(a)

\begin{align*} \text{Parallel since } & {-8 \choose 4} = 4{-2 \choose 1} \end{align*}

(b)

\begin{align*} {9 \choose 7} & = k {18 \choose 21} \\ {9 \choose 7} & = {18k \choose 21k} \\ \\ 9 = 18k \phantom{00} & \text{ or } \phantom{00} 7 = 21k \\ {9 \over 18} = k \phantom{0000} & \phantom{000} {7 \over 21} = k \\ {1 \over 2} = k \phantom{0000} & \phantom{0000} {1 \over 3} = k \\ \\ \therefore & \phantom{.} \text{Not parallel} \end{align*}

(c)

\begin{align*} {6 \choose -8} & = 2{3 \choose -4} \\ \\ {-3 \choose 4} & = -{3 \choose -4} \\ \\ \text{Parallel since } & {6 \choose -8} = -2 {-3 \choose 4} \end{align*}

(a)

\begin{align*} & \phantom{00} {8 \choose -7} \\ \\ 2{8 \choose -7} & = {16 \choose -14} \\ \\ -3{8 \choose -7} & = {-24 \choose 21} \end{align*}

(b)

\begin{align*} {3 \choose 9} & = 3{1 \choose 3} \\ \\ 2{1 \choose 3} & = {2 \choose 6} \\ \\ -{1 \choose 3} & = {-1 \choose -3} \end{align*}

(c)

\begin{align*} {-6 \choose -2} & = -2{3 \choose 1} \\ \\ -3{3 \choose 1} & = {-9 \choose -3} \\ \\ 0.5{3 \choose 1} & = {1.5 \choose 0.5} \end{align*}

(a)

\begin{align*} \textbf{p} + 2\textbf{q} & = {5 \choose 2} + 2 {6 \choose -3} \\ & = {5 \choose 2} + {12 \choose -6} \\ & = {5 + 12 \choose 2 - 6} \\ & = {17 \choose -4} \end{align*}

(b)

\begin{align*} 3 \textbf{p} - {1 \over 2} \textbf{q} & = 3 {5 \choose 2} - {1 \over 2} {6 \choose -3} \\ & = {15 \choose 6} - {3 \choose -1.5} \\ & = {15 - 3 \choose 6 + 1.5} \\ & = {12 \choose 7.5} \end{align*}

(c)

\begin{align*} 4 \textbf{p} - 3 \textbf{q} + \textbf{r} & = 4 {5 \choose 2} - 3 {6 \choose -3} + {-3 \choose -4} \\ & = {20 \choose 8} - {18 \choose -9} + {-3 \choose -4} \\ & = {20 - 18 \choose 8 + 9} + {-3 \choose -4} \\ & = {2 - 3 \choose 17 - 4} \\ & = {-1 \choose 13} \end{align*}

(a)

$$\overrightarrow{OA} = {4 \choose 7} $$

(b)

$$ \overrightarrow{OB} = {-2 \choose 5} $$

(c)

$$ \overrightarrow{OC} = {6 \choose -1} $$

(d)

$$ \overrightarrow{OD} = {-4 \choose -9} $$

(a)

\begin{align*} \overrightarrow{PQ} & = \overrightarrow{PO} + \overrightarrow{OQ} \\ & = -\overrightarrow{OP} + \overrightarrow{OQ} \\ & = - {3 \choose -2} + {2 \choose -4} \\ & = {-3 \choose 2} + {2 \choose -4} \\ & = {-1 \choose -2} \end{align*}

(b)

\begin{align*} \overrightarrow{QR} & = \overrightarrow{QO} + \overrightarrow{OR} \\ & = -\overrightarrow{OQ} + \overrightarrow{OR} \\ & = -{2 \choose -4} + {2 \choose 3} \\ & = {-2 \choose 4} + {2 \choose 3} \\ & = {0 \choose 7} \end{align*}

(c)

\begin{align*} \overrightarrow{RP} & = \overrightarrow{RO} + \overrightarrow{OP} \\ & = -\overrightarrow{OR} + \overrightarrow{OP} \\ & = - {2 \choose 3} + {3 \choose -2} \\ & = {-2 \choose -3} + {3 \choose -2} \\ & = {1 \choose -5} \end{align*}

(d)

\begin{align*} \overrightarrow{PR} & = \overrightarrow{PO} + \overrightarrow{OR} \\ & = -\overrightarrow{OP} + \overrightarrow{OR} \\ & = -{3 \choose -2} + {2 \choose 3} \\ & = {-3 \choose 2} + {2 \choose 3} \\ & = {-1 \choose 5} \end{align*}

(a)

\begin{align*} {6 \choose -3} & = 3{2 \choose -1} \\ \\ {-4 \choose 2} & = -2{2 \choose -1} \\ \\ \text{Parallel since } & {6 \choose -3} = -{3 \over 2} {-4 \choose 2} \end{align*}

(b)

\begin{align*} {-5 \choose 15} & = 5{-1 \choose 3} \\ \\ {-3 \choose 9} & = 3{-1 \choose 3} \\ \\ \text{Parallel since } & {-5 \choose 15} = {5 \over 3} {-3 \choose 9} \end{align*}

(c)

\begin{align*} {7 \choose -8} & = k{2 \choose -3} \\ {7 \choose -8} & = {2k \choose -3k} \\ \\ 7 = 2k \phantom{00} & \text{ or } \phantom{00} -8 = -3k \\ {7 \over 2} = k \phantom{000} & \phantom{0000000} 8 = 3k \\ & \phantom{000000.} {8 \over 3} = k \\ \\ \therefore & \phantom{.} \text{Not parallel} \end{align*}

(a)

\begin{align*} {20 \choose p} & = k{5 \choose 2} \\ {20 \choose p} & = {5k \choose 2k} \\ \\ 20 & = 5k \\ {20 \over 5} & = k \\ 4 & = k \\ \\ p & = 2k \\ p & = 2(4) \\ p & = 8 \end{align*}

(b)

\begin{align*} {h \choose 12} & = k{3 \choose -9} \\ {h \choose 12} & = {3k \choose -9k} \\ \\ 12 & = -9k \\ {12 \over -9} & = k \\ -{4 \over 3} & = k \\ \\ h & = 3k \\ h & = 3\left(- {4 \over 3}\right) \\ h & = -4 \end{align*}

(a)

\begin{align*} \textbf{a} + 2 \textbf{b} & = {8 \choose 9} \\ {x \choose y} + 2 {3 \choose 4} & = {8 \choose 9} \\ {x \choose y} + {6 \choose 8} & = {8 \choose 9} \\ {x + 6 \choose y + 8} & = {8 \choose 9} \\ \\ x + 6 & = 8 \\ x & = 8 - 6 \\ x & = 2 \\ \\ y + 8 & = 9 \\ y & = 9 - 8 \\ y & = 1 \end{align*}

(b)

\begin{align*} 4 \textbf{u} + \textbf{v} & = 2{ {1 \over 2} \choose 9} \\ 4 {2 \choose y} + {x \choose 2} & = {1 \choose 18} \\ {8 \choose 4y} + {x \choose 2} & = {1 \choose 18} \\ {8 + x \choose 4y + 2} & = {1 \choose 18} \\ \\ 8 + x & = 1 \\ x & = 1 - 8 \\ x & = -7 \\ \\ 4y + 2 & = 18 \\ 4y & = 18 - 2 \\ 4y & = 16 \\ y & = {16 \over 4} \\ y & = 4 \end{align*}

(c)

\begin{align*} 5 \textbf{p} - 2 \textbf{q} & = {3 \choose 23} \\ 5 {x \choose 5} - 2 {6 \choose y} & = {3 \choose 23} \\ {5x \choose 25} - {12 \choose 2y} & = {3 \choose 23} \\ {5x - 12 \choose 25 - 2y} & = {3 \choose 23} \\ \\ 5x - 12 & = 3 \\ 5x & = 3 + 12 \\ 5x & = 15 \\ x & = {15 \over 5} \\ x & = 3 \\ \\ 25 - 2y & = 23 \\ -2y & = 23 - 25 \\ -2y & = -2 \\ y & = {-2 \over -2} \\ y & = 1 \end{align*}

(a)

(b)

(c)

(d)

(e)

(f)

\begin{align*} \overrightarrow{OB} & = \overrightarrow{OA} + \overrightarrow{AB} \\ & = {-3 \choose 8} + {-2 \choose -4} \\ & = {-5 \choose 4} \\ \\ \therefore \phantom{.} & B(-5, 4) \end{align*}

(i)

\begin{align*} \overrightarrow{CD} & = {2 \over 3} \overrightarrow{AB} \\ & = {2 \over 3} {9 \choose -15} \\ & = {6 \choose -10} \end{align*}

(ii)

\begin{align*} \overrightarrow{OB} & = \overrightarrow{OA} + \overrightarrow{AB} \\ & = {-2 \choose 7} + {9 \choose -15} \\ & = {7 \choose -8} \\ \\ \therefore & \phantom{.} B(7, -8) \end{align*}

(iii)

\begin{align*} \overrightarrow{OC} & = \overrightarrow{OD} + \overrightarrow{DC} \\ & = \overrightarrow{OD} - \overrightarrow{CD} \\ & = {8 \choose -5} - {6 \choose -10} \\ & = {8 - 6 \choose -5 + 10} \\ & = {2 \choose 5} \\ \\ \therefore & \phantom{.} C(2, 5) \end{align*}

\begin{align*} {a \choose b} & = k {c \choose d} \\ {a \choose b} & = {kc \choose kd} \\ \\ a = kc \phantom{00} & \text{ or } \phantom{00} b = kd \\ {a \over c} = k \phantom{000} & \phantom{0000} {b \over d} = k \\ \\ \therefore k & = {a \over c} = {b \over d} \end{align*}

\begin{align*} \textbf{u} & = k \textbf{v} \\ {-15 \choose 8} & = k \textbf{v} \\ {1 \over k} {-15 \choose 8} & = \textbf{v} \\ {-{15 \over k} \choose {8 \over k}} & = \textbf{v} \\ \\ |\textbf{v}| & = \sqrt{ \left(-{15 \over k}\right)^2 + \left(8 \over k\right)^2} \\ 51 & = \sqrt{ {225 \over k^2} + {64 \over k^2}} \\ 51 & = \sqrt{ 289 \over k^2 } \\ 51 & = {17 \over k} \\ 51k & = 17 \\ k & = {17 \over 51} \\ k & = {1 \over 3} \\ \\ {-15 \choose 8} & = {1 \over 3} \textbf{v} \\ 3{-15 \choose 8} & = \textbf{v} \\ {-45 \choose 24} & = \textbf{v} \end{align*}

(i)

\begin{align*} 2 \overrightarrow{AB} + 5 \overrightarrow{CD} & = 2 {-3 \choose 5} + 5 {1 \choose 4} \\ & = {-6 \choose 10} + {5 \choose 20} \\ & = {-1 \choose 30} \end{align*}

(ii)

\begin{align*} \overrightarrow{AB} & = \lambda \overrightarrow{EF} \\ {-3 \choose 5} & = \lambda {k \choose 7.5} \\ {-3 \choose 5} & = {k \lambda \choose 7.5 \lambda} \\ \\ 7.5\lambda & = 5 \\ \lambda & = {5 \over 7.5} \\ \lambda & = {2 \over 3} \\ \\ -3 & = k \lambda \\ -3 & = k \left({2 \over 3}\right) \\ -3 \div {2 \over 3} & = k \\ -4.5 & = k \end{align*}

(iii)

\begin{align*} \overrightarrow{PQ} & = { {1 \over 4} \choose 1} \\ & = {1 \over 4} { 1 \choose 4} \\ & = {1 \over 4} \overrightarrow{CD} \\ \\ \text{Since } \overrightarrow{PQ} = {1 \over 4} \overrightarrow{CD} &, \text{ both vectors are parallel} \end{align*}

(i)

\begin{align*} \overrightarrow{LM} & = \overrightarrow{LO} + \overrightarrow{OM} \\ & = -\overrightarrow{OL} + {t \choose 6} \\ & = - {-3 \choose 2} + {t \choose 6} \\ & = {3 \choose -2} + {t \choose 6} \\ & = {3 + t \choose 4} \end{align*}

(ii)

\begin{align*} \overrightarrow{LM} & = k \textbf{p} \\ {3 + t \choose 4} & = k {8 \choose 1} \\ {3 + t \choose 4} & = {8k \choose k} \\ \\ 4 & = k \\ \\ 3 + t & = 8k \\ 3 + t & = 8(4) \\ 3 + t & = 32 \\ t & = 32 - 3 \\ t & = 29 \end{align*}

(iii)

\begin{align*} | \overrightarrow{LM} | & = \sqrt{ (3 + t)^2 + 4^2} \\ & = \sqrt{ 3^2 + 2(3)(t) + t^2 + 16 } \\ & = \sqrt{ 9 + 6t + t^2 + 16} \\ & = \sqrt{ t^2 + 6t + 25 } \\ \\ | \textbf{p} | & = \sqrt{ 8^2 + 1^2 } \\ & = \sqrt{ 64 + 1 } \\ & = \sqrt{ 65 } \\ \\ | \textbf{p} | & = | \overrightarrow{LM} | \\ \sqrt{65} & = \sqrt{t^2 + 6t + 25} \\ 65 & = t^2 + 6t + 25 \\ 0 & = t^2 + 6t + 25 - 65 \\ 0 & = t^2 + 6t - 40 \\ 0 & = (t + 10)(t - 4) \\ \\ t + 10 = 0 \phantom{000} & \text{ or } \phantom{00} t - 4 = 0 \\ t = -10 \phantom{.}& \phantom{00000000} t = 4 \end{align*}

(i)

\begin{align*} \overrightarrow{OQ} & = \overrightarrow{OP} + \overrightarrow{PQ} \\ & = {2 \choose -3} + {8 \choose -2} \\ & = {10 \choose -5} \\ \\ \therefore & \phantom{.} Q(10, - 5) \end{align*}

(ii)

\begin{align*} \text{Gradient} & = {y_1 - y_2 \over x_1 - x_2} \\ & = {-5 - (-3) \over 10 - 2} \\ & = {-2 \over 8} \\ & = -{1 \over 4} \end{align*}

(iii)

\begin{align*} \text{Gradient} & = {y \over x} \end{align*}

(iv)

$$ \overrightarrow{PQ} = k {x \choose y}, \text{ for some real value of } k $$