Here we will discuss about internal and external division of line segment.
To find the coordinates 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 coordinates of the points P and Q respectively referred to rectangular coordinate axes OX and OY and the point R divides the linesegment PQ internally in a given ratio m : n (say), i.e., PR : RQ = m : n. We are to find the coordinates of R.
Let, (x, y) be the required coordinate 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 = (mx_{2} + nx_{1})/(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 coordinates 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 coordinates of the points P and Q respectively referred to rectangular coordinate axes OX and OY and the point R divides the linesegment PQ externally in a given ratio m : n (say) i.e., PR : RQ = m : n. We are to find the coordinates of R.
Let, (x, y) be the required coordinates 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 coordinates of the point R are
((mx₂  nx₁)/(m  n), (my₂  ny₁)/(m  n))
Corollary: To find the coordinates of the middle point of a given line segment:
Let (x₁, y₁) and (x₂, y₂) he the coordinates of the points P and Q respectively and R, the midpoint of the line segment PQ. To find the coordinates R. Clearly, the point R divides the line segment PQ internally in the ratio 1 : 1; hence, the coordinates of R are ((x₁ + x₂)/2, (y₁ + y₂)/2). [Putting m = n the coordinates 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 coordinates.
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 coordinates of the centre?
Solution:
Clearly, the midpoint of the given diameter is the centre of the circle. Therefore, the required coordinates of the centre of the circle = the coordinates of the midpoint of the linesegment 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 coordinates of the point.
Solution:
Let (x, y) be the coordinates of the point which divides internally the linesegment 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 coordinates of the required point are (2, 7).
[Note: To get the coordinates 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 linesegment
AB externally in the ratio 4 : 3. Find the coordinates of C.
Solution:
Let (x, y) be the required coordinates of C. Since C divides the linesegment 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 coordinates of C are (16,  19).
[Note: To get the coordinate 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 linesegment joining the points (5,  4) and (2, 3) is divided by the xaxis.
Solution:
Let the given points be A (5,  4) and B (2, 3) and xaxis. intersects the linesegment ¯(AB )at P such that AP : PB = m : n. Then the coordinates of P are ((m ∙ 2 + n ∙ 5)/(m + n), (m ∙ 3 + n ∙ (4))/(m + n)). Clearly, the point P lies on the xaxis ; hence, y coordinate 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 xaxis divides the linesegment 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 linesegment 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 linesegment ¯BA externally in the ratio 3 : 2.
[Note: (i) A point divides a given linesegment 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)]
● Coordinate Geometry
11 and 12 Grade Math
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