ – mx₁

or, x ( m + n) = mx₂ - mx₁ + m x₁ + nx₁ = mx₂ + nx₁

Therefore, x = (mx₂ + nx₁)/(m + n)

Again, taking RS/QT = PR/PQ we get,

(y - y₁)/(y₂ - y₁) = m/(m + n)

or, ( m + n) y - ( m + n) y₁ = my₂ – my₁

or, ( m+ n)y = my₂ – my₁ + my₁ + ny₁ = my₂ + ny₁

Therefore, y = (my₂ + ny₁)/(m + n)

Therefore, the required co-ordinates of the point R are

((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

(ii) External Division of line segment:

Let (x₁, y₁) and (x₂, y₂) be the cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ externally in a given ratio m : n (say) i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

Let, (x, y) be the required co-ordinates of R. Draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM and RN at S and T respectively, Then,

PS = LM = OM - OL = x₂ – x₁;

PT = LN = ON – OL = x – x₁;

QT = QM – SM = QM – PL = y₂ – y₁

and RT = RN – TN = RN – PL = y — y₁

Again, PR/RQ = m/n

or, QR/PR = n/m

or, 1 - QR/PR = 1 - n/m

or, PR - RQ/PR = (m - n)/m

or, PQ/PR = (m - n)/m

Now, by construction, the triangles PQS and PRT are similar; hence,

PS/PT = QS/RT = PQ/PR

Taking, PS/PT = PQ/PR we get,

(x₂ - x₁)/(x - x₁) = (m - n)/m

or, (m – n)x - x₁(m – n) = m (x₂ - x₁)

or, (m - n)x = mx₂ – mx₁ + mx₁ - nx₁ = mx₂ - nx₁.

Therefore, x = (mx₂ - nx₁)/(m - n)

Again, taking QS/RT = PQ/PR we get,

(y₂ - y₁)/(y - y₁) = (m - n)/m

or, (m – n)y - (m – n)y₁ = m(y₂ - y₁)

or, (m - n)y = my₂ – my₁ + my₁ - ny₁ = my₂ - ny₁

Therefore, x = (my₂ - ny₁)/(m - n)

Therefore, the co-ordinates of the point R are

((mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n))

Corollary: To find the co-ordinates of the middle point of a given line segment:

Let (x₁, y₁) and (x₂, y₂) he the co-ordinates of the points P and Q respectively and R, the mid-point of the line segment PQ. To find the co-ordinates R. Clearly, the point R divides the line segment PQ internally in the ratio 1 : 1; hence, the co-ordinates of R are ((x₁ + x₂)/2, (y₁ + y₂)/2). [Putting m = n the co-ordinates or R of ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))]. This formula is also known as midpoint formula. By using this formula we can easily find the midpoint between the two co-ordinates.

Example on Division of Line Segment:

1. A diameter of a circle has the extreme points (7, 9) and (-1, -3). What would be the co-ordinates of the centre?

Solution:

Clearly, the mid-point of the given diameter is the centre of the circle. Therefore, the required co-ordinates of the centre of the circle = the co-ordinates of the mid-point of the line-segment joining the points (7, 9) and (- 1, - 3)

= ((7 - 1)/2, (9 - 3)/2) = (3, 3).

2. A point divides internally the line- segment joining the points (8, 9) and (-7, 4) in the ratio 2 : 3. Find the co-ordinates of the point.

Solution:

Let (x, y) be the co-ordinates of the point which divides internally the line-segment joining the given points. Then,

x = (2 ∙ (- 7) + 3 ∙ 8)/(2 + 3) = (-14 + 24)/5 = 10/5 = 2

And y = (2 ∙ 4 + 3 ∙ 9)/(2 + 3) = (8 + 27)/5 = 35/5 = 5

Therefore, the co-ordinates of the required point are (2, 7).

[Note: To get the co-ordinates of the point in question we have used formula, x = (mx₁ + n x₁)/(m + n) and y = my₂ + ny₁)/(m + n).

For the given problem, x₁ = 8, y₁ = 9, x₂ = -7, y₂ = 4, m = 2 and n = 3.]

3. A (4, 5) and B (7, - 1) are two given points and the point C divides the line-segment AB externally in the ratio 4 : 3. Find the co-ordinates of C.

Solution:

Let (x, y) be the required co-ordinates of C. Since C divides the line-segment AB externally in the ratio 4 : 3 hence,

x = (4 ∙ 7 - 3 ∙ 4)/(4 - 3) = (28 - 12)/1 = 16

And y = (4 ∙ (-1) - 3 ∙ 5)/(4 - 3) = (-4 - 15)/1 = -19

Therefore, the required co-ordinates of C are (16, - 19).

[Note: To get the co-ordinate of C we have used formula, 

x = (mx₁ + n x₁)/(m + n) and y = my₂ + ny₁)/(m + n).

In the given problem, x₁ = 4, y₁ = 5, x₂ = 7, y₂ = - 1, m = 4 and n = 3].

4. Find the ratio in which the line-segment joining the points (5, - 4) and (2, 3) is divided by the x-axis.

Solution:

Let the given points be A (5, - 4) and B (2, 3) and x-axis. intersects the line-segment ¯(AB )at P such that AP : PB = m : n. Then the co-ordinates of P are ((m ∙ 2 + n ∙ 5)/(m + n), (m ∙ 3 + n ∙ (-4))/(m + n)). Clearly, the point P lies on the x-axis ; hence, y co-ordinate of P must be zero.

Therefore, (m ∙ 3 + n ∙ (-4))/(m + n) = 0

or, 3m - 4n = 0

or, 3m = 4n

or, m/n = 4/3

Therefore, the x-axis divides the line-segment joining the given points internally in 4 : 3.

5. Find the ratio in which the point (- 11, 16) divides the '-line segment joining the points (- 1, 2) and (4, - 5).

Solution:

Let the given points be A (- 1, 2) and B (4, - 5) and the line-segment AB is divided in the ratio m : n at (- 11, 16). Then we must have,

-11 = (m ∙ 4 + n ∙ (-1))/(m + n)

or, -11m - 11n = 4m - n

or, -15m = 10n

or, m/n = 10/-15 = - 2/3

Therefore, the point (- 11, 16) divides the line-segment ¯BA externally in the ratio 3 : 2.

[Note: (i) A point divides a given line-segment internally or externally in a definite ratio according as the value of m:n is positive or negative.

(ii) See that we can obtain the same ratio m : n = - 2 : 3 using the condition 16 = (m ∙ (-5) +n ∙ 2)/(m + n)]

● Co-ordinate Geometry 

• What is Co-ordinate Geometry?
  
• Rectangular Cartesian Co-ordinates
  
• Polar Co-ordinates
  
• Relation between Cartesian and Polar Co-Ordinates
  
• Distance between Two given Points
  
• Distance between Two Points in Polar Co-ordinates
  
• Division of Line Segment: Internal & External
  
• Area of the Triangle Formed by Three co-ordinate Points
  
• Condition of Collinearity of Three Points
  
• Medians of a Triangle are Concurrent
  
• Apollonius' Theorem
  
• Quadrilateral form a Parallelogram 
  
• Problems on Distance Between Two Points 
  
• Area of a Triangle Given 3 Points
  
• Worksheet on Quadrants
  
• Worksheet on Rectangular – Polar Conversion
  
• Worksheet on Line-Segment Joining the Points
  
• Worksheet on Distance Between Two Points
  
• Worksheet on Distance Between the Polar Co-ordinates
  
• Worksheet on Finding Mid-Point
  
• Worksheet on Division of Line-Segment
  
• Worksheet on Centroid of a Triangle
  
• Worksheet on Area of Co-ordinate Triangle
  
• Worksheet on Collinear Triangle
  
• Worksheet on Area of Polygon
  
• Worksheet on Cartesian Triangle

11 and 12 Grade Math 

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