**Solving Word Problems Using Section Formula :**

In this section, we will see some example problems of solving word problems using the concept section formula.

**Example 1 :**

Find the coordinates of a point A, where AB is the diameter of a circle whose center is (2, -3) and B(1, 4)

**Solution :**

Midpoint of the diameter is center

Center = (2,-3)

Let the coordinate of point A be (a, b)

Here x_{1 } = a, y_{1} = b, x_{2} = 1, y_{2} = 4

Midpoint = (x_{1} + x_{2})/2 , (y_{1} + y_{2})/2

(a + 1)/2, (b + 4)/2 = (2, -3)

Equating the coordinates of x and y

(a + 1)/2 = 2 (b + 4)/2 = -3

a + 1 = 4 b + 4 = -6

a = 4 - 1 b = -6 -4

a = 3 b = -10

Therefore the coordinate of A are (3,-10).

**Example 2 :**

If A and B are (-2, -2) and (2, -4),respectively,find the coordinates of P such that AP = (3/7) AB and P lies on the line segment AB.

**Solution :**

AP = (3/7) AB

AP/AB = 3/7

AB is the line segment P is a point lies on the line segment.

AB = AP + PB

7 = 3 + PB

PB = 4

AP : PB = 3 : 4

x_{1} = -2, y_{1} = -2, x_{2} = 2, y_{2} = -4, m = 3 and n = 4

Coordinates of P

= [3(2) + 4(-2)]/(3 + 4) , [3(-4) + 4(-2)]/(3 + 4)

= (6 - 8)/7,(-12 - 8)/7

= -2/7, -20/7

**Example 3 :**

Find the coordinates of the points which divide the line segment joining A (-2, 2) and B(2, 8) into four equal parts.

**Solution :**

Let "C", "D" and "E" be the points which divides the line segment into four equal parts.

length of AC = 1 unit

length of CD = 1 unit

length of DE = 1 unit

then length of EB is also = 1 unit

So, C divides the line segment in the ratio 1 : 3

x_{1} = -2, y_{1} = 2, x_{2} = 2, y_{2} = 8, m = 1 and n = 3

= [1(2) + 3(-2)]/(1 + 3) , [1(8) + 3(2)]/(1 + 3)

= (2 - 6)/4, (8 + 6)/4

= -4/4, 14/4

= (-1,7/2)

D divides the line segment in the ratio 2 : 2

x_{1} = -2, y_{1} = 2, x_{2} = 2, y_{2} = 8, m = 2 and n = 2

= [2(2) + 2(-2)]/(2 + 2) , [2(8) + 2(2)]/(2 + 2)

= (4 - 4)/4, (16 + 4)/4

= 0/4, 20/4

= (0, 5)

E divides the line segment in the ratio 3 : 1

x_{1} = -2, y_{1} = 2, x_{2} = 2, y_{2} = 8, m = 3 and n = 1

= [3(2) + 1(-2)]/(3 + 1) , [3(8) + 1(2)]/(3 + 1)

= (6 - 2)/4, (24 + 2)/4

= 4/4, 26/4

= (1, 13/2)

**Example 4 :**

Find the area of a rhombus if its vertices are (3,0) (4,5) (-1,4) and (-2,-1) taken in order.[Hint:Area of rhombus = (1/2) product of its diagonals]

**Solution :**

Let A(3, 0) B(4, 5) C(-1, 4) and D(-2, -1) are the vertices of the rhombus.

length of diagonal AC = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

x_{1} = 3, y_{1} = 0, x_{2} = -1, y_{2} = 4

= √(-1 - 3)^{2} + (4 - 0)^{2}

= √(-4)^{2} + 4^{2}

= √16 + 16

= √32

x_{1} = 4, y_{1} = 5, x_{2} = -2, y_{2} = -1

= √(-2-4)² + (-1-5)²

= √(-6)² + (-6)²

= √36 + 36

= √72

Area of rhombus = (1/2) x √32 x √72

= (1/2) ⋅ 4 √2 ⋅ 6√2

** = 24 square units. **

After having gone through the stuff given above, we hope that the students would have understood, solving word problems using section formula.

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