(i)

\begin{align} \text{Mid-point } & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {4 + 6 \over 2}, {5 + 9 \over 2} \right) \\ & = (5, 7) \end{align}

(ii)

\begin{align} \text{Mid-point } & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-5 + 11 \over 2}, {3 + (-7) \over 2} \right) \\ & = (3, -2) \end{align}

(a)

\begin{align} \text{Let coordinates of } & A \text{ be } (x_A, y_A). \\ \\ \text{Mid-point, } M & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ (3, 5) & = \left( {x_A + 6 \over 2}, {y_A + 7 \over 2} \right) \\ \\ \text{Comparing the } & x \text{-coordinate}, \\ 3 & = {x_A + 6 \over 2} \\ 6 & = x_A + 6 \\ 6 - 6 & = x_A \\ 0 & = x_A \\ \\ \text{Comparing the } & y \text{-coordinate}, \\ 5 & = {y_A + 7 \over 2} \\ 10 & = y_A + 7 \\ 10 - 7 & = y_A \\ 3 & = y_A \\ \\ \therefore & \phantom{0} A(0, 3) \end{align}

(b)

\begin{align} \text{Let coordinates of } & A \text{ be } (x_A, y_A). \\ \\ \text{Mid-point, } M & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ (-2, 6) & = \left( {x_A + (-4) \over 2}, {y_A + (-8) \over 2} \right) \\ \\ \text{Comparing the } & x \text{-coordinate}, \\ -2 & = {x_A - 4 \over 2} \\ -4 & = x_A - 4 \\ -4 + 4 & = x_A \\ 0 & = x_A \\ \\ \text{Comparing the } & y \text{-coordinate}, \\ 6 & = {y_A - 8 \over 2} \\ 12 & = y_A - 8 \\ 12 + 8 & = y_A \\ 20 & = y_A \\ \\ \therefore & \phantom{0} A(0, 20) \end{align}

(a)

\begin{align} \text{Mid-point } & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {2a + 4a \over 2}, {-a + 5a \over 2} \right) \\ & = \left( {6a \over 2}, {4a \over 2} \right) \\ & = (3a, \phantom{/} 2a) \end{align}

(b)

\begin{align} \text{Mid-point } & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {2t + 4 \over 2}, {5 + (1 - 2t) \over 2} \right) \\ & = \left( {2(t + 2) \over 2}, {6 - 2t \over 2} \right) \\ & = \left( t + 2, {2(3 - t) \over 2} \right) \\ & = (t + 2, \phantom{/} 3 - t) \end{align}

(i)

\begin{align} \text{Coordinates of } D & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-1 + 3 \over 2}, {6 + 2 \over 2} \right) \\ & = (1, 4) \\ \\ \text{Coordinates of } E & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-1 + (-5) \over 2}, {6 + (-4) \over 2} \right) \\ & = (-3, 1) \end{align}

(ii)

\begin{align} \text{Gradient of } DE & = {y_2 - y_1 \over x_2 - x_1} \\ & = {4 - 1 \over 1 - (-3)} \\ & = {3 \over 4} \\ \\ y & = mx + c \\ y & = {3 \over 4}x + c \\ \\ \text{Using } D(1, 4), & \text{ let } x = 1 \text{ and } y = 4, \\ 4 & = {3 \over 4}(1) + c \\ 4 & = {3 \over 4} + c \\ 4 - {3 \over 4} & = c \\ {13 \over 4} & = c \\ \\ \therefore y & = {3 \over 4}x + {13 \over 4} \\ 4y & = 3x + 13 \end{align}

(i)

\begin{align} \text{Let coordinates} & \text{ of } B \text{ be } (x_B, y_B). \\ \\ \text{Centre, } C & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ (3, -4) & = \left( {9 + x_B \over 2} , {-6 + y_B \over 2} \right) \\ \\ \text{Comparing the } & x \text{-coordinate}, \\ 3 & = {9 + x_B \over 2} \\ 6 & = 9 + x_B \\ \\ x_B & = 6 - 9 \\ & = -3 \\ \\ \text{Comparing the } & y \text{-coordinate}, \\ -4 & = {-6 + y_B \over 2} \\ -8 & = -6 + y_B \\ \\ y_B & = -8 + 6 \\ & = - 2 \\ \\ \therefore B& \phantom{/} (-3, -2) \end{align}

(ii) Since AB is the diameter and C is the centre, AC or BC is the radius

\begin{align} \text{Length of } AC & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{ (9 - 3)^2 + [-6 - (-4)]^2 } \\ & = \sqrt{40} \\ & = \sqrt{4} \times \sqrt{10} \\ & = 2\sqrt{10} \text{ units} \end{align}

(i)

\begin{align} \text{Distance between two points } & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AB & = \sqrt{ [6 - (-2)]^2 + (-4 - 4)^2 } \\ & = \sqrt{128} \text{ units} \\ \\ BC & = \sqrt{ [5 - (-2)]^2 + (3 - 4)^2} \\ & = \sqrt{50} \text{ units} \\ \\ AC & = \sqrt{ (6 - 5)^2 + (-4 - 3)^2} \\ & = \sqrt{50} \text{ units} \\ \\ \text{Since } AC = BC, & \phantom{/} \triangle CBA \text{ is an isosceles triangle}. \end{align}

(ii)

\begin{align} \text{Mid-point of } AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {6 + (-2) \over 2}, {-4 + 4 \over 2} \right) \\ & = (2, 0) \end{align}

(iii)

\begin{align} \text{Height} & = \text{Distance from mid-point to } C \\ & = \sqrt{ (2 - 5)^2 + (0 - 3)^2} \\ & = \sqrt{18} \text{ units} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2} \times AB \times \sqrt{18} \\ & = {1 \over 2} \times \sqrt{128} \times \sqrt{18} \\ & = {1 \over 2} \times \sqrt{2304} \\ & = {1 \over 2} \times 48 \\ & = 24 \text{ sq. units} \end{align}

\begin{align} \text{Mid-point of }AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ (5, 1) & = \left( {p^2 + q^2 \over 2}, {p + q \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ 5 & = {p^2 + q^2 \over 2} \\ 10 & = p^2 + q^2 \phantom{000} \text{ --- (1)} \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ 1 & = {p + q \over 2} \\ 2 & = p + q \\ \\ p & = 2 - q \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 10 & = (2- q)^2 + q^2 \\ 10 & = 2^2 - 2(2)(q) + q^2 + q^2 \\ 10 & = 4 - 4q + 2q^2 \\ 0 & = -6 - 4q + 2q^2 \\ 0 & = 2q^2 - 4q - 6 \\ 0 & = q^2 - 2q - 3 \\ 0 & = (q - 3)(q + 1) \\ \end{align}
\begin{align} q - 3 & = 0 \phantom{00}&\text{or}\phantom{0000} q + 1 & = 0 \\ q & = 3 & q & = -1 \\ \\ \text{Substitute } & \text{into (2),} \\ p & = 2 - (3) & p & = 2 - (-1) \\ p & = -1 & p & = 3 \\ \\ \text{When } & p = -1 \text{ and } q = 3, \\ A & \phantom{.} ((-1)^2, -1) \\ A & \phantom{.} (1, - 1) \\ \\ B & \phantom{.} ((3)^2, 3) \\ B & \phantom{.} (9, 3) \\ \\ \\ \text{When } & p = 3 \text{ and } q = -1, \\ A & \phantom{.} ((3)^2, 3) \\ A & \phantom{.} (9, 3) \\ \\ B & \phantom{.} ((-1)^2, -1) \\ B & \phantom{.} (1, -1) \end{align}

\begin{align} 2x + y & = 1 \\ y & = 1 - 2x \phantom{000} \text{ --- (1)} \\ \\ y & = x^2 + 2x - 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 1 - 2x & = x^2 + 2x - 3 \\ 0 & = x^2 + 4x - 4 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(4) \pm \sqrt{4^2 - 4(1)(-4)} \over 2(1)} \\ & = {-4 \pm \sqrt{32} \over 2} \\ & = {-4 \pm \sqrt{16}\sqrt{2} \over 2} \\ & = {-4 \pm 4 \sqrt{2} \over 2} \\ & = -2 \pm 2\sqrt{2} \\ & = -2 + 2\sqrt{2} \phantom{0} \text{ or } \phantom{0} -2 - 2\sqrt{2} \end{align}
\begin{align} \text{Subs} & \text{titute into (1),} & \text{Subs} & \text{titute into (1),} \\ y & = 1- 2(-2 + 2\sqrt{2}) \phantom{00} & \text{or} \phantom{0000} y & = 1 - 2(-2 - 2\sqrt{2}) \\ & = 1 + 4 - 4\sqrt{2} & & = 1 + 4 + 4\sqrt{2} \\ & = 5 - 4\sqrt{2} & & = 5 + 4\sqrt{2} \end{align}
\begin{align} \therefore A(-2 + 2\sqrt{2}, 5 - 4\sqrt{2}) & \text{ and } B(-2 -2\sqrt{2}, 5 + 4\sqrt{2}) \\ \\ \text{Mid-point of } AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {(-2 + 2\sqrt{2}) + (-2 - 2\sqrt{2}) \over 2} , {(5 - 4\sqrt{2}) + (5 + 4\sqrt{2}) \over 2} \right) \\ & = \left( {-4 \over 2}, {10 \over 2} \right) \\ & = (-2, 5) \end{align}

\begin{align} \text{Let the coord} & \text{inates of } C \text{ be } (a, b). \\ \\ \text{Mid-point of } AC & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {3 + a \over 2}, {7 + b \over 2} \right) \\ \\ \text{Since the point lies on } & x \text{-axis, } \text{it's } y \text{-coordinate is } 0. \\ \therefore 0 & = {7 + b \over 2} \\ 0 & = 7 + b \\ -7 & = b \\ \\ \text{Mid-point of } BC & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-2 + a \over 2}, {5 + b \over 2} \right) \\ \\ \text{Since the point lies on } & y \text{-axis, } \text{it's } x \text{-coordinate is } 0. \\ \therefore 0 & = {-2 + a \over 2} \\ 0 & = -2 + a \\ -a & = -2 \\ a & = 2 \\ \\ \therefore C & \phantom{.} (2, - 7) \end{align}

(i)

\begin{align} \text{Mid-point} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ \text{Coordinate of } N & = \left( {-2 + 8 \over 2}, {-2 + (-4) \over 2} \right) \\ & = (3, -3) \\ \\ \text{Coordinate of } M & = \left( {6 + 8 \over 2} , {8 + (-4) \over 2} \right) \\ & = (7, 2) \\ \end{align}

Since P is the x-intercept of the line MN, form the equation of line MN and find it's x-intercept
\begin{align} \text{Gradient of } MN & = {y_2 - y_1 \over x_2 - x_1} \\ & = {-3 - 2 \over 3 - 7} \\ & = {-5 \over -4} \\ & = {5 \over 4} \\ \\ y & = mx + c \\ y & = {5 \over 4}x + c \\ \\ \text{Using } M(7, 2), & \text{ let } x = 7 \text{ and } y = 2, \\ 2 & = {5 \over 4}(7) + c \\ 2 & = {35 \over 4} + c \\ 2 - {35 \over 4} & = c \\ -{27 \over 4} & = c \\ \\ \therefore y & = {5 \over 4}x - {27 \over 4} \\ \\ \text{Equation of } MN: \phantom{0} 4y & = 5x - 27 \\ \\ \text{Let } & y = 0, \\ 4(0) & = 5x - 27 \\ 0 & = 5x - 27 \\ 27 & = 5x \\ {27 \over 5} & = x \\ \\ \therefore & \phantom{.} P \left( {27 \over 5}, 0 \right) \end{align}

(ii)

\begin{align} \require{cancel} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ MP & = \sqrt{ \left(7 - {27 \over 5}\right)^2 + (2 - 0)^2} \\ & = \sqrt{164 \over 25} \\ & = {\sqrt{164} \over \sqrt{25}} \\ & = {\sqrt{4} \sqrt{41} \over 5} \\ & = {2\sqrt{41} \over 5} \\ & = {2 \over 5} \sqrt{41} \\ \\ PN & = \sqrt{ \left({27 \over 5} - 3\right)^2 + (0 - (-3))^2} \\ & = \sqrt{369 \over 25} \\ & = {\sqrt{369} \over \sqrt{25}} \\ & = {\sqrt{9} \sqrt{41} \over 5} \\ & = {3 \sqrt{41} \over 5} \\ & = {3 \over 5} \sqrt{41} \\ \\ \\ MP &: PN \\ {2 \over 5}\cancel{\sqrt{41}} & : {3 \over 5} \cancel{\sqrt{41}} \\ {2 \over 5} & : {3 \over 5} \\ 2 & : 3 \end{align}

(i)

\begin{align} \text{Mid-point} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ M & \phantom{.} \left( {3 + 6 \over 2}, {-2 + 2 \over 2} \right) \\ M & \phantom{.} \left( {9 \over 2}, 0 \right) \end{align}

(ii) In a parallelogram, the diagonals bisect each other and have a common mid-point. Thus, M is the mid-point of the diagonals AC and BD

\begin{align} \text{Mid-point} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \left({9 \over 2}, 0\right) & = \left( {b + 7 \over 2}, {3 + d \over 2} \right) \\ \\ \text{Comparing the } & x \text{-coordinate}, \\ {9 \over 2} & = {b + 7 \over 2} \\ 9 & = b + 7 \\ 9 - 7 & = b \\ 2 & = b \\ \\ \text{Comparing the } & y \text{-coordinate}, \\ 0 & = {3 + d \over 2} \\ 0 & = 3 + d \\ -3 & = d \\ \\ \therefore b & = 2, d = -3 \end{align}

(i) All sides of a rhombus have the same length. Thus, OR = OP

\begin{align} \text{Distance between 2 points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ OR & = \sqrt{ (0 - 3)^2 + (0 - 4)^2} \\ & = 5 \\ \\ OP & = \sqrt{ (0 - 4)^2 + (0 - r)^2} \\ & = \sqrt{16 + r^2} \\ \\ OR & = OP \\ 5 & = \sqrt{16 + r^2} \\ 5^2 & = 16 + r^2 \\ 25 & = 16 + r^2 \\ 9 & = r^2 \\ \\ r & = \pm \sqrt{9} \\ & = \pm 3 \\ & = 3 \phantom{0} \text{ or } \phantom{0} -3 \text{ (Reject, since } P \text{ above } x \text{-axis}) \end{align}

(ii)

\begin{align} \text{Mid-point of } PR & \phantom{.} \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \text{Mid-point of } PR & \phantom{.} \left( {3 + 4 \over 2}, {4 + 3 \over 2} \right) \\ \text{Mid-point of } PR & \phantom{.} \left( {7 \over 2}, {7 \over 2} \right) \end{align}

(iii) In a rhombus, the diagonals have the same mid-point. Thus, the mid-point of PR is also the mid-point of OQ

\begin{align} \text{Let the coordi} & \text{nates of } Q \text{ be } (a, b). \\ \\ \text{Mid-point of } OQ & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \left({7 \over 2}, {7 \over 2} \right) & = \left( {0 + a \over 2}, {0 + b \over 2} \right) \\ \left({7 \over 2}, {7 \over 2} \right) & = \left( {a \over 2}, {b \over 2} \right) \\ \\ \text{Comparing the } & x \text{-coordinate}, \\ {7 \over 2} & = {a \over 2} \\ 7 & = a \\ \\ \text{Comparing the } & y \text{-coordinate}, \\ {7 \over 2} & = {b \over 2} \\ 7 & = b \\ \\ \therefore & \phantom{.} Q (7, 7) \end{align}

(a) In a parallelogram, the diagonals bisect each other and have a common mid-point. Thus, the diagonals PR and QS share the same mid-point

\begin{align} \text{Let the coord} & \text{inates of } S \text{ be } (a, b). \\ \\ \text{Mid-point} & = \left({x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ \text{Mid-point of }PR & = \left({1 + 6 \over 2}, {3 + 2 \over 2} \right) \\ & = \left( {7 \over 2}, {5 \over 2} \right) \\ \\ \text{Mid-point of }QS & = \left({3 + a \over 2}, {5 + b \over 2} \right) \\ \left({7 \over 2}, {5 \over 2}\right) & = \left({3 + a \over 2}, {5 + b \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate}, \\ {7 \over 2} & = {3 + a \over 2} \\ 7 & = 3 + a \\ 7 - 3 & = a \\ 4 & = a \\ \\ \text{Comparing the } & y\text{-coordinate}, \\ {5 \over 2} & = {5 + b \over 2} \\ 5 & = 5 + b \\ 5 - 5 & = b \\ 0 & = b \\ \\ \therefore & \phantom{.} S (4, 0) \end{align}

(b)

\begin{align} \text{Let the coord} & \text{inates of } T \text{ be } (c, d). \\ \\ \text{Since } T \text{ lies on} & \text{ the line } y = 2x, \\ d & = 2(c) \\ d & = 2c \\ \\ \therefore & \phantom{.} T(c, 2c) \\ \\ \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ PT & = \sqrt{ (1 - c)^2 + (3 - 2c)^2 } \\ & = \sqrt{ 1^2 - 2(1)(c) + c^2 + 3^2 - 2(3)(2c) + (2c)^2 } \\ & = \sqrt{ 1 - 2c + c^2 + 9 - 12c + 4c^2 } \\ & = \sqrt{ 10 - 14c + 5c^2 } \\ \\ QT & = \sqrt{ (3 - c)^2 + (5 - 2c)^2} \\ & = \sqrt{ 3^2 - 2(3)(c) + c^2 + 5^2 - 2(5)(2c) + (2c)^2 } \\ & = \sqrt{ 9 - 6c + c^2 + 25 - 20c + 4c^2 } \\ & = \sqrt{ 34 - 26c + 5c^2} \\ \\ PT & = QT \\ \sqrt{10 - 14c + 5c^2} & = \sqrt{34 - 26c + 5c^2} \\ 10 - 14c + 5c^2 & = 34 - 26c + 5c^2 \\ -14c + 26c & = 34 - 10 \\ 12c & = 24 \\ c & = {24 \over 12} \\ c & = 2 \\ \\ \therefore & \phantom{.} T(2, 4) \end{align}

\begin{align} \text{Let the coord} & \text{inates of the fourth vertex be } (e, f). \\ \\ \text{Let } A, B, C \text{ and } D & \text{ respectively represent the vertices } (-1, 6), (0, 0), (3, 1) \text{ and } (e, f). \end{align}

Case 1: Parallelogram ABCD, with diagonals AC and BD sharing the same mid-point \begin{align} \text{Mid-point} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ \text{Mid-point of } AC & = \left( {-1 + 3 \over 2}, {6 + 1 \over 2} \right) \\ & = \left( 1, {7 \over 2} \right) \\ \\ \text{Mid-point of } BD & = \left( {0 + e \over 2}, {0 + f \over 2} \right) \\ \left(1, {7 \over 2} \right) & = \left( {e \over 2}, {f \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate}, \\ 1 & = {e \over 2} \\ 2 & = e \\ \\ \text{Comparing the } & y\text{-coordinate}, \\ {7 \over 2} & = {f \over 2} \\ 7 & = f \\ \\ \therefore & \phantom{.} D (2, 7) \end{align}

Case 2: Parallelogram ACDB, with diagonals AD and CB sharing the same mid-point \begin{align} \text{Mid-point of } AD & = \left( {-1 + e \over 2}, {6 + f \over 2} \right) \\ & = \left( {e - 1 \over 2}, {6 + f \over 2} \right) \\ \\ \text{Mid-point of } CB & = \left( {3 + 0 \over 2}, {1 + 0 \over 2} \right) \\ \left( {e - 1 \over 2}, {6 + f \over 2} \right) & = \left( {3 \over 2}, {1 \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate}, \\ {e - 1 \over 2} & = {3 \over 2} \\ e - 1 & = 3 \\ e & = 3 + 1 \\ & = 4 \\ \text{Comparing the } & y\text{-coordinate}, \\ {6 + f \over 2} & = {1 \over 2} \\ 6 + f & = 1 \\ f & = 1 - 6 \\ & = -5 \\ \\ \therefore \phantom{.} & D(4, -5) \end{align}

Case 3: Parallelogram ACBD, with diagonals AB and CD sharing the same mid-point \begin{align} \text{Mid-point of } AB & = \left( {-1 + 0 \over 2}, {6 + 0 \over 2} \right) \\ & = \left( -{1 \over 2}, 3 \right) \\ \\ \text{Mid-point of } CD & = \left( {3 + e \over 2}, {1 + f \over 2} \right) \\ \left( -{1 \over 2}, 3 \right) & = \left( {3 + e \over 2}, {1 + f \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate}, \\ -{1 \over 2} & = {3 + e \over 2} \\ -1 & = 3 + e \\ -1 - 3 & = e \\ -4 & = e \\ \\ \text{Comparing the } & y\text{-coordinate}, \\ 3 & = {1 + f \over 2} \\ 6 & = 1 + f \\ 6 - 1 & = f \\ 5 & = f \\ \\ \therefore \phantom{.} & D(-4, 5) \end{align}

(i)

\begin{align} \text{Mid-point} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ \text{Mid-point of } PR, M & = \left( {-4 + r \over 2}, {2 + 5 \over 2} \right) \\ & = \left( {r - 4 \over 2}, {7 \over 2} \right) \\ \\ \text{Mid-point of } QS, N & = \left( {6 + 0 \over 2}, {-2 + s \over 2} \right) \\ & = \left( 3, {s - 2 \over 2} \right) \end{align}

(ii)

\begin{align} \text{Mid-point of } PR, M & = \text{Mid-point of } QS, N \\ \\ \left( {r - 4 \over 2}, {7 \over 2} \right) & = \left( 3, {s - 2 \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate}, \\ {r - 4 \over 2} & = 3 \\ r - 4 & = 6 \\ r & = 6 + 4 \\ & = 10 \\ \\ \text{Comparing the } & y\text{-coordinate}, \\ {7 \over 2} & = {s - 2 \over 2} \\ 7 & = s - 2 \\ 7 + 2 & = s \\ 9 & = s \\ \\ \therefore r & = 10, s = 9 \end{align}

(iii)

If the diagonals have the same mid-point, $PQRS$ is a parallelogram.

\begin{align} M & = \left({0 + b \over 2}, {a + 0 \over 2} \right) \\ & = \left({b \over 2}, {a \over 2} \right) \\ \\ OM & = \sqrt{ \left(0 - {b \over 2}\right)^2 + \left(0 -{a \over 2}\right)^2 } \\ & = \sqrt{ {b^2 \over 4} + {a^2 \over 4} } \\ \\ AM & = \sqrt{ \left(0 - {b \over 2}\right)^2 + \left(a - {a \over 2}\right)^2} \\ & = \sqrt{ {b^2 \over 4} + \left(a \over 2\right)^2 } \\ & = \sqrt{ {b^2 \over 4} + {a^2 \over 4} } \\ \\ BM & = \sqrt{ \left(b - {b \over 2}\right)^2 + \left(0 - {a \over 2}\right)^2 } \\ & = \sqrt{ \left(b \over 2\right)^2 + {a^2 \over 4} } \\ & = \sqrt{ {b^2 \over 4} + {a^2 \over 4} } \\ \\ \therefore M & \text{ is equidistant from } O, A \text{ and } B \end{align}

(i)

\begin{align} A & = \left( {0 + 2a \over 2}, {0 + 0 \over 2} \right) \\ & = (a, 0) \\ \\ B & = \left( {2a + 2b \over 2}, {0 + 2c \over 2} \right) \\ & = \left( {2(a + b) \over 2}, c \right) \\ & = (a + b, c) \\ \\ C & = \left( {2b + 2d \over 2}, {2c + 2e \over 2} \right) \\ & = \left( {2(b + d) \over 2}, {2(c + e) \over 2} \right) \\ & = (b + d, c + e) \\ \\ D & = \left( {2d + 0 \over 2}, {2e + 0 \over 2} \right) \\ & = (d, e) \end{align}

(ii)

\begin{align} m_{AB} & = {0 - c \over a - (a + b)} \\ & = {-c \over a - a - b} \\ & = {-c \over -b} \\ & = {c \over b} \\ \\ m_{DC} & = {c + e - e \over b + d - d} \\ & = {c \over b} \\ \\ \therefore & \phantom{.} AB \phantom{.} // \phantom{.} DC \\ \\ \\ m_{BC} & = {c + e - c \over b + d - (a + b)} \\ & = {e \over b + d - a - b} \\ & = {e \over d - a} \\ \\ m_{AD} & = {e - 0 \over d - a} \\ & = {e \over d - a} \\ \\ \therefore & \phantom{.} BC \phantom{.} // \phantom{.} AD \\ \\ \\ \text{Hence } & ABCD \text{ is a parallelogram} \end{align}

\begin{align} \text{Length of rectangle} & = m \text{ units} \\ \\ \text{Assume coordinates of } & A \text{ is } (m, 0) \\ \\ \text{Breadth of rectangle} & = n \text{ units} \\ \\ \text{Assume coordinates of } & C \text{ is } (0, n) \\ \\ \text{Coordinates of } & B \text{ is } (m, n) \\ \\ \text{Midpoint of } OB & = \left({0 + m \over 2}, {0 + n \over 2} \right) \\ & = \left({m \over 2}, {n \over 2}\right) \\ \\ \text{Midpoint of } AC & = \left({m + 0 \over 2}, {0 + n \over 2} \right) \\ & = \left({m \over 2}, {n \over 2}\right) \\ \\ \text{Since diagonals have a comm} & \text{on midpoint, they bisect each other} \end{align}

(a)(i) Since OABC is a square, diagonals OB and AC have the same midpoint

\begin{align} \text{Midpoint of } OB & = \left( {0 + 2 \over 2}, {0 + 4 \over 2} \right) \\ & = (1, 2) \end{align}

(a)(ii)

\begin{align} \text{Let } M & \text{ denote midpoint} \\ \\ OB & = \sqrt{(0 - 2)^2 + (0 - 4)^2} \\ & = \sqrt{20} \\ & = \sqrt{4} \sqrt{5} \\ & = 2 \sqrt{5} \\ \\ OM & = \sqrt{5} \\ \\ \text{Let } A & \text{ be } (x, y) \\ \\ AM & = \sqrt{(x - 1)^2 + (y - 2)^2} \\ \\ \text{Since } & AM = OM, \\ \sqrt{5} & = \sqrt{(x - 1)^2 + (y - 2)^2 } \\ 5 & = (x - 1)^2 + (y - 2)^2 \phantom{00} \text{--- (1)} \\ \\ \\ \text{Since } & OABC \text{ is a square,} \\ OA & = AB \\ \sqrt{(0 - x)^2 + (0 - y)^2} & = \sqrt{(x - 2)^2 + (y - 4)^2} \\ \sqrt{x^2 + y^2} & = \sqrt{(x - 2)^2 + (y - 4)^2} \\ x^2 + y^2 & = (x - 2)^2 + (y - 4)^2 \\ x^2 + y^2 & = x^2 - 4x + 4 + y^2 - 8y + 16 \\ 0 & = - 4x - 8y + 20 \\ 4x & = -8y + 20 \\ x & = - 2y + 5 \phantom{00} \text{--- (2)} \\ \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 5 & = (-2y + 5 - 1)^2 + (y - 2)^2 \\ 5 & = (4 - 2y)^2 + (y - 2)^2 \\ 5 & = 16 - 16y + 4y^2 + y^2 - 4y + 4 \\ 0 & = 5y^2 - 20y + 15 \\ 0 & = y^2 - 4y + 3 \\ 0 & = (y - 1)(y - 3) \end{align} \begin{align} y - 1 & = 0 && \text{ or } & y - 3 & = 0 \\ y & = 1 &&& y & = 3 \\ \\ \text{Substitute } & \text{ into (2),} &&& \text{Substitute } & \text{ into (2),} \\ x & = -2(1) + 5 &&& x & = -2(3) + 5 \\ x & = 3 &&& x & = -1 \\ \\ \therefore & \phantom{.} A(3, 1) &&& \therefore & \phantom{.} A(-1, 3) \\ \therefore & \phantom{.} C(-1, 3) &&& \therefore & \phantom{.} C(3, 1) \end{align}

(a)(iii)

\begin{align} &\text{(a)(i) } (4, 5) \\ \\ & \text{(a)(ii) } A(6, 4), C(2, 6) \text{ or } A(2, 6), C(6, 4) \end{align}

(b)

\begin{align} &\text{(a)(i) } (1 + m, 2 + m) \\ \\ & \text{(a)(ii) } A(3 + m, 1 + m), C(-1 + m, 3 + m) \text{ or } A(-1 + m, 3 + m), C(3 + m, 1 + m) \end{align}

\begin{align} \text{Let } a \text{ denote } & x \text{-coordinate of } A \\ \\ \text{Since } A \text{ lies on } & y = 2x, A(a, 2a) \\ \\ \text{Let } b \text{ denote } & x \text{-coordinate of } B \\ \\ \text{Since } B \text{ lies on } & y = x + 3, B(b, b + 3) \\ \\ \\ \text{Midpoint of } AB & = \left( {a + b \over 2}, {2a + b + 3 \over 2} \right) \\ (2, 6) & = \left( {a + b \over 2}, {2a + b + 3 \over 2} \right) \\ \\ 2 & = {a + b \over 2} \\ 2(2) & = a + b \\ 4 & = a + b \phantom{00} \text{--- (1)} \\ \\ 6 & = {2a + b + 3 \over 2} \\ 2(6) & = 2a + b + 3 \\ 12 & = 2a + b + 3 \\ 9 - 2a & = b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 4 & = a + 9 - 2a \\ -5 & = - a\\ 5 & = a \\ \\ \text{Substitute } & a = 5 \text{ into (2),} \\ b & = 9 - 2(5) \\ b & = -1 \\ \\ \therefore & \phantom{.} A(5, 10) \\ \therefore & \phantom{.} B(-1, 2) \end{align}