SECTION FORMULA EXAMPLE PROBLEMS WITH SOLUTIONS

Equating the coefficients of x and y,

(6l - 3m)/(l + m) = -1 ----(1)

(-8l + 10m)/(l + m) = 6  ----(2)

6l -  3m  =  -1 (l + m)

6l - 3m  =  -l - m

6l + l = -m + 3m

7l = 2m

l/m  =  2/7

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

x₁  =  1, y₁  =  -5, x₂  =  -4 , y₂  =  5

 (-4l + m)/(l + m)  =  x -----(1)

 (5l - 5m)/(l + m)  =  0-----(2)

(5l - 5m)/(l + m)  =  0

5l - 5m  =  0

5l = 5m

l/m = 5/5  ==>  l : m = 1 : 1

By applying the ration in (1), we get

x = (-4(1) + 1)/(1 + 1)

x = -3/2

So, the point of intersection is (-3/2, 0).

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  =  (x₁ + x₂)/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

So, the required fourth vertex is (6, 3).

Example 4 :

Find the ratio in which the line 2x + 3y - 5 = 0 divides the line segment joining the points (8, -9) and (2, 1). Also, find the co-ordinates of the point of division.

Solution :

Let the required ratio be m : n.

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

= (m(2) + n(8))/(m + n), (m(1) + n(-9))/(m + n)

= (2m + 8n)/(m + n), (m - 9n)/(m + n) ----(1)

The point which divides the line segment joining the points given above lies on the line 2x + 3y - 5 = 0.

2[(2m + 8n)/(m + n)] + 3[(m - 9n)/(m + n)] - 5 = 0

2(2m + 8n) + 3(m - 9n) - 5(m + n) = 0

4m + 16n + 3m - 27n - 5m - 5n = 0

2m - 16n = 0

2m = 16n

m/n = 16/2

m/n = 8/1

m : n = 8 : 1

So, the required ratio is 8 : 1.

Applying this ratio in (1), we get

(2(8) + 8(1))/(8 + 1), (8 - 9(1))/(8 + 1)

= (16 + 8)/9,  (8 - 9)/9

= (24/9, -1/9)

So, the required ratio is 8 : 1 and the required point is (24/9, -1/9).

Example 5 :

The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x - 5y + 15 = 0 ?

Solution :

Finding the point P :

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

= (2(-3) + 3(2))/(2 + 3), (2(6) + 3(1))/(2 + 3)

= (-6 + 6)/5, (12 + 3)/5

= 0/5, 15/5

= (0, 3)

The point P lies on the line x - 5y + 15 = 0

0 - 5(3) + 15  = 0

-15 + 15 = 0

0 = 0

S, the point lies on the line.

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