Section formula is a formula to calculate the coordinates of any point point P(x, y) which divides the line joining the points A(x1, y1) and B(x2, y2) in the ratio of m1:m2 internally or externally, which are as given below:

### Derivation of Section Formula

#### Internal Division

Let A(x1, y1) and B(x2, y2) be given two points. Let the point P(x, y) divides the line joining AB internally in the ratio of m1:m2.

Then, AP:PB = m1:m2

Draw perpendiculars AL, PN and BM from the points A, P and B respectively to the x-axis. Then, OL=x1, ON=x, OM=x2, AL=y1, NP=y and BM=y2. Again draw perpendicular AQ and PR from A and P to the line segments PN and BM respectively.

Then,   AQ = LN = ON – OL = x – x1
PR = MN = OM – ON = x2 – x
PQ = NP – QN = NP – AL = y – y1
BR = BM – MR = BM – NP = y2 – y
In Ds PQA and BRP,
i.      PQA = BRP --------> both right angles
ii.     QAP = RPB --------> corresponding angles
iii.    APQ = PBR --------> corresponding angles
DPQA ~ DBRP -------------------> by AAA axiom

#### External Division

Let A(x1, y1) and B(x2, y2) be given two points. Let the point P(x, y) divides the line joining AB externally in the ratio of m1:m2.

Then, AP:PB = m1:m2

Draw perpendiculars AL, BM and PN from A, B and P respectively to the x-axis. Then, OL = x1, OM = x2, ON = x, AL = y1, BM = y2 and PN = y. Again draw perpendiculars AQ from A to PN  and PR from P to BM produced.

Then, AQ = LN = ON – OL = x – x1
PR = NM = ON – OM = x – x2
PQ = PN – NQ = PN–AL = y – y1
BR = RM – BM = PN–BM = y – y2
In Ds 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 – m2 for m2.

#### Mid-point formula

If P(x, y) is the mid-point of the line joining A(x1, y1) and B(x2, y2), then the ratio of the division is 1:1. Then,

#### K-formula

If P(x, y) divides the line joining the points A(x1, y1) and B(x2, y2) 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(x1, y1), B(x2, y2) and C(x3, y3) 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

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,
x1 = -3             x2 = 2               m1 = 3
y1 = -2             y2 = 8               m2 = 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).

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