**Section formula** is a
formula to calculate the coordinates of any 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. Here are
those formulas:

(i) If P(x, y) divides AB internally in the ratio of m_{1}:m_{2},
then the section formula for internal division is given by,

(ii) If P(x, y) divides AB externally in the ratio of m_{1}:m_{2},
then the section formula for external division is given by,

**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 A,
P and B respectively to the x-axis. Then, OL = x_{1}, ON = x, OM = x_{2},
AL = y_{1}, PN = 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 =
NM = OM – ON = x_{2} – x

PQ =
PN – QN = PN – AL = y – y_{1}

BR =
BM – RM = BM – PN = y_{2} – y

Now,
from the figure, In ΔPQA and ΔBRP,

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(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 – QN = PN – AL = y – y_{1}

BR =
RM – BM = PN – BM = y – y_{2}

In ΔPQA
and ΔBPR,

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 point P(x, y) divides the line AB internally in the ratio
of 1:1. Therefore, by applying the section formula for internal division,

**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,

It is a section formula in terms of k,
which is called K-formula.

It is convenient to let the ratio as k:1 while
solving the problems to find the ratio when the points are given. In such
situations, we use the K-formula. Look at the worked out examples below.

**Centroid Formula**

Medians of a triangle intersect at a common point
which is called centroid of a triangle. If A(x_{1}, y_{1}), B(x_{2},
y_{2}) and C(x_{3}, y_{3}) are the vertices of a
triangle ABC and G is the median, then the coordinates of centroid G is given
by the following formula,

**Proof of 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. D, E
and F are the mid-points of BC, CA and AB respectively. And, medians AD, BE and
CF intersect at point G.

**Worked Out 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.*

*Example 2: What ratio is the line joining
(-3, 4) and (2, -6) divided by the point (-1, 0)?*

*Example 3: **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.*

*Example 4: **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).*

*Example 5: Find the
centroid of the triangle with vertices A(-2, 4), B(11, 15) and C(3, -1).*

*Solution:*

*Let G(x, y) be the
centroid of the triangle ABC with vertices A(-2, 4), B(11, 15) and C(3, -1). By
using formula of centroid,*

*Hence, the centroid of the
triangle is G(4, 6).*

*Do you have any question regarding the section
formula?*

*Do you have any question regarding the section formula?*

*You can leave a comment
below in the comment box.*

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