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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 – x_{1};
PT = LM = OM – OL = x_{2} - x_{1};
RS = RN – SN = RN – PL = y - y_{1};
and QT = QM – TM = QM – PL = y_{2} – y_{1}
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 - x_{1})/(x_{2} - x_{1}) = m/(m + n)
or, x (m + n) – x_{1} (m + n) = mx_{2} – mx_{1}
or, x ( m + n) = mx_{2} - mx_{1} + m x_{1} + nx_{1} = mx_{2} + nx_{1}
Therefore, x = (mx_{1} + n x_{1})/(m + n)
Again, taking RS/QT = PR/PQ we get,
(y - y_{1})/(y_{2} - y_{1}) = m/(m + n)
or, ( m + n) y - ( m + n) y_{1} = my_{2} – my_{1}
or, ( m+ n)y = my_{2} – my_{1} + my_{1} + ny_{1} = my_{2} + ny_{1}
Therefore, y = my_{2} + ny_{1})/(m + n)
Therefore, the required co-ordinates of the point R are
((mx_{2} + nx_{1})/(m + n), (my _{2} + ny_{1})/(m + n))
(ii) External Division of line segment:
Let (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 = x_{2} – x_{1};
PT = LN = ON – OL = x – x_{1};
QT = QM – SM = QM – PL = y_{2} – y_{1}
and RT = RN – TN = RN – PL = y — y_{1}
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,
(x_{2} - x_{1})/(x - x_{1}) = (m - n)/m
or, (m – n)x - x_{1}(m – n) = m (x_{2} - x_{1})
or, (m - n)x = mx_{2} – mx_{1} + mx_{1} - nx_{1} = mx_{2} - nx_{1}.
Therefore, x = (mx_{2} - nx_{1})/(m - n)
Again, taking QS/RT = PQ/PR we get,
(y_{2} - y_{1})/(y - y_{1}) = (m - n)/m
or, (m – n)y - (m – n)y_{1} = m(y_{2} - y_{1})
or, (m - n)y = my_{2} – my_{1} + my_{1} - ny_{1} = my_{2} - ny_{1}
Therefore, x = (my_{2} - ny_{1})/(m - n)
Therefore, the co-ordinates of the point R are
((mx_{2} - nx_{1})/(m - n), (my_{2} - ny_{1})/(m - n))
Corollary: To find the co-ordinates of the middle point of a given line segment:
Let (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 ((x_{1} + x_{2})/2, (y_{1} + y_{2})/2). [Putting m = n the co-ordinates or R of ((mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(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 = (mx_{1} + n x_{1})/(m + n) and y = my_{2} + ny_{1})/(m + n).
For the given problem, x_{1} = 8, y_{1} = 9, x_{2} = -7, y_{2} = 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 = (mx_{1} + n x_{1})/(m + n) and y = my_{2} + ny_{1})/(m + n).
In the given problem, x_{1} = 4, y_{1} = 5, x_{2} = 7, y_{2} = - 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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