NAME THE TYPE OF QUADRILATERAL FORMED BY CONNECTING THE POINTS

Question :

Name the type of quadrilateral formed. If any, by the following points and give reasons for your answer:

(i) (-1, -2) (1, 0) (-1, 2) (-3, 0)

Solution :

Let the given points as A(-1, -2)  B(1, 0)  C(-1, 2) and D (-3, 0)

Distance between two points  =  √(x₂ - x₁)² + (y₂ - y₁) ²

Length of the side AB :

Here, x₁  =  -1, y₁  =  -2, x₂  =  1  and  y₂  =  0

  =  √(1-(-1))² + (0-(-2))²

  =  √(1+1)² + 2²

  =  √4 + 4

  =  √8

Length of the side BC :

Here, x₁  =  1, y₁  =  0, x₂  =  -1  and  y₂  =  2

  =  √(-1-1)² + (2-0)²

  =  √(-2)² + 2²

  =  √4 + 4

  =  √8

Length of the side CD :

Here, x₁  =  -1, y₁  =  2, x₂  =  -3  and  y₂  =  0

  =  √(-3-(-1))² + (0-2)²

  =  √(-3+1)² + (-2)²

  =  √4 + 4

  =  √8

Length of the side DA :

  =  √(-1-(-3))² + (-2-0)²

  =  √(-1+3)² + (-2)²

  =  √4 + 4

  =  √8

AB = BC = CD = DA. Since length of all sides are equal, the given points are the vertices of square.

(ii) (-3, 5) (3, 1) (0, 3) (-1, -4)

Solution :

Let the given points as A(-3,5)  B(3,1)  C(0,3) and D (-1,-4)

Distance between two points  =  √(x₂ - x₁)² + (y₂ - y₁) ²

Length of the side AB :

Here, x₁  =  -3, y₁  =  5, x₂  =  3  and  y₂  =  1

  =  √(3-(-3))² + (1-5)²

  =  √(3+3)² + (-4)²

  =  √36 + 16

  =  √52

Length of the side BC :

Here, x₁  =  3, y₁  =  1, x₂  =  0  and  y₂  =  3

  =  √(0-3)² + (3-1)²

  =  √(-3)² + (2)²

  =  √9 + 4

  =  √13

Length of the side CD :

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

  =  √(-1-0)² + (-4-3)²

  =  √(-1)² + (-7)²

  =  √1 + 49

  =  √50

Length of the side DA :

Here, x₁  =  -1, y₁  =  -4, x₂  =  -3  and  y₂  =  5

  =  √(-3-(-1))² + (5-(-4))²

  =  √(-3+1)² + (5 +4)²

  =  √4 + 81

  =  √85

The length of all sides of quadrilateral are of different. Therefore, it can be observed that it is only a quadrilateral. 

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