**Midpoint of the line segment :**

To find the midpoint of the line segment, we use the formula given formula.

Here (x_{1}, y_{1}) and (x_{2}, y_{2}) stands for the endpoints of the line segment.

Let us look into some example problems to understand the above concept.

**Example 1 :**

Find the co-ordinates of the mid point of the line segment joining the points (1 ,-1) and (-5 , 3)

**Solution :**

Midpoint = (x₁+ x₂)/2 , (y₁ +y₂)/2

= (1 + (-5))/2 , (-1 + 3)/2

= (-4/2) , (2/2)

= (-2 , 1)

**Example 2 :**

Find the co-ordinates of the mid point of the line segment joining the points (0 ,0) and (0, 4 )

**Solution :**

Midpoint = (x₁+ x₂)/2 , (y₁ +y₂)/2

= (0 + 0)/2 , (0 + 4)/2

= (0/2) , (4/2)

= (0 , 2)

**Example 3 :**

The centre of the circle is at (-6 , 4). If one end of the diameter of the circle is at origin, then find the other end.

Here one end of the diameter is at origin that is A (0,0) and let B (a, b) is the required endpoint of the diameter

For any circle,

Midpoint of diameter = Center of circle

Midpoint = (x₁ + x₂)/2 , (y₁+y₂)/2

(-6 , 4) = (0 + a)/2 , (0 + b)/2

By equating x and y co-ordinates

(-6 , 4) = a/2 , b/2

a/2 = - 6 Multiplying 2 on bot sides, a = -6 (2) a = -12 |
b/2 = 4 Multiplying 2 on both sides, b = 4(2) b = 8 |

Hence the required end point of the diameter is (-12, 8).

**Example 4 :**

If the points A(2,-2) B(8,4) C(5,7) are the three vertices of a parallelogram ABCD taken in order,find the fourth vertex D.

**Solution :**

Let (a,b) be the required vertex D.

In any parallelogram the diagonals AC and BD bisect each other.That is, the midpoint of the diagonal AC will be equal to the midpoint of the diagonal BD

Midpoint of the diagonal AC = (x_{1}+x_{2})/2 , (y_{1}+y_{2})/2

A(2,-2) C(5,7)

x_{1} = 2, x_{2} = 5, y_{1} = -2 and y_{2} = 7

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

= (7/2 ,5/2) ----(1)

Midpoint of the diagonal BD = (x_{1}+x_{2})/2 , (y_{1}+y_{2})/2

B(8, 4) D(a, b)

x_{1} = 8, x_{2} = a, y_{1} = 4 and y_{2}= b

= (8 + a)/2 , (-4+b)/2 -----(2)

Midpoint of the diagonal AC = Midpoint of the diagonal BD

(7/2,5/2) = (8+a)/2, (-4+b)/2

(8 + a)/2 = 7/2 8 + a = 7 Subtract 8 on both sides 8 + a - 8 = 7 - 8 a = -1 |
(-4 + b)/2 = 5/2 -4 + b = 5 Add 4 on both sides -4 + b + 4 = 5 + 4 b = 9 |

So the required vertex is (-1,9).

After having gone through the stuff given above, we hope that the students would have understood "Midpoint of the line segment".

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