SOLVING WORD PROBLEMS USING SECTION FORMULA

Solving Word Problems Using Section Formula :

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

Solving Word Problems Using 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 x₁  =  a, y₁  =  b, x₂  =  1, y₂  =  4 

Midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/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

x₁  =  -2, y₁  =  -2, x₂  =  2, y₂  =  -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

x₁  =  -2, y₁  =  2, x₂  =  2, y₂  =  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

x₁  =  -2, y₁  =  2, x₂  =  2, y₂  =  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

x₁  =  -2, y₁  =  2, x₂  =  2, y₂  =  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 =  √(x₂ - x₁)² + (y₂ - y₁)²

x₁  =  3, y₁  =  0,  x₂  =  -1, y₂  =  4

  =  √(-1 - 3)² + (4 - 0)²

  =  √(-4)² + 4²

  =  √16 + 16

  =  √32

x₁  =  4, y₁  =  5,  x₂  =  -2, y₂  =  -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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