# SECTION FORMULA EXAMPLE PROBLEMS WITH SOLUTIONS

Section Formula Example Problems With Solutions :

We use the section formula to find the point which divides the line segment in a given ratio.

The point P which divides the line segment joining the two points A (x1,  y1) and B (x2, y2) internally in the ratio l : m is If P divides a line segment AB joining the two points A (x1,  y1) and B (x2, y2) externally in the ratio l : m is, ## Section Formula Example Problems

Example 1 :

Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6).

Solution : Here x1  =  -3, y1  =  10, x2  =  6, y2  =  -8

[m(6) + n(-3)]/(m + n), [m(-8) + n(10)]/(m + n)  =  (-1, 6)

(6m - 3n)/(m + n), (-8m + 10n)/(m + n) = (-1, 6)

Equating the coefficients of x and y,

(6m - 3n)/(m + n)  =  -1 ---(1)       (-8m + 10n)/(m + n)  =  6

From (1),

6m -  3n  =  -1 (m + n)

6m - 3n  =  -1 (m + n)

6m - 3n  =  -m - n

6m + m  =  -n + 3n

7m  =  2n

m/n  =  2/7

m : n  =  2 : 7

So, the point (-1,6) is dividing the line segment in the ratio 2 : 7.

Example 2 :

Find the ratio in which the line segment joining A (1, -5) and B(-4, 5) is divided by the x axis. Also find the coordinates of the point of division.

Solution :

Let m:n be the ratio which divides the line segment joining the points A(1, -5) and B(-4, 5)

x =  1, y1  =  -5, x2  =  -4 , y2  =  5

[m(-4) + n(1)]/(m + n) , [m(5) + n(-5)]/(m + n)  =  (x, 0)

(-4m + n) /(m + n) , (5m - 5n)/(m + n)  =  (x,0)

(-4m + n)/(m + n)  =  x      (5m - 5n)/(m + n)  =  0

(5m - 5n)/(m + n)  =  0

5 m - 5 n  =  0

5 m = 5 n

m/n = 5/5  ==>  m : n = 1 : 1

Example 3 :

If (1, 2), (4, y) (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y

Solution :

Let A (1, 2), B (4, y) C (x, 6) and D (3, 5) are the vertices of a parallelogram.

In a parallelogram midpoint of the diagonals will be equal.

Midpoint of diagonal AC  =  Midpoint of diagonal BD

Midpoint  =  (x1x2)/2 , (y1 + y2)/2

Midpoint of diagonal AC :

x1  =  1, y1  =  2, x2  =  x , y2  =  6

=  (1 + x)/2, (2 + 6)/2

=  (1 + x)/2, 8/2

=  (1 + x)/2, 4 ------(1)

Midpoint of diagonal BD :

x1  =  4, y1  =  y, x2  =  3 , y2  =  5

=  (4 + 3)/2, (y + 5)/2

=  7/2, (y + 5)/2 ------(2)

Equating x and y coordinates, we get

 (1 + x)/2, 4  =  7/21 + x  =  7x  =  7 - 1x  =  6 (y + 5)/2  =  44  =  (y + 5)/28  =  y + 5y  =  8 - 5  = 3 After having gone through the stuff given above, we hope that the students would have understood, section formula example problems with solutions.

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