What Type of Quadrilateral is Formed by Connecting the Points ?
In this section, we will see some examples to examine which type of quadrilateral formed by connecting the given 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 = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}
Length of the side AB :
Here, x_{1 }= 4, y_{1} = 5, x_{2} = 7 and y_{2} = 6
= √(7-4)^{2} + (6-5)^{2}
= √3² + 1²
= √9 + 1
= √10
Length of the side BC :
Here, x_{1 }= 7, y_{1} = 6, x_{2} = 4 and y_{2} = 3
= √(4-7)² + (3-6)²
= √(-3)² + (-3)²
= √9 + 9
= √18
Length of the side CD :
Here, x_{1 }= 4, y_{1} = 3, x_{2} = 1 and y_{2} = 2
= √(1-4)^{2} + (2-3)^{2}
= √(-3)^{2} + (-1)^{2}
= √9 + 1
= √10
Length of the side DA :
Here, x_{1 }= 1, y_{1} = 2, x_{2} = 4 and y_{2} = 5
= √(4-1)^{2} + (5-2)^{2}
= √3^{2} + 3^{2}
= √9 + 9
= √18
AB = CD, BC = DA. length of opposite sides are equal.
Length of AC :
Here, x_{1 }= 4, y_{1} = 5, x_{2} = 4 and y_{2} = 3
= √(4-4)² + (3-5)²
= √0² + (-2)²
= √4
= 2
Length of BD :
Here, x_{1 }= 7, y_{1} = 6, x_{2} = 1 and y_{2} = 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
= √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}
Here, x_{1} = x, y_{1 } = 0, x_{2} = 2 and y_{2} = -5
= √(2 - x)^{2} + (5 - 0)^{2}
= √(2 - x)^{2} + 25 ---------------(1)
Here, x_{1} = x, y_{1 } = 0, x_{2} = -2 and y_{2} = 9
= √(-2-x)^{2} + (9-0)^{2}
= √(2+x)^{2} + 9^{2}
= √(2+x)^{2} + 81
√(2 - x)^{2} + 25 = √(2 + x)^{2} + 81
Taking squares on both sides, we get
4 + x^{2} - 4 x + 25 = 4 + x^{2} + 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)
After having gone through the stuff given above, we hope that the students would have understood, what type of quadrilateral is formed by connecting the points.
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