WHAT TYPE OF QUADRILATERAL IS FORMED BY CONNECTING THE POINTS

Examining the Type of Quadrilateral is Formed by Connecting the Points

Example 1 :

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

(4,5) (7,6) (4,3) (1,2)

Solution :

Let the given points as A(4, 5)  B(7, 6)  C(4, 3) and D (1, 2)

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

Length of the side AB :

Here, x₁  =  4, y₁  =  5, x₂  =  7  and  y₂  =  6

  =  √(7-4)² + (6-5)²

  =  √3² + 1²

  =  √9 + 1

  =  √10

Length of the side BC :

Here, x₁  =  7, y₁  =  6, x₂  =  4  and  y₂  =  3

  =  √(4-7)² + (3-6)²

  =  √(-3)² + (-3)²

  =  √9 + 9

  =  √18

Length of the side CD :

Here, x₁  =  4, y₁  =  3, x₂  =  1  and  y₂  =  2

  =  √(1-4)² + (2-3)²

  =  √(-3)² + (-1)²

  =  √9 + 1

  =  √10

Length of the side DA :

Here, x₁  =  1, y₁  =  2, x₂  =  4  and  y₂  =  5

  =  √(4-1)² + (5-2)²

  =  √3² + 3²

  =  √9 + 9

  =  √18

AB  =  CD, BC  =  DA. length of opposite sides are equal.

Length of AC :

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

  =  √(4-4)² + (3-5)²

  =  √0² + (-2)²

  =  √4

  =  2

Length of BD :

Here, x₁  =  7, y₁  =  6, x₂  =  1  and  y₂  =  2

  =  √(1-7)² + (2-6)²

  =  √(-6)² + (-4)²

  =  √36 + 16

  =  √52

It can be observed that opposite sides of this quadrilateral are of the same length.

Since the diagonals are of different lengths, the given points are the vertices of the parallelogram.

Example 2 :

Find the points on the x-axis which is equidistant from (2, -5) and (-2, 9)

Solution :

We have to find the point on x-axis. So its y-coordinate will be 0.

Let the point on x axis be (x, 0)

Distance between (x, 0) and (2, -5) =  
Distance between (x, 0) and (-2, 9)

Distance between two points

= √(x₂ - x₁)² + (y₂ - y₁)²

Here, x₁  =  x, y₁  =  0, x₂  =  2  and  y₂  =  -5

                 = √(2 - x)² + (5 - 0)²

                 = √(2 - x)² + 25     ---------------(1)

Here, x₁  =  x, y₁  =  0, x₂  =  -2  and  y₂  =  9

  =  √(-2-x)² + (9-0)²

  =  √(2+x)² + 9²

  =  √(2+x)² + 81

√(2 - x)² + 25 = √(2 + x)² + 81

Taking squares on both sides, we get

 4 + x² - 4 x + 25 = 4 + x² + 4 x + 81

 x² - x² - 4 x - 4 x + 4 - 4 = 81 - 25

-8 x = 56

x = -7

So, the required point is (-7, 0)

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