**Section formula**is a formula to calculate the coordinates of any point point P(x, y) which divides the line joining the points A(x

_{1}, y

_{1}) and B(x

_{2}, y

_{2}) in the ratio of m

_{1}:m

_{2}internally or externally, which are as given below:

###
**Derivation of Section Formula**

####
**Internal Division**

Let A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) be given two points. Let the point P(x, y) divides the line joining AB internally in the ratio of m_{1}:m_{2}.
Then, AP:PB = m

_{1}:m_{2}_{}

Draw perpendiculars AL, PN and BM from the points A,
P and B respectively to the x-axis. Then, OL=x

_{1}, ON=x, OM=x_{2}, AL=y_{1}, NP=y and BM=y_{2}. Again draw perpendicular AQ and PR from A and P to the line segments PN and BM respectively.
Then, AQ = LN = ON – OL = x – x

_{1}
PR = MN = OM – ON = x

_{2}– x
PQ = NP – QN = NP – AL = y – y

_{1}
BR = BM – MR = BM – NP = y

_{2}– y
In D

^{s}PQA and BRP,
i. ∠PQA = ∠BRP
--------> both right angles

ii. ∠QAP = ∠RPB
--------> corresponding angles

iii. ∠APQ = ∠PBR
--------> corresponding angles

####
**External Division**

Let A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) be given two points. Let the point P(x, y) divides the line joining AB externally in the ratio of m_{1}:m_{2}.
Then, AP:PB = m

_{1}:m_{2}_{}

Draw perpendiculars AL, BM and PN from A,
B and P respectively to the x-axis. Then, OL = x

_{1}, OM = x_{2}, ON = x, AL = y_{1}, BM = y_{2}and PN = y. Again draw perpendiculars AQ from A to PN and PR from P to BM produced.
Then,
AQ = LN = ON – OL = x – x

_{1}
PR = NM = ON – OM = x – x

_{2}
PQ = PN – NQ = PN–AL = y – y

_{1}
BR = RM – BM = PN–BM = y – y

_{2}
In D

^{s}PQA and BPR,
i. ∠PQA = ∠BRP
--------> both right angles

ii. ∠QAP = ∠RPB
--------> corresponding angles

iii. ∠APQ = ∠PBR
--------> corresponding angles

∴ DPQA ~ DBRP
---------------> by AAA axiom

Note: The
coordinates in external division are obtained from those of the internal
division by writing – m

_{2}for m_{2}.####
**Mid-point formula**

If P(x,
y) is the mid-point of the line joining A(x

_{1}, y_{1}) and B(x_{2}, y_{2}), then the ratio of the division is 1:1. Then,####
**K-formula**

If P(x,
y) divides the line joining the points A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio of k:1, then,
For the problems in which it is required to find
the ratio when a given point divides the line joining the two given points, it is convenient to take the
ratio as k:1.

####
**Centroid formula**

Let A(x

_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are the vertices of a triangle ABC. Let D, E and F are the mid-points of sides BC, AC and AB respectively. Then, AD, BE and CF are called medians of the triangle ABC. If these median intersect each other at a point G, then G is called the centroid of the triangle ABC.
From the plane geometry, we know that the
centroid of a triangle divides the median in the ratio of 2:1. Now, if the
coordinates of G are (x, y), then by section formula.

###
*Workout
Examples*

*Workout Examples*

*Example 1:*

*Find the co-ordinates of the point which divides the line joining the points (-3, -2) and (2, 8) in the ratio of 3:2 internally.*

*Solution:*

*Let, P(x, y) divides the line joining A(-3, -2) and B(2, 8) in the ratio of 3:2 internally.*

*Here,*

*x*_{1}= -3 x_{2}= 2 m_{1}= 3

*y*_{1}= -2 y_{2}= 8 m_{2}= 2

*By using formula,*

*Hence the required point is (0,4).*

*Example 2:*

*If the points A (6, 4), B (a, 2) and C (1, -1) are three points on the same straight line, find the ratio in which the point B divides the line segment AC and also find the value of a.*

*Solution:*

*Suppose, the point B(a,2) divides the line segment joining A(6, 4) and C(1, -1) in the ratio of m1:m2. So the y co-ordinate of B is*

*Example 3:*

*Find the co-ordinates of the points of trisection of the line joining the points M (7, -3) and N (1, 3).*

*Solution:*

*Points of trisection of a line segment joining two points are those points which divides the given line segment into three equal parts.*

*Let, P(a, b) and Q(c, d) are the points of trisection of the line segment joining the points M(7, -3) and N(1, 3). So P(a, b) divides MN in the ratio of 1:2 and hence,*

*Hence, the required points of trisection are P(5, -1) and Q(3, 1).**You can comment your questions or problems regarding the*

**section formula**here.
## No comments: