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)
(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.
AB = \(\sqrt{(1+1)^2 + (0+2)^2}\)
= \(\sqrt{4+4}\)
= \(2\sqrt{2}\)
BC = \(\sqrt{(-1-1)^2 + (2-0)^2}\)
= \(\sqrt{4+4}\)
= \(2\sqrt{2}\)
CD = \(\sqrt{(-3+1)^2 + (0-2)^2}\)
= \(\sqrt{4+4}\)
= \(2\sqrt{2}\)
DA = \(\sqrt{(-3+1)^2 + (0-2)^2}\)
= \(\sqrt{4+4}\)
= \(2\sqrt{2}\)
AC = \(\sqrt{(-1+1)^2 + (2+2)^2}\)
= \(\sqrt{0+16}\)
= 4
BD = \(\sqrt{(-3-1)^2 + (0-0)^2}\)
= \(\sqrt{16+0}\)
= 4
Side length = AB = BC = CD = DA = \(2\sqrt{2}\)
Diagonal Measure = AC = BD = 4
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.
AB = \(\sqrt{(-3-3)^2 + (1-5)^2}\)
= \(\sqrt{36+16}\)
= \(2\sqrt{13}\)
BC = \(\sqrt{(0-3)^2 + (3-1)^2}\)
= \(\sqrt{9+4}\)
= \(\sqrt{13}\)
CD = \(\sqrt{(-1-0)^2 + (-4-3)^2}\)
= \(\sqrt{1+49}\)
= \(5\sqrt{2}\)
AD = \(\sqrt{(-1+3)^2 + (-4-5)^2}\)
= \(\sqrt{4+81}\)
= \(\sqrt{85}\)
Its also seen that points A, B and C are collinear.
So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral which has 4 sides.
Therefore, the given points cannot form a general quadrilateral.
(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.
AB = \(\sqrt{(7-4)^2 + (6-5)^2}\)
= \(\sqrt{9+1}\)
= \(\sqrt{10}\)
BC = \(\sqrt{(4-7)^2 + (3-6)^2}\)
= \(\sqrt{9+9}\)
= \(\sqrt{18}\)
CD = \(\sqrt{(1-4)^2 + (2-3)^2}\)
= \(\sqrt{9+1}\)
= \(\sqrt{10}\)
AD = \(\sqrt{(1-4)^2 + (2-5)^2}\)
= \(\sqrt{9+9}\)
= \(\sqrt{18}\)
AC =\(\sqrt{(4+4)^2 + (3-5)^2}\)
= \(\sqrt{0+4}\)
= 2
BD = \(\sqrt{(1-7)^2 + (2-6)^2}\)
= \(\sqrt{36+16}\)
= \(2\sqrt{13}\)
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.