(i)

\begin{align} \text{Distance between 2 points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ \\ AB & = \sqrt{ (- 2 - 10)^2 + (1 - 6)^2 } \\ & = \sqrt{169} \\ \\ BC & = \sqrt{ (10 - a)^2 + (6 - (-6))^2 } \\ & = \sqrt{ 10^2 - 2(10)(a) + a^2 + 12^2 } \\ & = \sqrt{ 100 - 20a + a^2 + 144} \\ & = \sqrt{ 244 - 20a + a^2 } \\ \\ AB & = BC \\ \sqrt{169} & = \sqrt{ 244 - 20a + a^2 } \\ 169 & = 244 - 20a + a^2 \\ 0 & = 75 - 20a + a^2 \\ 0 & = a^2 - 20 a + 75 \\ 0 & = (a - 15)(a - 5) \\ \\ a - 15 = 0 \phantom{00} & \phantom{00} a - 5 = 0 \\ a = 15 \phantom{0} & \phantom{00-0} a = 5 \end{align}

(ii)

Since $ABCD$ is a rhombus, the diagonals $AC$ and $BD$ have the same mid-point \begin{align} \text{Let the coor} & \text{dinates of } D \text{ be } (e, f). \\ \\ \text{Mid-point of } AC & = \left( {-2 + a \over 2}, {1 + (-6) \over 2} \right) \\ & = \left( {a - 2 \over 2}, -{5 \over 2} \right) \\ \\ \text{Mid-point of } BD & = \left( {10 + e \over 2}, {6 + f \over 2} \right) \\ \\ \\ \text{Mid-point of } AC & = \text{Mid-point of } BD \\ \left( {a - 2 \over 2}, -{5 \over 2} \right) & = \left( {10 + e \over 2}, {6 + f \over 2} \right) \\ \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ -{5 \over 2} & = {6 + f \over 2} \\ -5 & = 6 + f \\ \\ f & = -5 - 6 \\ & = - 11 \\ \\ \\ \text{Comparing the } & x\text{-coordinate and when } a = 15, \\ \\ {(15) - 2 \over 2} & = {10 + e \over 2} \\ {13 \over 2} & = {10 + e \over 2} \\ 13 & = 10 + e \\ \\ e & = 13 - 10 \\ & = 3 \\ \\ \therefore & \phantom{/} D (3, -11) \\ \\ \\ \text{Comparing the } & x\text{-coordinate and when } a = 5, \\ \\ {(5) -2 \over 2} & = {10 + e \over 2} \\ {3 \over 2} & = {10 + e \over 2} \\ 3 & = 10 + e \\ \\ e & = 3 - 10 \\ & = -7 \\ \\ \therefore & \phantom{/} D(-7, -11) \end{align}

(i)

If points $A$, $B$ and $C$ are collinear, they lie on a straight line. Thus lines $AB$, $BC$ and $AC$ have the same gradient \begin{align} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \\ \text{Gradient of } AB & = {1 - 3 \over (k + 1) - (2k + 1)} \\ & = {-2 \over k + 1 - 2k - 1} \\ & = {-2 \over -k} \\ & = {2 \over k} \\ \\ \text{Gradient of } BC & = {3 - 2k \over (2k + 1) - (2k + 2)} \\ & = {3 - 2k \over 2k + 1 - 2k - 2} \\ & = {3- 2k \over -1} \\ & = -(3- 2k) \\ & = -3 + 2k \\ \\ \text{Gradient of } AB & = \text{Gradient of } BC \\ {2 \over k} & = -3 + 2k \\ 2 & = k(-3 + 2k) \\ 2 & = -3k + 2k^2 \\ 0 & = 2k^2 - 3k - 2\\ 0 & = (2k + 1)(k - 2) \\ \\ 2k + 1 = 0 \phantom{0-} & \phantom{000} k - 2 = 0 \\ 2k = -1 \phantom{0} & \phantom{00-00} k = 2 \\ k = -{1 \over 2} \end{align}

(ii)

\begin{align} \text{Gradient of } AC & = {1 - 2k \over (k + 1) - (2k + 2)} \\ & = {1 - 2k \over k + 1 - 2k - 2} \\ & = {1 - 2k \over -k - 1} \\ \\ \\ \text{When } & k = 2, \\ \\ \text{Gradient of } AC & = {1 - 2(2) \over -(2) - 1} \\ & = 1 \\ \\ \\ \text{Gradient of } BD \times \text{Gradient of } AC & = -1 \\ \text{Gradient of } BD & = {-1 \over \text{Gradient of } AC} \\ & = {-1 \over (1)} \\ & = -1 \end{align}

Since $D$ lies on the $x$-axis, the $y$-coordinate of $D$ is 0
\begin{align} \text{Let the coor} & \text{dinates of } D \text{ be } (e, 0). \\ \\ \text{Gradient of } BD & = {3 - 0 \over 2k + 1 - e} \\ -1 & = {3 \over 2k + 1 - e} \\ -1 & = {3 \over 2(2) + 1 - e} \phantom{00000} (\text{When } k = 2) \\ -1 & = {3 \over 5 - e} \\ -(5 - e) & = 3 \\ -5 + e & = 3 \\ e & = 3 + 5 \\ & = 8 \\ \\ \therefore & \phantom{/} D (8, 0) \end{align}

Plotting the points $B (5, 3)$ and $D (8, 0)$ in a rough sketch:
\begin{align} \text{Let } \theta \text{ represent the acute} & \text{ angle } BD \text{ makes with the } x\text{-axis}. \\ \\ \tan \theta & = {3 \over 3} \\ & = 1 \\ \theta & = \tan^{-1} (1) \\ & = 45^\circ \\ \\ \therefore \text{Obtuse angle} & = 180^\circ - 45^\circ \\ & = 135^\circ \end{align}

(i)

\begin{align} \text{Distance between 2 points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ BC & = \sqrt{ (0 - 9)^2 + (1 - 4)^2} \\ & = \sqrt{ 90 } \\ & = \sqrt{9} \sqrt{10} \\ & = 3\sqrt{10} \text{ units} \end{align}

(ii)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 6 & 0 & 9 & 6 \\ 7 & 1 & 4 & 7 \end{matrix} \right| \\ & = {1 \over 2} (6 \times 1 + 0 \times 4 + 9 \times 7 - 7 \times 0 - 1 \times 9 - 4 \times 6) \\ & = {1 \over 2} (36) \\ & = 18 \text{ sq. units} \end{align}

(iii)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \times BC \times AF \\ 18 & = {1 \over 2}(3\sqrt{10})(AF) \\ 36 & = 3\sqrt{10}(AF) \\ \\ AF & = {36 \over 3\sqrt{10}} \\ & = {12 \over \sqrt{10}} \\ & = {12 \over \sqrt{10}} \times {\sqrt{10} \over \sqrt{10}} \phantom{00000000} [\text{Rationalise denominator}] \\ & = {12\sqrt{10} \over 10} \\ & = {6 \sqrt{10} \over 5} \text{ units} \end{align}

(a)

\begin{align} 2x + y & = 8 \\ y & = -2x + 8 \\ \\ \text{Comparing with } & y = mx + c, \\ \\ \therefore \text{Gradient} & = -2 \\ \\ \\ \text{Gradient of line} \times (-2) & = -1 \\ \text{Gradient of line} & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ \therefore a & = {1 \over 2} \end{align}

$b$ is $y$-intercept of the line $x + y + 3 = 0$
\begin{align} \text{When } & x = 0, \\ (0) + y + 3 & = 0 \\ y + 3 & = 0 \\ y & = -3 \\ \\ \therefore b & = -3 \end{align}

(b)

Form the equation of the foot of the perpendicular from $A$ to $BC$ \begin{align} y & = x - 1 \\ \\ \text{Comparing with } & y = mx + c, \\ \\ \therefore \text{Gradient} & = 1 \\ \\ \\ \text{Gradient of foot of } \perp \times (1) & = -1 \\ \text{Gradient of foot of } \perp & = -1 \end{align}

The foot of the perpendicular has a gradient of $-1$ and passes through the point $(3, 2)$
\begin{align} y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & (3, 2), \\ 2 & = -3 + c \\ 5 & = c \\ \\ y & = -x + 5 \phantom{000} \text{ --- (1)} \\ \\ \text{Equation of AB: } \phantom{0} y & = 3x + 1 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -x + 5 & = 3x + 1 \\ -4x & = 1 - 5 \\ & = -4 \\ x & = {-4 \over -4} \\ & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into (1),} \\ y & = -(1) + 5 \\ & = 4 \\ \\ \therefore & \phantom{/} A(1, 4) \end{align}

(i)

\begin{align} y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & B(-1, -3), \\ -3 & = {1 \over 2}(-1) + c \\ -3 & = -{1 \over 2} + c \\ -{5 \over 2} & = c \\ \\ y & = {1 \over 2}x - {5 \over 2} \\ \therefore 2y & = x - 5 \end{align}

(ii)

\begin{align} \text{Gradient of } AB & = {5 - (-3) \over 3 - (-1)} \\ & = 2 \\ \\ \text{Gradient of } AC \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } AC & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over 2} \\ & = -{1 \over 2} \\ \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & A(3, 5), \\ 5 & = -{1 \over 2}(3) + c \\ 5 & = -{3 \over 2} + c \\ {13 \over 2} & = c \\ \\ y & = -{1 \over 2}x + {13 \over 2} \\ 2y & = -x + 13 \\ \therefore 2y + x & = 13 \end{align}

(iii)

\begin{align} \text{Equation of } BC: \phantom{0} 2y & = x - 5 \phantom{000} \text{ --- (1)} \\ \\ \text{Equation of } AC: \phantom{0} 2y + x & = 13 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (x - 5) + x & = 13 \\ 2x & = 13 + 5 \\ & = 18 \\ x & = {18 \over 2} \\ & = 9 \\ \\ \text{Substitute } & x = 9 \text{ into (1),} \\ 2y & = (9) - 5 \\ & = 4 \\ y & = {4 \over 2} \\ & = 2 \\ \\ \therefore & \phantom{/} C(9,2) \end{align}

(iv)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 3 & -1 & 9 & 3 \\ 5 & -3 & 2 & 5 \end{matrix} \right| \\ & = {1 \over 2}[ 3 \times (-3) + (-1) \times 2 + 9 \times 5 - 5 \times (-1) - (-3) \times 9 - 2 \times 3] \\ & = {1 \over 2}(60) \\ & = 30 \text{ sq. units} \\ \\ BC & = \sqrt{ (-1 - 9)^2 + (-3 - 2)^2} \\ & = \sqrt{125} \\ & = \sqrt{25}\sqrt{5} \\ & = 5\sqrt{5} \text{ units} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2}(BC)(AD) \\ 30 & = {1 \over 2}(5\sqrt{5})(AD) \\ 60 & = (5\sqrt{5})(AD) \\ \\ AD & = {60 \over 5\sqrt{5}} \\ & = {12 \over \sqrt{5}} \\ & = {12 \over \sqrt{5}} \times {\sqrt{5} \over \sqrt{5}} \\ & = {12\sqrt{5} \over 5} \text{ units} \end{align}

(i)

$A$ is the $x$-intercept of lines $AB$ and $AD$ \begin{align} \text{When } & y = 0, \\ \\ 5(0) + 6 & = 3x \\ 6 & = 3x \\ \\ x & = {6 \over 3} \\ & = 2 \\ \\ \therefore & \phantom{/} A(2,0) \end{align}

$D$ is the $y$-intercept of line $AD$ \begin{align} \text{When } & x = 0, \\ 3(0) + 2y & = 6 \\ 2y & = 6 \\ y & = {6 \over 2} \\ & = 3 \\ \\ \therefore & \phantom{/} D (0, 3) \end{align}

(ii)

\begin{align} \text{Let coordinates} & \text{ of } E \text{ be } (f, g) \\ \\ \text{Mid-point of } AE, D & = \left({2 + f \over 2}, {0 + g \over 2} \right) \\ (0, 3) & = \left({2 + f \over 2}, {g \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ 0 & = {2 + f \over 2} \\ 0 & = 2 + f \\ -2 & = f \\ \\ \text{Comparing the } &y\text{-coordinate,} \\ 3 & = {g \over 2} \\ 2(3) & = g \\ 6 & = g \\ \\ \therefore & \phantom{.} E(-2, 6) \end{align}

Since $BC$ is parallel to the $x$-axis, the gradient of $BC$ (and $BE$) is $0$. Thus the equation of $BE$ is $y = 6$
This means the $y$-coordinates of $B$ and $C$ are $6$ as well
\begin{align} \text{When } & y = 6, \\ 5(6) + 6 & = 3x \\ 30 + 6 & = 3x \\ 36 & = 3x \\ {36 \over 3} & = x \\ 12 & = x \\ \\ \therefore & \phantom{/} B (12, 6) \\ \\ \\ \text{Let the coor} & \text{dinates of } C \text{ be } (h, 6) \\ 5y + 6 & = 3x \\ 5y & = 3x - 6 \\ y & = {3 \over 5}x - {6 \over 5} \\ \\ \text{Gradient of } AB & = {3 \over 5} \\ & = \text{Gradient of } DC \phantom{000} [AB \phantom{0} // \phantom{0} DC] \\ \\ {3 \over 5} & = {3 - 6 \over 0 - h} \\ {3 \over 5} & = {-3 \over -h} \\ 3(-h) & = 5(-3) \\ -3h & = -15 \\ h & = {-15 \over -3} \\ & = 5 \\ \\ \therefore & \phantom{/} C(5, 6) \end{align}
\begin{align} \text{Area of } ABCD & = {1 \over 2} \left| \begin{matrix} 2 & 12 & 5 & 0 & 2 \\ 0 & 6 & 6 & 3 & 0 \end{matrix} \right| \\ & = {1 \over 2}(2 \times 6 + 12 \times 6 + 5 \times 3 + 0 \times 0 - 0 \times 12 - 6 \times 5 - 6 \times 0 - 3 \times 2) \\ & = {1 \over 2}(63) \\ & = 31.5 \text{ sq. units} \end{align}

(a)

If $A$, $B$ and $C$ lie on the same line, lines $AB$ and $BC$ have the same gradient \begin{align} \text{Gradient of } AB & = {2 - 10 \over -1 - 3} \\ & = 2 \\ \\ \text{Gradient of } BC & = {10 - 8 \over 3 - p} \\ & = {2 \over 3 - p} \\ \\ \text{Gradient of } AB & = \text{Gradient of } BC \\ 2 & = {2 \over 3 - p} \\ 2(3 - p) & = 2 \\ 6 - 2p & = 2 \\ -2p & = 2 - 6 \\ & = -4 \\ p & = {-4 \over -2} \\ & = 2 \end{align}

(b)

\begin{align} \text{Gradient of } AC & = {2 - 8 \over -1 - p} \\ & = {-6 \over -1 - p} \\ \\ \text{Gradient of } BC & = {10 - 8 \over 3 - p} \\ & = {2 \over 3 - p} \\ \\ \text{Gradient of } AC \times \text{Gradient of } BC & = -1 \\ \left(-6 \over -1 -p\right) \left(2 \over 3 -p \right) & = -1 \\ {-6(2) \over (-1-p)(3 - p)} & = -1 \\ -6(2) & = -(-1 - p)(3 - p) \\ -12 & = (1 + p)(3 - p) \\ -12 & = 3 - p + 3p - p^2 \\ -12 & = 3 + 2p - p^2 \\ 0 & = 15 + 2p - p^2 \\ 0 & = -p^2 + 2p + 15 \\ 0 & = p^2 - 2p - 15 \\ 0 & = (p - 5)(p + 3) \\ \\ p - 5 = 0 \phantom{00} &\text{or} \phantom{00} p + 3 = 0 \\ p = 5 \phantom{00} & \phantom{or00+3} p = -3 \end{align}

(i)

\begin{align} x - y - 2 & = 0 \\ x & = y + 2 \phantom{00} \text{ --- (1)} \\ \\ 2x - 5y - 7 & = 0 \phantom{00} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(y + 2) - 5y - 7 & = 0 \\ 2y + 4 - 5y - 7 & = 0 \\ - 3y - 3 & = 0 \\ -3y & = 3 \\ y & = {3 \over -3} \\ & = -1 \\ \\ \text{Substitute } & y = - 1 \text{ into (1),} \\ x & = (-1) + 2 \\ & = 1 \\ \\ \therefore & \phantom{/} P(1, -1) \end{align}

(ii)

The line through $P$ has a gradient of $2$ and passes through $P (1, -1)$ \begin{align} y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & P(1, -1), \\ -1 & = 2(1) + c \\ -1 & = 2 + c \\ -3 & = c \\ \\ \therefore \text{Equation of line: } & \phantom{/} y = 2x - 3 \end{align}

Point $A$ is the $x$-intercept of the line
\begin{align} \text{When } & y = 0, \\ 0 & = 2x - 3 \\ 3 & = 2x \\ {3 \over 2} & = x \\ \\ \therefore & \phantom{/} A \left({3 \over 2}, 0\right) \end{align}

Point $B$ is the $y$-intercept of the line
\begin{align} \text{When } & x = 0, \\ y & = 2(0) - 3 \\ & = -3 \\ \\ \therefore & \phantom{/} B (0, -3) \end{align}

(iii)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AP & = \sqrt{ \left({3 \over 2} - 1\right)^2 + (0 - (-1))^2 } \\ & = \sqrt{5 \over 4} \\ & = {\sqrt{5} \over \sqrt{4}} \\ & = {\sqrt{5} \over 2} \text{ units} \\ \\ PB & = \sqrt{ (1 - 0)^2 + (-1 - (-3))^2} \\ & = \sqrt{5} \text{ units} \\ \\ AP & : PB \\ {\sqrt{5} \over 2} & : \sqrt{5} \\ {1 \over 2} & : 1 \\ 1 & : 2 \end{align}

(i) I think there's an error with the question - it should be 8 - 3a < 0

\begin{align} 8 - 3a & < 0 \\ 8 & < 3a \\ {8 \over 3} & < a \\ 2{2 \over 3} & < a \end{align}

Since the coordinates of $C$ is $(a, 8 - 3a)$, $C$ is between points $A$ and $B$. Thus, taking the points in an anti-clockwise direction ($A$, $C$, $B$)
\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 2 & a & 6 & 2 \\ 6 & 8 - 3a & -2 & 6 \end{matrix} \right| \\ 10 & = {1 \over 2} [ 2 \times (8 - 3a) + a \times (-2) + 6 \times 6 - 6 \times a - (8 - 3a) \times 6 - (-2) \times 2 ] \\ 10 & = {1 \over 2} ( 16 - 6a - 2a + 36 - 6a - 48 + 18a + 4 ) \\ 10 & = {1 \over 2} (8 + 4a) \\ 20 & = 8 + 4a \\ 20 - 8 & = 4a \\ 12 & = 4a \\ {12 \over 4} & = a \\ 3 & = a \text{ (Shown)} \\ \\ \therefore & \phantom{/} C(3, -1) \end{align}

(ii)

\begin{align} \text{Gradient of } AB & = {6 - (-2) \over 2- 6} \\ & = - 2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & B(6, -2), \\ -2 & = -2(6) + c \\ -2 & = -12 + c \\ 10 & = c \\ \\ y & = -2x + 10 \\ \therefore y + 2x & = 10 \end{align}

(iii)

\begin{align} \text{Gradient of line} \times \text{Gradient of } AB & = -1 \\ \text{Gradient of line} & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & C(3, -1), \\ -1 & = {1 \over 2}(3) + c \\ -1 & = {3 \over 2} + c \\ -{5 \over 2} & = c \\ \\ y & = {1 \over 2}x - {5 \over 2} \\ \therefore 2y & = x - 5 \end{align}

(iv)

The foot of the perpendicular is the point of intersection of lines $AB$ and the line found in part (iii) \begin{align} y + 2x & = 10 \\ y & = 10 - 2x \phantom{000} \text{ --- (1)} \\ \\ 2y & = x - 5 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(10 - 2x) & = x - 5 \\ 20 - 4x & = x - 5 \\ -4x - x & = - 5 - 20 \\ -5x & = -25 \\ x & = {-25 \over -5} \\ & = 5 \\ \\ \text{Substitute } & x = 5 \text{ into (1),} \\ y & = 10 - 2(5) \\ & = 0 \\ \\ \therefore \text{Foot of perpendicular is } & (5, 0) \end{align}

(i)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ \\ AB & = \sqrt{ (2 - 7)^2 + (4 - 9)^2 } \\ & = \sqrt{50} \\ \\ BC & = \sqrt{ (7 - 8)^2 + (9 - 2)^2 } \\ & = \sqrt{50} \\ & = AB \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{Mid-point of AC, M} & = \left({2 + 8 \over 2}, {4 + 2 \over 2}\right) \\ & = (5, 3) \end{align}

(iii)

\begin{align} \text{Gradient of } BM & = {9 - 3 \over 7 - 5} \\ & = 3 \\ \\ y & = mx + c \\ y & = 3x + c \\ \\ \text{Using } & B(7, 9), \\ 9 & = 3(7) + c \\ 9 & = 21 + c \\ -12 & = c \\ \\ \therefore y & = 3x - 12 \end{align}

(iv)

\begin{align} \text{Gradient of } AB & = {4 - 9 \over 2 - 7} \\ & = 1 \\ & = \text{Gradient of } CD \phantom{000} [AB \phantom{0} // \phantom{0} CD] \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & C(8, 2), \\ 2 & = 8 + c \\ -6 & = c \\ \\ \therefore y & = x - 6 \end{align}

(v)

Point $D$ is the point of intersection between lines $BM$ and $CD$ \begin{align} \text{Equation of } BM: \phantom{0} y & = 3x - 12 \phantom{00} \text{ --- (1)} \\ \\ \text{Equation of } CD: \phantom{0} y & = x - 6 \phantom{00} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3x - 12 & = x - 6 \\ 3x - x & = - 6 + 12 \\ 2x & = 6 \\ x & = {6 \over 2} \\ & = 3 \\ \\ \text{Substitute } & x = 3 \text{ into (1),} \\ y & = 3(3) - 12 \\ & = -3 \\ \\ \therefore & \phantom{/} D(3, -3) \end{align}

(i)

\begin{align} \text{Distance between 2 points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ PQ & = \sqrt{ (1 - 0)^2 + (0 - 2)^2 } \\ & = \sqrt{5} \text{ units} \\ \\ QR & = \sqrt{ (0 - 2)^2 + (2 - 3)^2} \\ & = \sqrt{5} \text{ units} \\ & = PQ \\ \\ \therefore \triangle PQR & \text{ is an isosceles } \triangle. \end{align}

(ii)

\begin{align} \text{Area of } \triangle PQR & = {1 \over 2} \left| \begin{matrix} 1 & 2 & 0 & 1 \\ 0 & 3 & 2 & 0 \end{matrix} \right| \\ & = {1 \over 2} (1 \times 3 + 2 \times 2 + 0 \times 0 - 0 \times 2 - 3 \times 0 - 2 \times 1) \\ & = {1 \over 2} (5) \\ & = 2.5 \text{ sq. units} \end{align}

(iii)

In a kite, the diagonals $PR$ and $QS$ are perpendicular \begin{align} \text{Gradient of } PR & = {0 - 3 \over 1 - 2} \\ & = 3 \\ \\ \text{Gradient of } QS \times \text{Gradient of } PR & = -1 \\ \text{Gradient of } QS & = {-1 \over \text{Gradient of } PR} \\ & = {-1 \over 3} \\ \\ y & = mx + c \\ y & = -{1 \over 3} x + c \\ \\ \text{Using } & Q(0, 2), \\ 2 & = -{1 \over 3}(0) + c \\ 2 & = c \\ \\ y & = -{1 \over 3}x + 2 \\ 3y & = -x + 6 \\ \therefore 3y + x & = 6 \end{align}

(iv)

\begin{align} 3y + x & = 6 \phantom{00} \text{ --- (1)} \\ \\ x - y & = 5 \\ x & = y + 5 \phantom{00} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3y + (y + 5) & = 6 \\ 4y & = 6 - 5 \\ & = 1 \\ y & = {1 \over 4} \\ \\ \text{Substitute } & y = {1 \over 4} \text{ into (2),} \\ x & = {1 \over 4} + 5 \\ & = {21 \over 4} \\ \\ \therefore & \phantom{/} S \left({21 \over 4}, {1 \over 4}\right) \end{align}

(i)

\begin{align} \text{Gradient of } AC & = {30 - (-10) \over 12 - (-3)} \\ & = {8 \over 3} \\ \\ \tan \theta & = {8 \over 3} \\ \theta & = \tan^{-1} \left(8 \over 3\right) \\ & = 69.443 \\ & \approx 69.4^\circ \text{ (1 d.p.)} \end{align}

(ii)

\begin{align} \text{Gradient of } BC & = {30 - 2 \over 12 - (-12)} \\ & = {7 \over 6} \\ \\ \tan \theta & = {7 \over 6} \\ \theta & = \tan^{-1} \left(7 \over 6\right) \\ & = 49.398 \\ & \approx 49.4^\circ \text{ (1 d.p.)} \end{align}

(iii)

\begin{align} \angle e & = 180^\circ - 69.443^\circ \phantom{00} (\angle \text{s on a straight line)} \\ & = 110.557^\circ \\ \\ \angle ACB & = \angle f \\ & = 180^\circ - 49.398^\circ - 110.557^\circ \phantom{00} (\angle \text{ sum of } \triangle) \\ & = 20.045^\circ \\ & \approx 20.0^\circ \text{ (1 d.p.)} \end{align}

(iv)

Since $BC = BD$, $B$ is the mid-point of $DC$ \begin{align} \text{Let the coor} & \text{dinates of } D \text{ be } (g, h). \\ \\ \text{Mid-point of } DC, B & = \left( {g + 12 \over 2}, {h + 30 \over 2} \right) \\ (-12, 2) & = \left( {g + 12 \over 2}, {h + 30 \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ -12 & = {g + 12 \over 2} \\ 2(-12) & = g + 12 \\ -24 & = g + 12 \\ -36 & = g \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ 2 & = {h + 30 \over 2} \\ 2(2) & = h + 30 \\ 4 & = h + 30 \\ -26 & = h \\ \\ \therefore & \phantom{/} D(-36, -26) \end{align}

(v)

\begin{align} \text{Gradient of } AD & = {-10 - (-26) \over -3 - (-36)} \\ & = {16 \over 33} \\ \\ y & = mx + c \\ y & = {16 \over 33}x + c \\ \\ \text{Using } & D(-36, -26), \\ -26 & = {16 \over 33}(-36) + c \\ -26 & = -{192 \over 11} + c \\ -{94 \over 11} & = c \\ \\ y & = {16 \over 33}x - {94 \over 11} \\ \therefore 33y & = 16x - 282 \end{align}

(vi)

Find the angle that line $AD$ makes with the horizontal \begin{align} \tan \angle i & = \text{Gradient of } AD \\ & = {16 \over 33} \\ \\ \angle i & = \tan^{-1} \left(16 \over 33\right) \\ & = 25.866^\circ \\ \\ \angle j & = 49.398^\circ \phantom{0} (\text{Corresponding } \angle s) \\ \\ \angle ADB & = 49.398^\circ - 25.866^\circ \\ & = 23.532^\circ \\ & \approx 23.5^\circ \end{align}