**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 (x_{1}, y_{1}) and B (x_{2}, y_{2}) internally in the ratio l : m is

If P divides a line segment AB joining the two points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) externally in the ratio l : m is,

**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 x_{1} = -3, y_{1} = 10, x_{2} = 6, y_{2} = -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 } = 1, y_{1} = -5, x_{2} = -4 , y_{2} = 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 = (x_{1} + x_{2})/2 , (y_{1} + y_{2})/2

**Midpoint of diagonal AC :**

x_{1} = 1, y_{1} = 2, x_{2} = x , y_{2} = 6

= (1 + x)/2, (2 + 6)/2

= (1 + x)/2, 8/2

= (1 + x)/2, 4 ------(1)

**Midpoint of diagonal BD :**

x_{1} = 4, y_{1} = y, x_{2} = 3 , y_{2} = 5

= (4 + 3)/2, (y + 5)/2

= 7/2, (y + 5)/2 ------(2)

Equating x and y coordinates, we get

(1 + x)/2, 4 = 7/2 1 + x = 7 x = 7 - 1 x = 6 |
(y + 5)/2 = 4 4 = (y + 5)/2 8 = y + 5 y = 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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