Q. 64.1( 216 Votes )

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)

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

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

Answer :

To Find: Type of quadrilateral formed

(i) Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x1, y1) and (x2, y2).

D = √(x2 - x1)2 + (y2 - y1)2

AB = [(-1- 1)2 + (-2- 0)2]1/2


= =


= 2√2


BC = √[(1 + 1)2 + (0- 2)2]


= =


= 2√2


CD = √[(-1 + 3)2 + (2- 0)2]


= =


= 2√2


AD = √[(-1+ 3)2 + (-2- 0)2]


= =


= 2√2


Diagonal AC = √[(-1 + 1)2 + (-2 - 2)2]


=


= 4


Diagonal BD = √[(1 + 3)2 + (0- 0)2]


=


= 4


It is clear that all sides of this quadrilateral are of the same length and the diagonals are of the same length. Therefore, the given points are the vertices of a square


(ii)Let the points (- 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively


The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x1, y1) and (x2, y2).

D = √(x2 - x1)2 + (y2 - y1)2

AB = √[(-3- 3)2 + (5- 1)2]


= =


= 2√13


BC = √[(3- 0)2 + (1- 3)2]


=


=


CD = √[(0 + 1)2 + (3+ 4)2]


= =


= 5√2


AD = √[(-3+ 1)2 + (5+ 4)2]


=


=


We can observe that all sides of this quadrilateral are of different lengths.


Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc


(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x1, y1) and (x2, y2).

D = √(x2 - x1)2 + (y2 - y1)2


AB = √[(4- 7)2 + (5 - 6)2]


=


=


BC = √[(7- 4)2 + (6 - 3)2]


=


=


CD = √[(4- 1)2 + (3 - 2)2]


=


=


AD = √[(4- 1)2 + (5 - 2)2]


=


=


Diagonal AC =√[(4 - 4)2 + (5- 3)2]


=


= 2


Diagonal BD = √[(7 - 1)2 + (6 - 2)2]


=


=


= 2√13


We can observe that opposite sides of this quadrilateral are of the same length.


However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram

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