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.

Internal Division of line segment


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 = OMOL = x2 - x1;
RS = RNSN = RNPL = y - y1;
and QT = QMTM = QMPL = 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 = (mx1 + n x1)/(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), (my 2 + 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.

External Division of line segment


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 = ONOL = x – x1;
QT = QMSM = QMPL = y2 – y1
and RT = RNTN = RNPL = 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:

midpoint formula


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

    From Division of Line Segment to HOME PAGE