We studied bisection and trisection of a given line segment. These are only particular cases of the general problem of dividing a line segment joining two points ( x1, y1 ) and ( x 2 , y2 ) in the ratio m : n.

**Section Formula**

We studied
bisection and trisection of a given line segment. These are only particular cases
of the general problem of dividing a line segment joining two points (
*x*_{1,} *y*_{1} ) and (*
x *_{2}* *,* y*_{2}* *)* *in the ratio* m *:* n*.

Given a segment
*AB* and a positive real number *r*.

We wish to
find the coordinate of point *P* which divides *AB* in the ratio *r*
:1.

This means AP/PB = r/1 or AP = r(PB).

This means
that *x* – *x*_{1} = *r*(*x*_{2} – *x*)

Solving this,* x *= ( r*x*_{2} + *x*_{1})
/ (r + 1) …..
(1)

We can use
this result for points on a line to the general case as follows.

Taking AP
: PB = r :1 , we get A′ P′ : P′B′ = r :1 .

Therefore
A′P′ = r(P′B′)

Thus, (*x* − *x*_{1}
) = *r*( *x*_{2} − *x*)

which gives* x *= ( *rx*_{2} + *x*_{1}
) / (r + 1) …
[see (1)]

Precisely
in the same way we can have* y *= *ry*_{2} + *y*_{1} / r + 1

If *P* is between A and B, and AP/PB = r , then
we have the formula,

If *r* is taken as *m/n* , then the section formula is , which is the standard form.

**Thinking Corner **

(i) What happens when m =*
n *= 1? Can you identify it with a result already proved?

(ii) AP : PB = 1 : 2 and AQ : QB = 2:1. What is AP : AB? What is
AQ : AB?

**Note**

**• **The line joining the points** **(*x*_{1}** **,** ***y*_{1})** **and** **(*x*_{2}** **,** ***y*_{2}** **)** **is divided by** ***x*-axis in the ratio *-y*_{1}/*
y*_{2} and by *y*-axis in the ratio −*x*_{1}/* x*_{1}.

**• **If three points are collinear,
then one of the points divide the line segment joining** **the other two points in
the ratio *r* : 1.

**• **Remember that the section
formula can be used only when the given three points** **are collinear.

**• **This formula is helpful
to find the centroid, incenter and excenters of a triangle.** **It has applications in
physics too; it helps to find the center of mass of systems, equilibrium points
and many more.

**Example 5.17**

Find the
coordinates of the point which divides the line segment joining the points (3,5)
and (8,−10) internally in the ratio 3:2.

*Solution*

Let *A*(3,5),
*B*(8,−10) be the given points and let the point *P*(*x*,*y*)
divides the line segment *AB *internally in the ratio 3:2.

By section
formula, P ( x,* y *) =

Here *x*_{1} = 3, *y*_{1} = 5, *x*_{2}
= 8, *y*_{2} = −10 and *m* = 3,* n *= 2

Therefore
P ( x,* y *) =

= *P* ( [3(8) + 2(3)] / [3+2], [3(−10) + 2(5)]
/ [3+2] )

= *P* ( [24 + 6]/5 , [−30 + 10]/5) = P(6, − 4
)

**Example 5.17**

In what ratio
does the point P(–2, 4) divide the line segment joining the points A(–3, 6) and
B(1, –2) internally?

*Solution *

Given points
are A(–3, 6) and B(1, –2). P(–2, 4) divide AB internally in* *the ratio m : n.

By section formula,

m :* n *= 1 : 3

Hence P divides
AB internally in the ratio 1:3.

**Note**

We may arrive at the same result by also equating the y-coordinates.
Try it.

**Example 5.19**

What are
the coordinates of *B* if point *P*(−2,3) divides the line segment joining
*A*(−3,5) and *B* internally in the ratio 1:6?

*Solution*

Let *A*(−3,5)
and *B*(*x*_{2} , *y*_{2} ) be the given two points.

Given *P*(−2,3)
divides *AB* internally in the ratio 1:6.

Therefore,
the coordinate of *B* is (4, −9)

Tags : Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths , 9th Maths : UNIT 5 : Coordinate Geometry

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9th Maths : UNIT 5 : Coordinate Geometry : Section Formula | Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths

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