(a)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 2 & 5 & -1 & 2 \\ 3 & 6 & 4 & 3 \end{matrix} \right| \\ & = {1 \over 2}[2 \times 6 + 5 \times 4 + (-1) \times 3 - 3 \times 5 - 6 \times (-1) - 4 \times 2] \\ & = {1 \over 2}(12) \\ & = 6 \text{ sq. units} \end{align}

(b)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 5 & 1 & -2 & 5 \\ 2 & 6 & -3 & 2 \end{matrix} \right| \\ & = {1 \over 2} [5 \times 6 + 1 \times (-3) + (-2) \times 2 - 2 \times 1 - 6 \times (-2) - (-3) \times 5] \\ & = {1 \over 2} (48) \\ & = 24 \text{ sq. units} \end{align}

Sketch out the points and remember to select them in anti-clockwise order

(a)

\begin{align} \text{Area} & = {1 \over 2} \left| \begin{matrix} 2 & 6 & 3 & 2 \\ 4 & 2 & 5 & 4 \end{matrix} \right| \\ & = {1 \over 2} (2 \times 2 + 6 \times 5 + 3 \times 4 - 4 \times 6 - 2 \times 3 - 5 \times 2) \\ & = {1 \over 2} (6) \\ & = 3 \text{ sq. units} \end{align}

(b)

\begin{align} \text{Area} & = {1 \over 2} \left| \begin{matrix} -4 & 6 & -2 & -4 \\ -2 & 0 & 4 & -2 \end{matrix} \right| \\ & = {1 \over 2} [(-4) \times 0 + 6 \times 4 + (-2) \times (-2) - (-2) \times 6 - 0 \times (-2) - 4 \times (-4)] \\ & = {1 \over 2} (56) \\ & = 28 \text{ sq. units} \end{align}

(i)

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

(ii)

$$ \text{Points } A, B \text{ and } C \text{ lie on a straight line}$$

(a)

\begin{align} \text{Area of } OABC & = {1 \over 2} \left| \begin{matrix} 0 & 4 & 6 & -2 & 0 \\ 0 & 1 & 4 & 4 & 0 \end{matrix} \right| \\ & = {1 \over 2} [0 \times 1 + 4 \times 4 + 6 \times 4 + (-2) \times 0 - 0 \times 4 - 1 \times 6 - 4 \times (-2) - 4 \times 0 ] \\ & = {1 \over 2} (42) \\ & = 21 \text{ sq. units} \end{align}

(b)

\begin{align} \text{Area of } PQRS & = {1 \over 2} \left| \begin{matrix} 1 & -4 & 1 & 4 & 1 \\ 4 & 3 & -2 & 0 & 4 \end{matrix} \right| \\ & = {1 \over 2} [ 1 \times 3 + (-4) \times (-2) + 1 \times 0 + 4 \times 4 - 4 \times (-4) - 3 \times 1 - (-2) \times 4 - 0 \times 1 ] \\ & = {1 \over 2} (48) \\ & = 24 \text{ sq. units} \end{align}

(i)

\begin{align} \text{Area of } ABC & = {1 \over 2} \left| \begin{matrix} 2 & 3 & 2 & 2 \\ -3 & -1 & 0 & -3 \end{matrix} \right| \\ & = {1 \over 2} \left[ 2 \times (-1) + 3 \times 0 + 2 \times (-3) - (-3) \times 3 - (-1) \times 2 - 0 \times 2 \right] \\ & = {1 \over 2}(3) \\ & = {3 \over 2} \text{ sq. units} \\ \\ \\ \text{Area of } ACD & = {1 \over 2} \left| \begin{matrix} 2 & 2 & -1 & 2 \\ -3 & 0 & 1 & -3 \end{matrix} \right| \\ & = {1 \over 2} \left[ 2 \times 0 + 2 \times 1 + (-1) \times (-3) - (-3) \times 2 - 0 \times (-1) - 1 \times 2 \right] \\ & = {1 \over 2} (9) \\ & = {9 \over 2} \text{ sq. units} \\ \\ \\ \text{Area of } ADE & = {1 \over 2} \left| \begin{matrix} 2 & -1 & -2 & 2 \\ -3 & 1 & -1 & -3 \end{matrix} \right| \\ & = {1 \over 2} \left[ 2 \times 1 + (-1) \times (-1) + (-2) \times (-3) - (-3) \times (-1) - 1 \times (-2) - (-1) \times 2 \right] \\ & = {1 \over 2} (10) \\ & = 5 \text{ sq. units} \end{align}

(Ii) The triangles in (i) combine to form the pentagon

\begin{align} \text{Area of pentagon} & = {3 \over 2} + {9 \over 2} + 5 \\ & = 11 \text{ sq. units} \end{align}

(i) Sketch out the points and remember to select them in anti-clockwise order

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 4 & 9 & 5 & 4 \\ -1 & -1 & 1 & -1 \end{matrix} \right| \\ & = {1 \over 2} [ 4 \times (-1) + 9 \times 1 + 5 \times (-1) - (-1) \times 9 - (-1) \times 5 - 1 \times 4 ] \\ & = {1 \over 2} (10) \\ & = 5 \text{ sq. units} \end{align}

(ii)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 - (y_2 - y_1)^2} \\ \\ AB & = \sqrt{ (4 - 5)^2 + (-1 - 1)^2} \\ & = \sqrt{5} \text{ units (Shown)} \end{align}

(iii)

\begin{align} AC & = \sqrt{ (4 - 9)^2 + (-1 - (-1))^2} \\ & = \sqrt{25} \\ & = 5 \text{ units} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2}ab \sin C \\ 5 & = {1 \over 2} (AB)(AC) \sin \angle BAC \\ 5 & = {1 \over 2} (\sqrt{5})(5)\sin \angle BAC \\ 1 & = {1 \over 2} (\sqrt{5}) \sin \angle BAC \\ 1 & = {\sqrt{5} \over 2} \sin \angle BAC \\ \\ \sin \angle BAC & = {1 \over {\sqrt{5} \over 2} } \\ & = {2 \over \sqrt{5}} \text{ (Shown)} \end{align}

(i)

Taking the points in the order of $A$, $C$, $B$: \begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 3 & k & -5 & 3 \\ 5 & k + 2 & 9 & 5 \end{matrix} \right| \\ 18 & = {1 \over 2} [ 3 \times (k + 2) + k \times 9 + (-5) \times 5 - 5 \times k - (k + 2) \times (-5) - 9 \times 3 ] \\ 18 & = {1 \over 2} ( 3k + 6 + 9k - 25 - 5k + 5k + 10 - 27 ) \\ 18 & = {1 \over 2} (12k -36) \\ 36 & = 12k - 36 \\ 36 + 36 & = 12k \\ 72 & = 12k \\ \\ k & = {72 \over 12} \\ & = 6 \end{align}

Taking the points in the order of $A$, $B$, $C$: \begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 3 & -5 & -k & 3 \\ 5 & 9 & k + 2 & 5 \end{matrix} \right| \\ 18 & = {1 \over 2} [ 3 \times 9 + (-5) \times (k + 2) + (-k) \times 5 - 5 \times (-5) - 9 \times (-k) - (k+2) \times 3 ] \\ 18 & = {1 \over 2} ( 27 - 5k -10 - 5k + 25 + 9k - 3k - 6 ) \\ 18 & = {1 \over 2} (36 - 4k) \\ 36 & = 36 - 4k \\ 36 - 36 & = -4k \\ 0 & = -4k \\ \\ k & = {0 \over -4} \\ & = 0 \end{align}

(ii)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AB & = \sqrt{ (3 - (-5))^2 + (5 - 9)^2} \\ & = \sqrt{80} \\ & = \sqrt{16} \sqrt{5} \\ & = 4\sqrt{5} \text{ units} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2} \times \text{Base} \times \text{Height} \\ 18 & = {1 \over 2} (AB)(h) \\ 36 & = (4 \sqrt{5})(h) \\ \\ h & = {36 \over 4 \sqrt{5}} \\ & = {9 \over \sqrt{5}} \\ & = {9 \over \sqrt{5}} \times {\sqrt{5} \over \sqrt{5}} \\ & = {9\sqrt{5} \over 5} \\ \\ \therefore \text{Perpendicular distance} & = {9\sqrt{5} \over 5} \text{ units} \end{align}

(i)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 2 & 3 + t & 3 & 2 \\ t & 2 & 4 & t \end{matrix} \right| \\ & = {1 \over 2} [ 2 \times 2 + (3 + t) \times 4 + 3 \times t - t \times (3 + t) - 2 \times 3 - 4 \times 2 ] \\ & = {1 \over 2} (4 + 12 + 4t + 3t - 3t - t^2 - 6 - 8) \\ & = {1 \over 2} (2 + 4t - t^2) \text{ sq. units} \end{align}

(ii)

\begin{align} 2{1 \over 2} & = {1 \over 2} (2 + 4t - t^2) \\ 5 & = 2 + 4t - t^2 \\ 0 & = -3 + 4t - t^2 \\ 0 & = t^2 - 4t + 3 \\ 0 & = (t - 3)(t - 1) \\ \\ t - 3 = 0 \phantom{00} & \phantom{000} t - 1 = 0 \\ t = 3 \phantom{00} & \phantom{00000-} t = 1 \end{align}

(i)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 1 & 5 & 3 & 1 \\ 3 & 1 & r & 3 \end{matrix} \right| \\ 4 & = {1 \over 2} (1 \times 1 + 5 \times r + 3 \times 3 - 3 \times 5 - 1 \times 3 - r \times 1) \\ 4 & = {1 \over 2} (1 + 5r + 9 - 15 - 3 - r) \\ 4 & = {1 \over 2} (4r - 8) \\ 8 & = 4r - 8 \\ 16 & = 4r \\ \\ r & = {16 \over 4} \\ & = 4 \end{align}

(ii)

\begin{align} \text{Area of } \triangle ACB & = {1 \over 2} \left| \begin{matrix} 1 & 3 & 5 & 1 \\ 3 & r & 1 & 3 \end{matrix} \right| \\ 4 & = {1 \over 2} (1 \times r + 3 \times 1 + 5 \times 3 - 3 \times 3 - r \times 5 - 1 \times 1) \\ 4 & = {1 \over 2} (r + 3 + 15 - 9 - 5r - 1) \\ 4 & = {1 \over 2} (8 - 4r) \\ 8 & = 8 - 4r \\ 0 & = -4r \\ \\ r & = {0 \over -4} \\ & = 0 \end{align}

(i)

The diagonals $DB$ and $AC$ share the same mid-point \begin{align} \text{Mid-point of } AC & = \left( {2 + 1 \over 2}, {1 + 4 \over 2} \right) \\ & = (1.5, 2.5) \\ \\ \\ \text{Let coordinates } & \text{of } B \text{ be } (e, f). \\ \\ \text{Mid-point of } DB & = \left( {0 + e \over 2}, {2 + f \over 2} \right) \\ (1.5, 2.5) & = \left( {e \over 2}, {2 + f \over 2} \right) \\ \\ \text{Comparing the } & x\text{-coordinate,} \\ 1.5 & = {e \over 2} \\ 2(1.5) & = e \\ \\ e & = 3 \\ \\ \text{Comparing the } & y\text{-coordinate,} \\ 2.5 & = {2 + f \over 2} \\ 2(2.5) & = 2 + f \\ 5 & = 2 + f \\ \\ f & = 5 - 2 \\ & = 3 \\ \\ \therefore & \phantom{/} B(3, 3) \end{align}

(ii)

\begin{align} \text{Area}& = {1 \over 2} \left| \begin{matrix} 2 & 3 & 1 & 0 & 2 \\ 1 & 3 & 4 & 2 & 1 \end{matrix} \right| \\ & = {1 \over 2} (2 \times 3 + 3 \times 4 + 1 \times 2 + 0 \times 1 - 1 \times 3 - 3 \times 1 - 4 \times 0 - 2 \times 2) \\ & = {1 \over 2} (10) \\ & = 5 \text{ sq. units} \end{align}

(i)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 3 & -4 & 6 & 3 \\ 4 & 2 & -1 & 4 \end{matrix} \right| \\ & = {1 \over 2} [3 \times 2 + (-4) \times (-1) + 6 \times 4 - 4 \times (-4) - 2 \times 6 - (-1) \times 3] \\ & = {1 \over 2} (41) \\ & = 20.5 \text{ sq. units} \end{align}

(ii)

\begin{align} \text{Area of } ABCD & = {1 \over 2} \left| \begin{matrix} 3 & -4 & 6 & p & 3 \\ 4 & 2 & -1 & 3 & 4 \end{matrix} \right| \\ & = {1 \over 2} [ 3 \times 2 + (-4) \times (-1) + 6 \times 3 + p \times 4 - 4 \times (-4) - 2 \times 6 - (-1) \times p - 3 \times 3 ] \\ & = {1 \over 2} ( 6 + 4 + 18 + 4p + 16 - 12 + p - 9) \\ & = {1 \over 2} (23 + 5p) \text{ sq. units} \end{align}

(iii)

\begin{align} \text{Area of } ABCD & = 3 \times \text{Area of } \triangle ABC \\ {1 \over 2} (23 + 5p) & = 3 \times 20.5 \\ {1 \over 2} (23 + 5p) & = 61.5 \\ 23 + 5p & = 2(61.5) \\ 23 + 5p & = 123 \\ 5p & = 123 - 23 \\ & = 100 \\ p & = {100 \over 5} \\ & = 20 \end{align}

(i)

\begin{align} 3x - 2y & = 9 \\ -2y & = -3x + 9 \\ 2y & = 3x - 9 \\ y & = {3 \over 2}x - {9 \over 2} \\ \\ \text{Comparing with } & y = mx + c, \\ \\ \therefore \text{Gradient} & = {3 \over 2} \\ \\ \\ \text{Gradient of } AD \times \left(3 \over 2\right) & = -1 \\ \text{Gradient of } AD & = {-1 \over {3 \over 2}} \\ & = -{2 \over 3} \\ \\ \text{Gradient of } BC & = \text{Gradient of } AD \\ & = -{2 \over 3} \end{align}

The gradient of $BC$ is $-{2 \over 3}$ and it passes through the point $B (4, 7)$
\begin{align} y & = mx + c \\ y & = -{2 \over 3}x + c \\ \\ \text{Using } & B(4, 7), \\ 7 & = -{2 \over 3}(4) + c \\ 7 & = -{8 \over 3} + c \\ {29 \over 3} & = c \\ \\ y & = -{2 \over 3}x + {29 \over 3} \\ 3y & = -2x + 29 \\ \therefore 3y + 2x & = 29 \end{align}

(ii)

The lines $BC$ and $DE$ meet at point $E$ \begin{align} \text{Gradient of } DE & = \text{Gradient of line } 3x - 2y = 9 \\ & = {3 \over 2} \\ \\ y & = mx + c \\ y & = {3 \over 2}x + c \\ \\ \text{Using } & D(-3, 3), \\ 3 & = {3 \over 2}(-3) + c \\ 3 & = -{9 \over 2} + c \\ {15 \over 2} = c \\ \\ \text{Equation of } DE: \phantom{00} y & = {3 \over 2}x + {15 \over 2} \phantom{000} \text{ --- (1)} \\ \\ \text{Equation of } BC: \phantom{00} 3y & + 2x = 29 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{ (1) into (2),} \\ 3\left({3 \over 2}x + {15 \over 2}\right) + 2x & = 29 \\ {9 \over 2}x + {45 \over 2} + 2x & = 29 \\ 9x + 45 + 4x & = 58 \\ 13x & = 58 - 45 \\ & = 13 \\ x & = {13 \over 13} \\ & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into (1),} \\ y & = {3 \over 2}(1) + {15 \over 2} \\ & = 9 \\ \\ \therefore & \phantom{/} E(1, 9) \end{align}

(iii)

Since $ABFD$ is a parallelogram, lines $AD$ and $BF$ are parallel. Since $AD$ and $BC$ are parallel as well, $F$ lies on the line $BC$ and the point is actually the $y$-intercept of the line \begin{align} \text{When } & x = 0, \\ 3y + 2(0) & = 29 \\ 3y & = 29 \\ y & = {29 \over 3} \\ \\ \therefore & \phantom{/} F \left(0, {29 \over 3}\right) \end{align}

Lines $DF$ and $AB$ are parallel sides of the parallelogram, so they have the same gradient
\begin{align} \text{Gradient of } DF & = {3 - {29 \over 3} \over -3 - 0} \\ & = {20 \over 9} \\ \\ \text{Gradient of } AB & = \text{Gradient of } DF \\ & = {20 \over 9} \\ \\ y & = mx + c \\ y & = {20 \over 9}x + c \\ \\ \text{Using } & B(4, 7), \\ 7 & = {20 \over 9}(4) + c \\ 7 & = {80 \over 9} + c \\ -{17 \over 9} & = c \\ \\ y & = {20 \over 9}x - {17 \over 9} \\ \text{Equation of } AB: \phantom{0} 9y & = 20x - 17 \phantom{00} \text{ --- (3)} \\ \\ \\ \text{Gradient of } AD & = -{2 \over 3} \\ \\ y & = mx + c \\ y & = -{2 \over 3}x + c \\ \\ \text{Using } & D(-3, 3), \\ 3 & = -{2 \over 3}(-3) + c \\ 3 & = 2 + c \\ 1 & = c \\ \\ \text{Equation of } AD: \phantom{0} y & = -{2 \over 3}x + 1 \phantom{00} \text{ --- (4)} \\ \\ \\ \text{Substitute } & \text{(4) into (3),} \\ 9\left(-{2 \over 3}x + 1\right) & = 20x - 17 \\ -6x + 9 & = 20x - 17 \\ -26x & = -26 \\ x & = {-26 \over -26} \\ & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into (4),} \\ y & = -{2 \over 3}(1) + 1 \\ & = {1 \over 3} \\ \\ \therefore & \phantom{/} A \left(1, {1 \over 3} \right) \end{align}

(iv)

\begin{align} \text{Area of } ABFD & = {1 \over 2} \left| \begin{matrix} 1 & 4 & 0 & -3 & 1 \\ {1 \over 3} & 7 & {29 \over 3} & 3 & {1 \over 3} \end{matrix} \right| \\ & = {1 \over 2} \left[ 1 \times 7 + 4 \times {29 \over 3} + 0 \times 3 + (-3) \times {1 \over 3} - {1 \over 3} \times 4 - 7 \times 0 - {29 \over 3} \times (-3) - 3 \times 1 \right] \\ & = {1 \over 2} \left(208 \over 3\right) \\ & = {104 \over 3} \text{ sq. units} \end{align}

(i)

\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ AB & = \sqrt{ (0 - 3a)^2 + (1 - (4a + 1))^2} \\ 5 & = \sqrt{ (-3a)^2 + (1 - 4a - 1)^2} \\ 5 & = \sqrt{ 9a^2 + (-4a)^2} \\ 5 & = \sqrt{ 9a^2 + 16a^2 } \\ 5 & = \sqrt{ 25a^2} \\ 5 & = 5a \\ \\ a & = {5 \over 5} \\ & = 1 \end{align}

(ii)

\begin{align} & \phantom{/} A (3, 5) \\ \\ \text{Gradient of } AB & = {5 -1 \over 3 - 0} \\ & = {4 \over 3} \\ \\ \\ \text{Gradient of } BC \times \text{Gradient of } AB & = -1 \\ \text{Gradient of } BC & = {-1 \over \text{Gradient of } AB} \\ & = {-1 \over {4 \over 3}} \\ & = -{3 \over 4} \end{align}

The line $BC$ has a gradient of $-{3 \over 4}$ and passes through the point $B (0, 1)$
\begin{align} y & = mx + c \\ y & = -{3 \over 4}x + c \\ \\ \text{Using } & B(0, 1), \\ 1 & = -{3 \over 4}(0) + c \\ 1 & = c \\ \\ y & = -{3 \over 4}x + 1 \\ 4y & = -3x + 4 \\ \text{Equation of } BC: \phantom{00} 4y + 3x & = 4 \end{align}

Point $C$ is the $x$-intercept of the line $BC$
\begin{align} \text{When } & y =0, \\ 4(0) + 3x & = 4 \\ 3x & = 4 \\ x & = {4 \over 3} \\ \\ \therefore & \phantom{/} C \left({4 \over 3}, 0\right) \end{align}

(iii)

\begin{align} \text{Mid-point of } AB & = \left( {0 +3 \over 2}, {1 +5 \over 2} \right) \\ & = \left( {3 \over 2} , 3 \right) \\ \\ \text{Gradient of } \perp \text{bisector} & = \text{Gradient of } BC \\ & = -{3 \over 4} \\ \\ y & = mx + c \\ y & = -{3 \over 4}x + c \\ \\ \text{Using } & \text{mid-point } \left({3 \over 2}, 3\right), \\ 3 & = -{3 \over 4}\left(3 \over 2\right) + c \\ 3 & = -{9 \over 8} + c \\ {33 \over 8} + c \\ \\ y & = -{3 \over 4}x + {33 \over 8} \\ 4y & = -3x + {33 \over 2} \\ \text{Equation of } \perp \text{bisector}: \phantom{0} 4y + 3x & = {33 \over 2} \end{align}

The point $D$ is the $x$-intercept
\begin{align} \text{When } & y = 0, \\ 4(0) + 3x & = {33 \over 2} \\ 3x & = {33 \over 2} \\ x & = {33 \over 2} \div 3 \\ & = {11 \over 2} \\ \\ \therefore & \phantom{/} D \left({11 \over 2}, 0 \right) \end{align}
\begin{align} \text{Area of } ABCD & = {1 \over 2} \left| \begin{matrix} 3 & 0 & {4 \over 3}& {11 \over 2} & 3 \\ 5 & 1 & 0 & 0 & 5 \end{matrix} \right| \\ & = {1 \over 2} \left[ 3 \times 1 + 0 \times 0 + {4 \over 3} \times 0 + {11 \over 2} \times 5 - 5 \times 0 - 1 \times {4 \over 3} - 0 \times {11 \over 2} - 0 \times 3 \right] \\ & = {1 \over 2} \left(175 \over 6 \right) \\ & = {175 \over 12} \\ & = 14{7 \over 12} \text{ units} \end{align}