Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) Square
(ii) No quadrilateral
(iii) Parallelogram
Let the points be \(A(-1, -2)\), \(B(1, 0)\), \(C(-1, 2)\) and \(D(-3, 0)\).
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), find the side lengths.
\(AB = \sqrt{(1 + 1)^2 + (0 + 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} \).
\(BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} \).
\(CD = \sqrt{(-3 + 1)^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} \).
\(DA = \sqrt{(-1 + 3)^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} \).
So all four sides are equal.
Now find diagonals: \(AC = \sqrt{(-1 + 1)^2 + (2 + 2)^2} = \sqrt{0^2 + 4^2} = 4\) and \(BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2} = 4\).
All sides are equal and both diagonals are equal, so \(ABCD\) is a square.
Let the points be \(P(-3, 5)\), \(Q(3, 1)\), \(R(0, 3)\) and \(S(-1, -4)\).
Find slopes to check collinearity. Slope of \(PQ\): \( m_{PQ} = \frac{1 - 5}{3 + 3} = \frac{-4}{6} = -\frac{2}{3} \).
Slope of \(PR\): \( m_{PR} = \frac{3 - 5}{0 + 3} = \frac{-2}{3} = -\frac{2}{3} \).
Slope of \(QR\): \( m_{QR} = \frac{3 - 1}{0 - 3} = \frac{2}{-3} = -\frac{2}{3} \).
Since \( m_{PQ} = m_{PR} = m_{QR} \), the points \(P, Q, R\) lie on the same straight line.
Three collinear points cannot form a quadrilateral, so these four points do not determine a quadrilateral.
Let the points be \(A(4, 5)\), \(B(7, 6)\), \(C(4, 3)\) and \(D(1, 2)\).
First, find the slopes of opposite sides.
Slope of \(AB\): \( m_{AB} = \frac{6 - 5}{7 - 4} = \frac{1}{3} \).
Slope of \(CD\): \( m_{CD} = \frac{2 - 3}{1 - 4} = \frac{-1}{-3} = \frac{1}{3} \), so \(AB \parallel CD\).
Slope of \(BC\): \( m_{BC} = \frac{3 - 6}{4 - 7} = \frac{-3}{-3} = 1 \).
Slope of \(AD\): \( m_{AD} = \frac{2 - 5}{1 - 4} = \frac{-3}{-3} = 1 \), so \(BC \parallel AD\).
Now check lengths: \(AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{10} \) and \(CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10} \).
Also, \(BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} \) and \(AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} \).
Opposite sides are equal and parallel, so \(ABCD\) is a parallelogram.