# EXAMPLE PROBLEMS USING USING SECTION FORMULA

## About "Example Problems Using Section Formula"

Example Problems Using Section Formula :

Here we are going to see some example problems using the 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 (x1, y1) and (x2 , y2) in the ratio m : n.

Section formula (internally) :

(mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

Section formula (externally) :

(mx2 - nx1)/(m - n), (my2 - ny1)/(m - n)

## Example Problems Using Section Formula - Practice questions

Question 1 :

Find the coordinates of the point which divides the line segment joining the points A(4,−3) and B(9, 7) in the ratio 3:2.

Solution :

=  (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

=  (3(9) + 2(4))/(3 + 2), (3(7) + 2(-3))/(3 + 2)

=  (27 + 8)/5, (21 - 6))/5

=  35/5, 15/5

=  (7, 3)

Question 2 :

In what ratio does the point P(2,−5) divide the line segment joining A(−3, 5) and B(4, −9).

Solution :

=  (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

=  (m(4) + n(-3))/(m + n), (m(-9) + n(5))/(m + n)

(4m - 3n)/(m + n), (-9m + 5n)/(m + n)  =  (2, -5)

By equating x and y-coordinates, we get

(4m - 3n)/(m + n)  =  2

4m - 3n  =  2m + 2n

4m - 2m  =  2n + 3n

2m  =  5n

m/n  =  5/2

m : n  =  5 : 2

Hence the required ratio is 5 : 2

Question 3 :

Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = (2/5) AB.

Solution :

AP = (2/5) AB.

AP/AB  =  2/5

AB  =  5

AP + PB  =  AB

2 + PB  =  5

PB  =  5 - 2  =  3

So, the point P divides the line segment in the ratio 2 : 3.

A(1, 2) and B(6, 7)

P  =  (2(6) + 3(1))/(2 + 3), (2(7) + 3(2))/(2 + 3)

P  =  (15/5, 20/5)

P  =  (3, 4)

Hence the required point is (3, 4). After having gone through the stuff given above, we hope that the students would have understood, "Example Problems Using Section Formula"

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