\begin{align} \text{Gradient of } AB & = {7 - 1 \over 3 - 6} \\ & = - 2 \\ \\ \text{Gradient of } BC & = {1 - 8 \over 6 - 20} \\ & = {1 \over 2} \\ \\ \text{Gradient of } AB \times \text{Gradient of } BC & = -2 \times {1 \over 2} \\ & = - 1 \\ \\ \therefore AB & \perp BC \end{align}

\begin{align} \text{Gradient of } AB & = {-1 - 4 \over 2 - 5} \\ & = {5 \over 3} \\ \\ \text{Gradient of } BC & = {4 - (-2) \over 5 - 15} \\ & = -{3 \over 5} \\ \\ \text{Gradient of } AB \times \text{Gradient of } BC & = {5 \over 3} \times -{3 \over 5} \\ & = - 1 \\ \\ \therefore AB \perp BC, & \phantom{0} \angle ABC = 90^\circ \end{align}

\begin{align} \text{Gradient of } AB & = {3 - (-3) \over a - 2} \\ & = {6 \over a - 2} \\ \\ \text{Gradient of } BC & = {-3 - 1 \over 2 - 10} \\ & = {1 \over 2} \\ \\ \text{Gradient of } AB \times \text{Gradient of } BC & = {6 \over a - 2} \times {1 \over 2} \\ -1 & = {6 \over 2(a - 2)} \\ -2(a - 2) & = 6 \\ -2a + 4 & = 6 \\ -2a & = 6 - 4 \\ -2a & = 2 \\ a & = {2 \over -2} \\ & = -1 \end{align}

(i) If $\angle ABC = 90^\circ$, lines $AB$ and $BC$ are perpendicular

\begin{align} \text{Gradient of } AB & = {2 - 0 \over 1 - 9} \\ & = -{1 \over 4} \\ \\ \text{Gradient of } BC & = {0 - t \over 9 - 6} \\ & = -{t \over 3} \\ \\ \text{Gradient of } AB \times \text{Gradient of } BC & = - 1 \\ -{1 \over 4} \times -{t \over 3} & = -1 \\ {t \over 12} & = -1 \\ t & = -12 \end{align}

(ii)

\begin{align} \text{Gradient of } AC & = {2 - t \over 1 - 6} \\ & = {2 - t \over -5} \\ \\ \text{Gradient of } AC \times \text{Gradient of } BC & = -1 \\ {2 - t \over -5} \times -{t \over 3} & = - 1 \\ {-t(2 - t) \over -15} & = - 1 \\ -t(2 - t) & = (-15)(-1) \\ -t(2 - t) & = 15 \\ -2t + t^2 & = 15 \\ t^2 - 2t - 15 & = 0 \\ (t - 5)(t + 3) & = 0 \\ \\ t - 5 = 0 \phantom{00} & \text{or} \phantom{000} t + 3 = 0 \\ t = 5 \phantom{00} & \phantom{or000+3} t = - 3 \end{align}

\begin{align} x + 2y - 2 & = 0 & \\ 2y & = -x + 2 & y - kx + 4 & = 0 \\ y & = -{1 \over 2}x + 1 & y & = kx - 4 \\ \\ \implies \text{Gradient} & = -{1 \over 2} & \text{Gradient} & = k \\ \\ \therefore -{1 \over 2} \times k & = -1 \\ -{1 \over 2}k & = -1 \\ k & = {-1 \over -{1 \over 2}} \\ & = 2 \end{align}

\begin{align} y - 3x & = 4 \\ y & = 3x + 4 \\ \\ \implies \text{Gradient} & = 3 \\ \\ \text{Gradient of line} \times (3) & = -1 \\ \text{Gradient of line} & = -{1 \over 3} \\ \\ \therefore a & = -{1 \over 3} \\ \\ y & = -{1 \over 3}x + b \\ \\ \text{Using } (1, -2), & \text{ let } x = 1 \text{ and } y = -2, \\ -2 & = -{1 \over 3}(1) + b \\ -2 + {1 \over 3} & = b \\ -{5 \over 3} & = b \\ \\ \therefore a & = -{1 \over 3}, b = -{5 \over 3} \end{align}

\begin{align} x + 2y - 4 & = 0 \\ 2y & = - x + 4 \\ y & = -{1 \over 2}x + 2 \\ \\ \implies \text{Gradient} & = -{1 \over 2} \\ \\ \text{Gradient of line} \times \left(-{1 \over 2}\right) & = - 1 \\ \text{Gradient of line} & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } (4, 5), & \text{ let } x = 4 \text{ and } y = 5, \\ 5 & = 2(4) + c \\ 5 & = 8 + c \\ 5 - 8 & = c \\ -3 & = c \\ \\ \therefore \text{Equation of line: } \phantom{0} y & = 2x - 3 \end{align}

\begin{align} y & = x - 1 \phantom{0} \text{ --- (1)} \\ y & = x^2 - x \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{ (1) into (2)}, \\ x - 1 & = x^2 - x \\ 0 & = x^2 - 2x + 1 \\ 0 & = (x - 1)^2 \\ 0 & = (x - 1) \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into (1)}, \\ y & = 1 - 1 \\ & = 0 \\ \\ \therefore A & \phantom{.}(1, 0) \\ \\ y & = x - 1 \\ \\ \implies \text{Gradient} & = 1 \\ \\ \text{Gradient of line} \times (1) & = -1 \\ \text{Gradient of line} & = -1 \\ \\ \text{Equation of line: } \phantom{0} y & = mx + c \\ y & = (-1)x + c \\ y & = -x + c \\ \\ \text{Using } A(1, 0), & \text{ let } x = 1 \text{ and } y = 0, \\ 0 & = -(1) + c \\ 0 & = -1 + c \\ 1 & = c \\ \\ \therefore \text{Equation of line: } y & = -x + 1 \end{align}

\begin{align} \text{Mid-point of }AB & = \left( {3 + 7 \over 2}, {3 + 3 \over 2}\right) \\ & = (5, 3) \\ \\ \text{Gradient of } AB & = {3 - 3 \over 3 - 7} \\ & = 0 \end{align}

Since $AB$ is a horizontal line, the perpendicular bisector is a vertical line that passes through the mid-point $(5, 3)$
$$ \text{Equation: } \phantom{0} x = 5 $$

\begin{align} x + y & = 8 \\ x & = 8 - y \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 6x - 32 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ y^2 & = 6(8 - y) - 32 \\ y^2 & = 48 - 6y - 32 \\ y^2 + 6y - 16 & = 0 \\ (y + 8)(y - 2) & = 0 \\ \\ y + 8 = 0 \phantom{00} & \phantom{00} y - 2 = 0 \\ y = - 8 \phantom{0} & \phantom{00000} y = 2 \\ \\ \text{Substitute } & \text{back into (1),} \\ x = 8 - (-8) \phantom{00} & \phantom{00} x = 8 - (2) \\ = 16 \phantom{0000000} & \phantom{00x} = 6 \\ \\ \therefore A (16, -8), & \phantom{.} B(6,2) \\ \\ \text{Mid-point of } AB & = \left({16 + 6 \over 2}, {-8 + 2 \over 2}\right) \\ & = (11, -3) \\ \\ \text{Gradient of } AB & = {-8 - 2 \over 16 - 6} \\ & = -1 \\ \\ \text{Gradient of } \perp \text{bisector} \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } \perp \text{bisector} \times (-1) & = -1 \\ \text{Gradient of } \perp \text{bisector} & = {-1 \over -1} \\ & = 1 \\ \\ y & = mx + c \\ y & = (1)x + c \\ y & = x + c \\ \\ \text{Using mid-point } (11, -3), & \text{ let } x = 11 \text{ and } y = -3, \\ -3 & = 11 + c \\ -14 & = c \\ \\ \therefore \text{Equation of } \perp \text{ bisector: } & y = x - 14 \end{align}

(i)

\begin{align} \text{Gradient of } BC & = {-1 - 7 \over 2 - 6} \\ & = 2 \\ \\ \text{Gradient of } BC \times \text{Gradient of } AF & = -1 \\ (2) \times \text{Gradient of } AF & = -1 \\ \text{Gradient of } AF & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } A(3, 6), & \text{ let } x = 3 \text{ and } y = 6, \\ 6 & = -{1 \over 2}(3) + c \\ 6 & = -{3 \over 2} + c \\ 6 + {3 \over 2} & = c \\ {15 \over 2} & = c \\ \\ \therefore y & = -{1 \over 2}x + {15 \over 2} \\ 2y & = -x + 15 \\ 2y + x & = 15 \end{align}

(ii)

Form the equation of $BC$. Note from the previous part, gradient of $BC$ is $2$
\begin{align} y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } B(2, -1), & \text{ let } x = 2 \text{ and } y = -1, \\ -1 & = 2(2) + c \\ -1 & = 4 + c \\ -1 - 4 & = c \\ -5 & = c \\ \\ y & = 2x - 5 \end{align}

$F$ is the point of intersection between lines $AF$ and $BC$
\begin{align} y & = 2x - 5 \phantom{00} \text{ --- (1)} \\ \\ 2y + x & = 15 \phantom{00} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2)}, \\ 2(2x - 5) + x & = 15 \\ 4x - 10 + x & = 15 \\ 5x & = 15 + 10 \\ 5x & = 25 \\ x & = {25 \over 5} \\ x & = 5 \\ \\ \text{Substitute } & x = 5 \text{ into (1)}, \\ y & = 2(5) - 5 \\ & = 5 \\ \\ \therefore & \phantom{/} F(5,5) \end{align}

(iii)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{ (3 - 5)^2 + (6 - 5)^2} \\ & = \sqrt{5} \text{ units} \end{align}

(i)

\begin{align} x - 4y & = 8 \\ -4y & = -x + 8 \\ 4y & = x - 8 \\ y & = {1 \over 4}x - 2 \\ \\ \implies \text{Gradient} & = {1 \over 4} \\ \\ \text{Gradient of } AP \times {1 \over 4} & = -1 \\ \text{Gradient of } AP & = {-1 \over {1 \over 4}} \\ & = -4 \\ \\ y & = mx + c \\ y & = -4x + c \\ \\ \text{Using } A(3, 1), & \text{ let } x = 3 \text{ and } y = 1, \\ 1 & = -4(3) + c \\ 1 & = -12 + c \\ 13 & = c \\ \\ \therefore y & = -4x + 13 \end{align}

(ii) $P$ is the $x$-intercept of line $AP$ while $Q$ is the $y$-intercept of line $AP$

\begin{align} \text{When } & y = 0, \\ 0 & = -4x + 13 \\ 4x & = 13 \\ x & = {13 \over 4} \\ \\ \therefore & \phantom{/} P \left( {13 \over 4}, 0 \right) \end{align} \begin{align} \text{When } & x = 0, \\ y & = -4(0) + 13 \\ & = 13 \\ \\ \therefore & \phantom{/} Q \left( 0, 13 \right) \end{align}

(iii)

\begin{align} \require{cancel} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ PA & = \sqrt{ \left({13 \over 4} - 3\right)^2 + (0 - 1)^2} \\ & = \sqrt{17 \over 16} \\ & = {\sqrt{17} \over \sqrt{16}} \\ & = {\sqrt{17} \over 4} \\ \\ AQ & = \sqrt{ (3 - 0)^2 + (1 - 13)^2} \\ & = \sqrt{153} \\ & = \sqrt{9} \sqrt{17} \\ & = 3\sqrt{17} \\ \\ PA & : AQ \\ {\cancel{\sqrt{17}} \over 4} & : 3\cancel{\sqrt{17}} \\ {1 \over 4} & : 3 \\ 1 & : 12 \end{align}

(i)

\begin{align} \text{Gradient of } AB & = {13 - 3 \over 4 - 9} \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } B(9,3), & \text{ let } x = 9 \text{ and } y = 3, \\ 3 & = -2(9) + c \\ 3 & = -18 + c \\ 21 & = c \\ \\ \therefore y & = -2x + 21 \end{align}

(ii)

\begin{align} y - 4x & = 5 \\ y & = 4x + 5 \\ \\ \implies \text{Gradient} & = 4 \\ \\ \text{Gradient of line through } C \times 4 & = -1 \\ \text{Gradient of line through } C & = {-1 \over 4} \\ & = -{1 \over 4} \\ \\ y & = mx + c \\ y & = -{1 \over 4}x + c \\ \\ \text{Using } C(10, 8), & \text{ let } x = 10 \text{ and } y = 8, \\ 8 & = -{1 \over 4}(10) + c \\ 8 & = -2.5 + c \\ 10.5 & = c \\ \\ \therefore y & = -{1 \over 4}x + 10.5 \\ 4y & = -x + 42 \end{align}

(iii)

\begin{align} y & = -2x + 21 \phantom{000} \text{ --- (1)} \\ \\ 4y & = -x + 42 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4(-2x + 21) & = -x + 42 \\ -8x + 84 & = -x + 42 \\ -7x & = -42 \\ x & = {-42 \over -7} \\ & = 6 \\ \\ \text{Substitute } & x = 6 \text{ into (1),} \\ y & = -2(6) + 21 \\ & = 9 \\ \\ \therefore & \phantom{/} P (6, 9) \end{align}

(i)

\begin{align} y & = 2x - 5 \\ \\ \implies \text{Gradient of } AB & = 2 \\ \\ \text{Gradient of } DM \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } DM \times (2) & = -1 \\ \text{Gradient of } DM & = {-1 \over 2} \\ & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } D(-2, 6), & \text{ let } x = -2 \text{ and } y = 6, \\ 6 & = -{1 \over 2}(-2) + c \\ 6 & = 1 + c \\ 5 & = c \\ \\ \therefore y & = -{1 \over 2}x + 5 \\ 2y & = -x + 10 \\ 2y + x & = 10 \end{align}

(ii) $M$ is the point of intersection between line $DM$ and $AB$

\begin{align} y & = 2x - 5 \phantom{000} \text{ --- (1)} \\ \\ 10 & = 2y + x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 10 & = 2(2x - 5) + x \\ 10 & = 4x - 10 + x \\ 20 & = 5x \\ {20 \over 5} & = x \\ 4 & = x \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 2(4) - 5 \\ & = 3 \\ \\ \therefore & \phantom{/} M (4, 3) \end{align}

$M$ is the mid-point of $AB$
\begin{align} \text{Let the coor} & \text{dinates of } B \text{ be } (e, f). \\ \\ \text{Coordinates of } M & = \left( {3 + e \over 2}, {1 + f \over 2} \right) \\ (4, 3) & = \left( {3 + e \over 2}, {1 + f \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ 4 & = {3 + e \over 2} \\ 4(2) & = 3 + e \\ 8 & = 3 + e \\ 8 - 3 & = e \\ 5 & = e \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ 3 & = {1 + f \over 2} \\ 3(2) & = 1 + f \\ 6 & = 1 + f \\ 6 - 1 & = f \\ 5 & = f \\ \\ \therefore & \phantom{/} B (5, 5) \end{align}

(iii)

\begin{align} \require{cancel} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AM & = \sqrt{ (3 - 4)^2 + (1 - 3)^2} \\ & = \sqrt{ 5} \\ \\ MD & = \sqrt{ (4 - (-2))^2 + (3 - 6)^2} \\ & = \sqrt{45} \\ & = \sqrt{9} \sqrt{5} \\ & = 3\sqrt{5} \\ \\ AM & : MD \\ \cancel{\sqrt{5}} & : 3\cancel{\sqrt{5}} \\ 1 & : 3 \end{align}

(i)

\begin{align} \text{Mid-point of } AB & = \left( {5 + 3 \over 2}, {4 + (-2) \over 2} \right) \\ & = (4, 1) \\ \\ \text{Gradient of } AB & = {4 - (-2) \over 5 - 3} \\ & = 3 \\ \\ \text{Gradient of } \perp \text{bisector} \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } \perp \text{bisector} & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over 3} \\ & = -{1 \over 3} \\ \\ y & = mx + c \\ y & = -{1 \over 3}x + c \\ \\ \text{Using mid-point } (4, 1), & \text{ let } x = 4 \text{ and } y = 1, \\ 1 & = -{1 \over 3}(4) + c \\ 1 & = -{4 \over 3} + c \\ 1 + {4 \over 3} & = c \\ {7 \over 3} & = c \\ \\ \therefore y & = -{1 \over 3}x + {7 \over 3} \\ 3y & = -x + 7 \\ 3y + x & = 7 \end{align}

(ii) If $P$ is equidistant from $A$ and $B$ (i.e. $PA = PB$), $P$ must lie on the perpendicular bisector of $AB$.

Thus point $P$ is the point of intersection between the perpendicular bisector of $AB$ and the line $y = x + 5$

\begin{align} 3y + x & = 7 \phantom{000} \text{ --- (1)} \\ \\ y & = x + 5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (2) into (1),} \\ 3(x + 5) + x & = 7 \\ 3x + 15 + x & = 7 \\ 4x & = -8 \\ x & = {-8 \over 4} \\ & = -2 \\ \\ \text{Substitute } & x = -2 \text{ into (2),} \\ y & = (-2) + 5 \\ & = 3 \\ \\ \therefore & \phantom{/} P (-2, 3) \end{align}

(i)

\begin{align} \text{Mid-point of } AB & = \left( {3 + 5 \over 2}, {6 + 2 \over 2} \right) \\ & = (4, 4) \\ \\ \text{Gradient of } AB & = {6 - 2 \over 3 - 5} \\ & = -2 \\ \\ \text{Gradient of } \perp \text{bisector} \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } \perp \text{bisector} & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using mid-point } (4,4), & \text{ let } x = 4 \text{ and } y = 4, \\ 4 & = {1 \over 2}(4) + c \\ 4 & = 2 + c \\ 4 - 2 & = c \\ 2 & = c \\ \\ \therefore y & = {1 \over 2}x + 2 \\ 2y & = x + 4 \end{align}

(ii) Since both points $P$ and $Q$ are equidistant from $A$ and $B$, they lie on the perpendicular bisector of $AB$.

$P$ is the $x$-intercept of the perpendicular bisector
\begin{align} 2y & = x + 4, \\ \\ \text{Let } & y = 0, \\ 2(0) & = x + 4 \\ 0 & = x + 4 \\ -4 & = x \\ \\ \therefore & \phantom{/} P(-4, 0) \end{align}

$Q$ is the $y$-intercept of the perpendicular bisector
\begin{align} \text{Let } & x = 0, \\ 2y & = (0) + 4 \\ 2y & = 4 \\ y & = {4 \over 2} \\ & = 2 \\ \\ \therefore & \phantom{/} Q(0, 2) \end{align}

(i)

\begin{align} \text{Mid-point of } PQ & = \left( {3 + 5 \over 2}, {5 + 9 \over 2} \right) \\ & = (4, 7) \\ \\ \text{Gradient of } PQ & = {5 - 9 \over 3 - 5} \\ & = 2 \\ \\ \text{Gradient of } \perp \text{bisector} \times \text{Gradient of } PQ & = -1 \\ \text{Gradient of } \perp \text{bisector} & = {-1 \over \text{Gradient of } PQ} \\ & = {-1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using mid-point } (4, 7), & \text{ let } x = 4 \text{ and } y = 7, \\ 7 & = -{1 \over 2}(4) + c \\ 7 & = -2 + c \\ 7 + 2 & = c \\ 9 & = c \\ \\ \therefore y & = -{1 \over 2}x + 9 \\ 2y & = -x + 18 \\ 2y + x & = 18 \end{align}

(ii) Since the point is equidistant from $P$ and $Q$, it must lie on the perpendicular bisector of $PQ$

\begin{align} 2y + x & = 18 \phantom{000} \text{ --- (1)} \\ \\ y & = 6x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2(6x) + x & = 18 \\ 12x + x & = 18 \\ 13x & = 18 \\ x & = {18 \over 13} \\ \\ \text{Substitute } & x = {18 \over 13} \text{ into (2),} \\ y & = 6\left(18 \over 13\right) \\ & = {108 \over 13} \\ \\ \therefore & \phantom{/} \left({18 \over 13}, {108 \over 13}\right) \end{align}

(i)

\begin{align} y & = 5x \\ \\ \implies \text{Gradient} & = 5 \end{align}

Since $BC$ is parallel to line $y = 5x$, $BC$ has a gradient of 5. Point $C$ lies on the $y$-axis
\begin{align} y & = mx + c \\ y & = 5x + c \\ \\ \text{Using } B(-1, 1), & \text{ let } x = -1 \text{ and } y = 1, \\ 1 & = 5(-1) + c \\ 1 & = -5 + c \\ 1 + 5 & = c \\ 6 & = c \\ \\ \text{Equation of } BC: \phantom{.} y & = 5x + 6 \\ \text{Let } & x = 0, \\ y & = 5(0) + 6 \\ & = 6 \\ \\ \therefore & \phantom{/} C (0, 6) \end{align}

(ii)

\begin{align} \text{Gradient of } BD & = {1 - 7 \over -1 - 8} \\ & = {2 \over 3} \\ \\ \text{Gradient of } AC \times \text{Gradient of } BD & = -1 \\ \text{Gradient of } AC & = {-1 \over \text{Gradient of } BD} \\ & = {-1 \over {2 \over 3}} \\ & = -{3 \over 2} \end{align}

$AC$ has a gradient of $-{3 \over 2}$ and passes through point $C(0, 6)$
\begin{align} y & = mx + c \\ y & = -{3 \over 2}x + c \\ \\ \text{Using } C(0,6), & \text{ let } x = 0 \text{ and } y = 6, \\ 6 & = -{3 \over 2}(0) + c \\ 6 & = c \\ \\ \therefore \text{Equation of } AC \phantom{.} y & = -{3 \over 2}x + 6 \end{align}

(iii) Point $A$ lies on the $x$-axis

\begin{align} \text{When } & y = 0, \\ 0 & = -{3 \over 2}x + 6 \\ {3 \over 2}x & = 6 \\ x & = {6 \over {3 \over 2}} \\ & = 4 \\ \\ \therefore & \phantom{/} A(4, 0) \\ \\ \text{Mid-point of } AC, M \phantom{.} & \left( {0 + 4 \over 2}, {6 + 0 \over 2} \right) \\ & (2, 3) \end{align}

(iv) To find the area of the quadrilateral, find the sum of the areas of $\triangle ACD$ and $\triangle ACB$

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AC & = \sqrt{ (4 - 0)^2 + (0 - 6)^2} \\ & = \sqrt{52} \text{ units} \\ \\ MD & = \sqrt{ (2 - 8)^2 + (3 - 7)^2} \\ & = \sqrt{52} \text{ units} \\ \\ MB & = \sqrt{ (2 - (-1))^2 + (3 - 1)^2} \\ & = \sqrt{13} \text{ units} \\ \\ \text{Area of } ABCD & = \text{Area of } \triangle ACD + \text{Area of } \triangle ACB \\ & = {1 \over 2}(AC)(MD) + {1 \over 2}(AC)(MB) \\ & = {1 \over 2}(\sqrt{52})(\sqrt{52}) + {1 \over 2}(\sqrt{52})(\sqrt{13}) \\ & = 26 + 13 \\ & = 39 \text{ sq. units} \end{align}

(i)

\begin{align} \text{Mid-point of } AB & = \left( {2 + 6 \over 2}, {3 + 7 \over 2} \right) \\ & = (4, 5) \\ \\ \text{Gradient of } AB & = {3 - 7 \over 2 - 6} \\ & = 1 \\ \\ \text{Gradient of } CD \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } CD & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over 1} \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using mid-point } (4, 5), & \text{ let } x = 4 \text{ and } y = 5, \\ 5 & = -(4) + c \\ 5 + 4 & = c \\ 9 & = c \\ \\ \therefore y & = -x + 9 \\ y + x & = 9 \end{align}

(ii) Point $C$ lies on the line $CD$

\begin{align} \text{Using } C(7, t), & \text{ let } x = 7 \text{ and } y = t, \\ (t) + (7) & = 9 \\ t + 7 & = 9 \\ t & = 9 - 7 \\ & = 2 \end{align}

(iii) Since $AB$ is the perpendicular bisector of $CD$, both lines have the same mid-point $(4, 5)$

\begin{align} \text{Let the coor} & \text{dinates of } D \text{ be } (e,f). \\ \\ \text{Mid-point of } CD & = \left({7 + e \over 2}, {2 + f \over 2} \right) \\ (4, 5) & = \left({7 + e \over 2}, {2 + f \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ 4 & = {7 + e \over 2} \\ 2(4) & = 7 + e \\ 8 & = 7 + e \\ 8 - 7 & = e \\ 1 & = e \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ 5 & = {2 + f \over 2} \\ 2(5) & = 2 +f \\ 10 & = 2 + f \\ 10 - 2 & = f \\ 8 & = f \\ \\ \therefore & \phantom{/} D (1, 8) \end{align}

(i)

\begin{align} \text{Gradient of } AB & = {-1 - 3 \over 1 - 5} \\ & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } A(1, -1), & \text{ let } x = 1 \text{ and } y = -1, \\ -1 & = 1 + c \\ -2 & = c \\ \\ \therefore \text{Equation of } AB: \phantom{.} y & = x - 2 \\ \\ \\ \text{Mid-point of } AB, M & = \left( {1 + 5 \over 2}, {-1 + 3 \over 2} \right) \\ & = (3, 1) \end{align}

(ii)

\begin{align} \text{Gradient of } PQ \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } PQ & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over 1} \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } M(3, 1), & \text{ let } x = 3 \text{ and } y = 1, \\ 1 & = -(3) + c \\ 1 & = -3 + c \\ 1 + 3 & = c \\ 4 & = c \\ \\ \therefore \text{Equation of } PQ: \phantom{.} y & = -x + 4 \\ y + x & = 4 \end{align} \begin{align} \text{Using } P(p, 3.5), & \text{ let } x = p \text{ and } y = 3.5, \\ (3.5) + (p) & = 4 \\ 3.5 + p & = 4 \\ p & = 4 - 3.5 \\ & = 0.5 \end{align} \begin{align} \text{Using } Q(7, q), & \text{ let } x = 7 \text{ and } y = q, \\ (7) + (q) & = 4 \\ 7 + q & = 4 \\ q & = 4 - 7 \\ & = -3 \end{align}

(iii) To find the area of the quadrilateral, find the sum of the areas of $\triangle APQ$ and $\triangle BQP$

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ MA & = \sqrt{ (3 - 1)^2 + (1 - (-1))^2} \\ & = \sqrt{8} \text{ units} \\ \\ PQ & = \sqrt{ (0.5 - 7)^2 + (3.5 - (-3))^2} \\ & = \sqrt{84.5} \text{ units} \\ \\ MB & = \sqrt{ (3 - 5)^2 + (1 - 3)^2} \\ & = \sqrt{8} \text{ units} \\ \\ \text{Area of } APBQ & = \text{Area of } \triangle APQ + \text{Area of } \triangle BQP \\ & = {1 \over 2}(PQ)(MA) + {1 \over 2}(PQ)(MB) \\ & = {1 \over 2}(\sqrt{84.5})(\sqrt{8}) + {1 \over 2}(\sqrt{84.5})(\sqrt{8}) \\ & = 13 + 13 \\ & = 26 \text{ sq. units} \end{align}

(i)

\begin{align} \text{Midpoint of } AB & = \left( {-4 + 8 \over 2}, {3 + (-3) \over 2} \right) \\ & = (2, 0) \\ \\ \text{Gradient of } AB & = {-3 - 3 \over 8 - (-4)} \\ & = -{1 \over 2} \\ \\ \text{Gradient of bisector} & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & \text{midpoint } (2, 0), \\ 0 & = 2(2) + c \\ 0 & = 4 + c \\ -4 & = c \\ \\ \therefore y & = 2x - 4 \end{align}

(ii)

\begin{align} y & = 2x - 4 \\ \\ \text{When } & x = 10, \\ y & = 2(10) - 4 \\ y & = 16 \\ \\ \therefore (10, 16) \text{ lies on } & \text{the perpendicular bisector} \end{align}

(iii)

\begin{align} y & = 2x - 4 \\ \\ \text{Since } C(p, q) \text{ lies on } & \text{the line, } q = 2p - 4 \\ \\ & \phantom{.} C(p, 2p - 4) \\ \\ AB & = \sqrt{(-4 - 8)^2 + [3 - (-3)]^2} \\ & = \sqrt{180} \\ \\ \text{Let midpoint } & \text{of } AB \text{ be } M \\ \\ & \phantom{.} M(2, 0) \\ \\ MC & = \sqrt{(p - 2)^2 + (2p - 4 - 0)^2} \\ & = \sqrt{(p - 2)^2 + (2p - 4)^2} \\ & = \sqrt{(p - 2)^2 + [2(p - 2)]^2} \\ & = \sqrt{(p - 2)^2 + 4(p - 2)^2} \\ & = \sqrt{5(p - 2)^2} \\ \\ \text{Area of triangle} & = {1 \over 2} \times AB \times MC \\ & = {1 \over 2} \sqrt{180} \times \sqrt{5(p - 2)^2} \\ & = {1 \over 2} \sqrt{900(p - 2)^2} \\ \\ {1 \over 2} \sqrt{900(p - 2)^2} & = 6 \\ \sqrt{900(p - 2)^2} & = 12 \\ 900(p - 2)^2 & = 12^2 \\ (p - 2)^2 & = {144 \over 900} \\ (p - 2)^2 & = {4 \over 25} \\ p - 2 & = \pm \sqrt{4 \over 25} \\ p - 2 & = \pm {2 \over 5} \\ p & = {12 \over 5} \text{ or } {8 \over 5} \\ \\ \text{If } & p = {12 \over 5}, \phantom{.} C \left({12 \over 5}, {4 \over 5}\right) \\ \\ \text{If } & p = {8 \over 5}, \phantom{.} C \left({8 \over 5}, -{4 \over 5}\right) \end{align}

(i)

\begin{align} & \text{Since } O \text{ is the centre, } OA, OB \text{ and } OC \text{ are the radii of the circle} \\ & \text{Since }OA = OC = OB, \text{ both triangles are isosceles triangles} \end{align}

(ii)

\begin{align} \text{Let } & \angle AOC = \theta \\ \\ \angle OCA & = {180^\circ - \theta \over 2} \phantom{0000000000000} [\text{Isosceles triangle } OAC] \\ & = 90^\circ - {1 \over 2} \theta \\ \\ \angle BOC & = 180^\circ - \theta \phantom{00000000000000} [\text{Adjacent angles on a straight line}] \\ \\ \angle OCB & = {180^\circ - (180^\circ - \theta) \over 2} \phantom{000000} [\text{Isosceles triangle } OBC] \\ & = {180^\circ - 180^\circ + \theta \over 2} \\ & = {\theta \over 2} \\ & = {1 \over 2} \theta \\ \\ \angle ACB & = 90^\circ - {1 \over 2} \theta + {1 \over 2} \theta \\ & = 90^\circ \\ \\ \therefore & \phantom{.} AC \text{ perpendicular to } BC \end{align}

(iii)

\begin{align} \text{Midpoint of } AB & = \left({1 + 5 \over 2}, {0 + 2 \over 2} \right) \\ & = (3, 1) \phantom{00000000000000} [\text{This is the centre, } O] \\ \\ \text{Radius} & = OA \\ & = \sqrt{(3 - 1)^2 + (1 - 0)^2} \\ & = \sqrt{5} \\ \\ \text{Radius} & = OC \\ & = \sqrt{(a - 3)^2 + (b - 1)^2} \\ \\ \sqrt{5} & = \sqrt{(a - 3)^2 + (b - 1)^2} \\ 5 & = (a - 3)^2 + (b - 1)^2 \phantom{00} \text{ (Shown)} \end{align}

(iv)

\begin{align} \text{Let } C \text{ denote } & \text{the point } (2, 3) \\ \\ OC & = \sqrt{(3 - 2)^2 + (1 - 3)^2} \\ & = \sqrt{5} \\ \\ \text{Since } OC = \text{Radius}, & \text{ the point } (2, 3) \text{ lies on the circle} \end{align}