(a)

\begin{align*} \text{Magnitude of } {5 \choose -12} & = \sqrt{ 5^2 + (-12)^2 } \\ & = 13 \text{ units} \end{align*}

(b)

\begin{align*} \text{Magnitude of } {-6 \choose 8} & = \sqrt{ (-6)^2 + 8^2 } \\ & = 10 \text{ units} \end{align*}

(c)

\begin{align*} \text{Magnitude of } {5 \choose 2} & = \sqrt{ 5^2 + 2^2 } \\ & = 5.3851 \\ & \approx 5.39 \text{ units} \end{align*}

(d)

\begin{align*} \text{Magnitude of } {-7 \choose -1} & = \sqrt{ (-7)^2 + (-1)^2 } \\ & = 7.071 \\ & \approx 7.07 \text{ units} \end{align*}

(e)

\begin{align*} \text{Magnitude of } {0 \choose -3} & = \sqrt{ 0^2 + (-3)^2 } \\ & = 3 \text{ units} \end{align*}

\begin{align*} | \overrightarrow{XY} | & = \sqrt{ p^2 + (-2)^2 } \\ 5 & = \sqrt{ p^2 + 4 } \\ 5^2 & = p^2 + 4 \\ 25 & = p^2 + 4 \\ 25 - 4 & = p^2 \\ 21 & = p^2 \\ \pm \sqrt{21} & = p \\ \\ \therefore p & = \pm 4.5825 \\ & \approx \pm 4.58 \end{align*}

(i)

\begin{align*} | \overrightarrow{AB} | & = \sqrt{ (-4)^2 + 2^2 } \\ & = \sqrt{ 20} \\ & = 4.4721 \\ & \approx 4.47 \text{ units} \end{align*}

(ii)

\begin{align*} | \overrightarrow{CD} | & = \sqrt{ p^2 + (-12)^2 } \\ & = \sqrt{ p^2 + 144 } \\ \\ | \overrightarrow{CD} | & = 3 | \overrightarrow{AB} | \\ \sqrt{ p^2 + 144 } & = 3 \sqrt{ 20 } \\ \left( \sqrt{ p^2 + 144 } \right)^2 & = \left( 3 \sqrt{ 20 } \right)^2 \\ p^2 + 144 & = 180 \\ p^2 & = 180 - 144 \\ p^2 & = 36 \\ p & = \pm \sqrt{36} \\ p & = \pm 6 \\ \\ \therefore \text{Positive value of } p & = 6 \end{align*}

\begin{align*} \textbf{a} & = \textbf{b} \\ {p + q \choose p} & = {3 \choose q + 1} \\ \\ p + q & = 3 \phantom{000} \text{--- (1)} \\ p & = q + 1 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (q + 1) + q & = 3 \\ q + 1 + q & = 3 \\ 2q & = 3 - 1 \\ 2q & = 2 \\ q & = {2 \over 2} \\ q & = 1 \\ \\ \text{Substitute } & q = 1 \text{ into (2),} \\ p & = (1) + 1 \\ p & = 2 \end{align*}

(a)(i)

$$ \overrightarrow{HK}, \overrightarrow{GL} $$

(a)(ii)

$$ \overrightarrow{CL}, \overrightarrow{DE} $$

(a)(iii)

$$ \overrightarrow{GH}, \overrightarrow{FG} $$

(a)(iv)

$$ \overrightarrow{AB}, \overrightarrow{CD} $$

(a)(v)

$$ \overrightarrow{CE}, \overrightarrow{KG} $$

(b)(i)

$$ \overrightarrow{FL}$$

(b)(ii)

$$ \overrightarrow{KJ}$$

(b)(iii)

$$ \overrightarrow{KB}$$

(c)(i)

$$ \text{Both vectors have different magnitude and direction.} $$

(c)(ii)

$$ \text{Both vectors have different magnitude and direction.} $$

(a)

\begin{align*} \overrightarrow{NM} & = \overrightarrow{KL} \end{align*}

(b)

\begin{align*} \overrightarrow{RQ} & = \overrightarrow{TU} \\ \\ \overrightarrow{QP} & = \overrightarrow{ST} \end{align*}

(c)

\begin{align*} \overrightarrow{AB} & = \overrightarrow{DC} \\ \\ \overrightarrow{BC} & = \overrightarrow{AD} \end{align*}

(d)

\begin{align*} \overrightarrow{LM} & = \overrightarrow{QP} \\ \\ \overrightarrow{MN} & = \overrightarrow{RQ} \\ \\ \overrightarrow{NO} & = \overrightarrow{SR} \\ \\ \overrightarrow{OP} & = \overrightarrow{LS} \end{align*}

(i)

(ii)

(iii)

(i)

\begin{align*} \overrightarrow{AD} + \overrightarrow{DC} & = \overrightarrow{AC} \end{align*}

(ii)

\begin{align*} \overrightarrow{AB} + \overrightarrow{BD} & = \overrightarrow{AD} \end{align*}

(iii)

\begin{align*} \overrightarrow{AC} + \overrightarrow{CB} + \overrightarrow{BD} & = \overrightarrow{AB} + \overrightarrow{BD} \\ & = \overrightarrow{AD} \end{align*}

(iv)

\begin{align*} \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} & = \overrightarrow{AC} + \overrightarrow{CA} \\ & = \textbf{0} \end{align*}

(i)

\begin{align*} \overrightarrow{SO} + \overrightarrow{OP} & = \overrightarrow{SP} \\ \\ \textbf{x} & = \overrightarrow{OP} \end{align*}

(ii)

\begin{align*} \overrightarrow{PO} + \overrightarrow{OR} & = \overrightarrow{PR} \\ \\ \textbf{x} & = \overrightarrow{OR} \end{align*}

(iii)

\begin{align*} \overrightarrow{RS} + \overrightarrow{SQ} & = \overrightarrow{RQ} \\ \\ \textbf{x} & = \overrightarrow{RS} \end{align*}

(iv)

\begin{align*} \overrightarrow{PR} + \overrightarrow{RP} & = \textbf{0} \\ \\ \textbf{x} & = \overrightarrow{RP} \end{align*}

(v)

\begin{align*} \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS} & = \overrightarrow{PR} + \overrightarrow{RS} \\ & = \overrightarrow{PS} \\ \\ \textbf{x} & = \overrightarrow{QR} \end{align*}

(vi)

\begin{align*} \overrightarrow{QR} + \overrightarrow{RS} + \overrightarrow{PQ} & = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS} \\ & = \overrightarrow{PR} + \overrightarrow{RS} \\ & = \overrightarrow{PS} \\ \\ \textbf{x} & = \overrightarrow{PQ} \end{align*}

(i)

\begin{align*} \textbf{a} - \textbf{b} & = {4 \choose 3} - {2 \choose 1} \\ & = {2 \choose 2} \end{align*}

(ii)

\begin{align*} \textbf{b} - \textbf{c} & = {2 \choose 1} - {-3 \choose 1} \\ & = {2 + 3 \choose 1 - 1} \\ & = {5 \choose 0} \end{align*}

(iii)

\begin{align*} \textbf{a} - ( \textbf{b} + \textbf{c}) & = {4 \choose 3} - \left[ {2 \choose 1} + {-3 \choose 1} \right] \\ & = {4 \choose 3} - {-1 \choose 2} \\ & = {4 + 1 \choose 3 - 2} \\ & = {5 \choose 1} \end{align*}

(iv)

\begin{align*} \textbf{a} - ( \textbf{b} - \textbf{c}) & = {4 \choose 3} - {5 \choose 0} \phantom{00000} [\text{from part ii}] \\ & = {-1 \choose 3} \end{align*}

\begin{align*} \textbf{u} & = k \textbf{v} \\ {4 \choose -3} & = k \textbf{v} \\ {1 \over k} {4 \choose -3} & = \textbf{v} \\ { {4 \over k} \choose -{3 \over k} } & = \textbf{v} \\ \\ | \textbf{v} | & = \sqrt{ \left(4 \over k\right)^2 + \left(-{3 \over k} \right)^2 } \\ 20 & = \sqrt{ {16 \over k^2} + {9 \over k^2} } \\ 20 & = \sqrt{ 25 \over k^2} \\ 20 & = {5 \over k} \\ 20k & = 5 \\ k & = {5 \over 20} \\ k & = {1 \over 4} \\ \\ \textbf{v} & = { {4 \over k} \choose -{3 \over k} } \\ & = \left( \begin{matrix} {4 \over {1 \over 4}} \\ -{3 \over {1 \over 4}} \end{matrix} \right) \\ & = {16 \choose -12} \end{align*}

(a)

\begin{align*} 2 \textbf{p} + 3 \textbf{q} & = 2 {5 \choose -12} + 3 {2 \choose 3} \\ & = {10 \choose -24} + {6 \choose 9} \\ & = {10 + 6 \choose -24 + 9 } \\ & = {16 \choose -15} \end{align*}

(b)(i)

\begin{align*} | \textbf{p} | & = \sqrt{ 5^2 + (-12)^2 } \\ & = 13 \text{ units} \end{align*}

(b)(ii)

\begin{align*} - \textbf{p} + 2 \textbf{q} & = - {5 \choose -12} + 2 {2 \choose 3} \\ & = {-5 \choose 12} + {4 \choose 6} \\ & = {-5 + 4 \choose 12 + 6} \\ & = {-1 \choose 18} \\ \\ | - \textbf{p} + 2 \textbf{q} | & = \sqrt{ (-1)^2 + 18^2 } \\ & = 18.0277 \\ & \approx 18 \text{ units (nearest whole number)} \end{align*}

(c)

\begin{align*} \textbf{r} & = k \textbf{p} \\ {20 \choose m} & = k {5 \choose -12} \\ {20 \choose m} & = {5k \choose -12k} \\ \\ 20 & = 5k \\ {20 \over 5} & = k \\ 4 & = k \\ \\ m & = -12k \\ m & = -12(4) \\ m & = -48 \end{align*}

(i)

\begin{align*} \overrightarrow{OB} & = \overrightarrow{OA} + \overrightarrow{AB} \\ & = {-5 \choose 3} + {6 \choose 1} \\ & = {-5 + 6 \choose 3 + 1} \\ & = {1 \choose 4} \\ \\ \therefore & \phantom{.} B(1, 4) \\ \\ \\ \overrightarrow{OD} & = \overrightarrow{OA} + \overrightarrow{AD} \\ & = {-5 \choose 3} + {8 \choose -9} \\ & = {-5 + 8 \choose 3 - 9} \\ & = {3 \choose -6} \\ \\ \therefore & \phantom{.} D(3, -6) \end{align*}

(ii)

\begin{align*} \overrightarrow{BC} & = \overrightarrow{BO} + \overrightarrow{OC} \\ & = - \overrightarrow{OB} + \overrightarrow{OC} \\ & = -{1 \choose 4} + {7 \choose 4} \\ & = {-1 \choose -4} + {7 \choose 4} \\ & = {-1 + 7 \choose -4 + 4} \\ & = {6 \choose 0} \\ \\ \overrightarrow{CD} & = \overrightarrow{CO} + \overrightarrow{OD} \\ & = -\overrightarrow{OC} + \overrightarrow{OD} \\ & = - {7 \choose 4} + {3 \choose -6} \\ & = {-7 \choose -4} + {3 \choose -6} \\ & = {-7 + 3 \choose -4 - 6} \\ & = {-4 \choose -10} \end{align*}

(i)

\begin{align*} \overrightarrow{OB} & = \overrightarrow{OA} + \overrightarrow{AB} \\ \textbf{b} & = \textbf{a} + {-3 \choose 2} \\ \textbf{b} & = {7 \choose 4} + {-3 \choose 2} \\ \textbf{b} & = {7 -3 \choose 4 + 2} \\ \textbf{b} & = {4 \choose 6} \end{align*}

(ii)

\begin{align*} \overrightarrow{OC} & = \overrightarrow{BA} \\ & = -\overrightarrow{AB} \\ & = - {-3 \choose 2} \\ & = {3 \choose -2} \\ \\ \therefore & \phantom{.} C (3, -2) \end{align*}

(i)

\begin{align*} AD & : BC \\ {2 \over 3} & : 1 \\ 2 & : 3 \\ \\ \overrightarrow{AC} & = \overrightarrow{AB} + \overrightarrow{BC} \\ & = \overrightarrow{AB} + {3 \over 2} \overrightarrow{AD} \\ & = \textbf{u} + {3 \over 2} \textbf{v} \end{align*}

(ii)

\begin{align*} \overrightarrow{DC} & = \overrightarrow{DA} + \overrightarrow{AC} \\ & = - \overrightarrow{AD} + \overrightarrow{AC} \\ & = -\textbf{v} + \left( \textbf{u} + {3 \over 2} \textbf{v} \right) \\ & = -\textbf{v} + \textbf{u} + {3 \over 2} \textbf{v} \\ & = \textbf{u} + {1 \over 2} \textbf{v} \end{align*}

(iii)

\begin{align*} DQ & : DC \\ {1 \over 4} & : 1 \\ 1 & : 4 \\ \\ \overrightarrow{AQ} & = \overrightarrow{AD} + \overrightarrow{DQ} \\ & = \overrightarrow{AD} + {1 \over 4} \overrightarrow{DC} \\ & = \textbf{v} + {1 \over 4} \left( \textbf{u} + {1 \over 2} \textbf{v} \right) \\ & = \textbf{v} + {1 \over 4} \textbf{u} + {1 \over 8} \textbf{v} \\ & = {1 \over 4} \textbf{u} + {9 \over 8} \textbf{v} \\ & = {1 \over 8} (2 \textbf{u} + 9 \textbf{v} ) \end{align*}

(iv)

\begin{align*} \overrightarrow{PQ} & = \overrightarrow{PA} + \overrightarrow{AQ} \\ & = -{1 \over 2} \overrightarrow{BA} + \overrightarrow{AQ} \\ & = -{1 \over 2} \textbf{u} + {1 \over 4} \textbf{u} + {9 \over 8} \textbf{v} \\ & = -{1 \over 4} \textbf{u} + {9 \over 8} \textbf{v} \\ & = {1 \over 8} (-2 \textbf{u} + 9 \textbf{v} ) \end{align*}

(a)(i)

\begin{align*} \overrightarrow{NM} & = \overrightarrow{NP} + \overrightarrow{PM} \\ & = - \overrightarrow{PN} + \overrightarrow{PM} \\ & = - 2 \textbf{a} + 2 \textbf{b} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{NL} & = {1 \over 2} \overrightarrow{NM} \\ & = {1 \over 2} ( - 2 \textbf{a} + 2 \textbf{b} ) \\ & = - \textbf{a} + \textbf{b} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{PK} & = {12 \over 7} \overrightarrow{PL} \\ & = {12 \over 7} \left( \overrightarrow{PN} + \overrightarrow{NL} \right) \\ & = {12 \over 7} [ 2 \textbf{a} + (- \textbf{a} + \textbf{b} ) ] \\ & = {12 \over 7} ( 2 \textbf{a} - \textbf{a} + \textbf{b} ) \\ & = {12 \over 7} ( \textbf{a} + \textbf{b} ) \end{align*}

(a)(iv)

\begin{align*} {PN} & : PR \\ {2 \over 3} & : 1 \\ 2 & : 3 \\ \\ \overrightarrow{PR} & = {3 \over 2} \overrightarrow{PN} \\ & = {3 \over 2} (2 \textbf{a} ) \\ & = 3 \textbf{a} \end{align*}

(a)(v)

\begin{align*} \overrightarrow{PQ} & = 2 \overrightarrow{PM} \\ & = 2 (2 \textbf{b} ) \\ & = 4 \textbf{b} \end{align*}

(b)

\begin{align*} \overrightarrow{RQ} & = \overrightarrow{RP} + \overrightarrow{PQ} \\ & = - \overrightarrow{PR} + \overrightarrow{PQ} \\ & = - 3 \textbf{a} + 4 \textbf{b} \end{align*}

(c)

\begin{align*} \overrightarrow{KR} & = \overrightarrow{KP} + \overrightarrow{PR} \\ & = - \overrightarrow{PK} + \overrightarrow{PR} \\ & = - {12 \over 7} ( \textbf{a} + \textbf{b} ) + 3 \textbf{a} \\ & = -{12 \over 7} \textbf{a} - {12 \over 7} \textbf{b} + 3 \textbf{a} \\ & = {9 \over 7} \textbf{a} - {12 \over 7} \textbf{b} \\ & = {3 \over 7} (3 \textbf{a} - 4 \textbf{b}) \phantom{000} (\text{Shown)} \end{align*}

(d)

\begin{align*} \overrightarrow{QR} & = -\overrightarrow{RQ} \\ & = - (-3 \textbf{a} + 4 \textbf{b} ) \phantom{00000} [\text{from part b}] \\ & = 3 \textbf{a} - 4 \textbf{b} \\ \\ \overrightarrow{KR} & = {3 \over 7} (3 \textbf{a} - 4 \textbf{b} ) \\ & = {3 \over 7} \overrightarrow{QR} \\ \\ KR & : QR \\ {3 \over 7} & : 1 \\ 3 & : 7 \\ \\ \therefore {KR \over QR} & = {3 \over 7} \end{align*}

(e)(i) Triangles PKR and PQR share a common height

\begin{align*} { \text{Area of } \triangle PKR \over \text{Area of } \triangle PQR } & = { {1 \over 2} \times KR \times h \over {1 \over 2} \times QR \times h} \\ & = {KR \over QR} \\ & = {3 \over 7} \end{align*}

(e)(ii) Triangles PKN and PKR share a common height

\begin{align*} { \text{Area of } \triangle PKN \over \text{Area of } \triangle PKR } & = { {1 \over 2} \times PN \times h \over {1 \over 2} \times PR \times h} \\ & = {PN \over PR} \\ & = {2 \over 3} \\ \\ \triangle PKN & : \triangle PKR : \triangle PQR \\ & : \phantom{00} 3 \phantom{000} : \phantom{000} 7 \phantom{00000} [\text{From part ei}] \\ 2 \phantom{00} & : \phantom{00} 3 \phantom{000} : \phantom{000} 7 \\ \\ \therefore { \text{Area of } \triangle PKN \over \text{Area of } \triangle PQR } & = {2 \over 7} \end{align*}

\begin{align*} \overrightarrow{OM} & = {1 \over 2} \overrightarrow{OA} \\ & = {1 \over 2} \textbf{a} \\ \\ \overrightarrow{MB} & = \overrightarrow{MO} + \overrightarrow{OB} \\ & = -\overrightarrow{OM} + \overrightarrow{OB} \\ & = - {1 \over 2} \textbf{a} + \textbf{b} \\ \\ BP & : PM \\ 3 & : 1 \\ \\ \overrightarrow{OP} & = \overrightarrow{OM} + \overrightarrow{MP} \\ & = \overrightarrow{OM} + {1 \over 4} \overrightarrow{MB} \\ & = {1 \over 2} \textbf{a} + {1 \over 4} \left( - {1 \over 2} \textbf{a} + \textbf{b} \right) \\ & = {1 \over 2} \textbf{a} - {1 \over 8} \textbf{a} + {1 \over 4} \textbf{b} \\ & = {3 \over 8} \textbf{a} + {1 \over 4} \textbf{b} \\ & = {1 \over 8} ( 3 \textbf{a} + 2 \textbf{b} ) \end{align*}

(i)

\begin{align*} \overrightarrow{OQ} & = \overrightarrow{OP} + \overrightarrow{PQ} \\ & = 4 \textbf{a} + 4 \textbf{b} \end{align*}

(ii)

\begin{align*} \overrightarrow{OX} & = \overrightarrow{OR} + \overrightarrow{RX} \\ & = \overrightarrow{PQ} + {1 \over 2} \overrightarrow{OP} \\ & = 4 \textbf{b} + {1 \over 2} (4 \textbf{a} ) \\ & = 4 \textbf{b} + 2 \textbf{a} \end{align*}

(iii)

\begin{align*} OR & : RS \\ {1 \over 2} & : 1 \\ 1 & : 2 \\ \\ \overrightarrow{QS} & = \overrightarrow{QR} + \overrightarrow{RS} \\ & = - \overrightarrow{RQ} + 2 \overrightarrow{OR} \\ & = - 4 \textbf{a} + 2 (4 \textbf{b}) \\ & = -4 \textbf{a} + 8 \textbf{b} \end{align*}

(a)(i)

\begin{align*} OQ & : QB \\ 2 & : 1 \\ \\ \overrightarrow{OQ} & = {2 \over 3} \overrightarrow{ OB} \\ & = {2 \over 3} \textbf{b} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{PQ} & = \overrightarrow{PO} + \overrightarrow{OQ} \\ & = - \overrightarrow{OP} + \overrightarrow{OQ} \\ & = - 2 \textbf{a} + {2 \over 3} \textbf{b} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{OM} & = \overrightarrow{OA} + \overrightarrow{AM} \\ & = \textbf{a} + {1 \over 2} \overrightarrow{AB} \\ & = \textbf{a} + {1 \over 2} (\overrightarrow{AO} + \overrightarrow{OB}) \\ & = \textbf{a} + {1 \over 2} (- \textbf{a} + \textbf{b} ) \\ & = \textbf{a} - {1 \over 2} \textbf{a} + {1 \over 2} \textbf{b} \\ & = {1 \over 2} \textbf{a} + {1 \over 2} \textbf{b} \\ & = {1 \over 2} (\textbf{a} + \textbf{b}) \end{align*}

(a)(iv)

\begin{align*} \overrightarrow{QM} & = \overrightarrow{QO} + \overrightarrow{OM} \\ & = -\overrightarrow{OQ} + \overrightarrow{OM} \\ & = - {2 \over 3} \textbf{b} + {1 \over 2} \textbf{a} + {1 \over 2} \textbf{b} \\ & = {1 \over 2} \textbf{a} - {1 \over 6} \textbf{b} \end{align*}

(b)

\begin{align*} \overrightarrow{QP} & = -\overrightarrow{PQ} \\ & = - \left( - 2 \textbf{a} + {2 \over 3} \textbf{b} \right) \phantom{00000} [\text{from part aii}]\\ & = 2 \textbf{a} - {2 \over 3} \textbf{b} \\ & = {4 \over 2} \textbf{a} - {4 \over 6} \textbf{b} \\ & = 4 \left( {1 \over 2} \textbf{a} - {1 \over 6} \textbf{b} \right) \\ & = 4 \overrightarrow{QM} \\ \\ QP & : QM \\ 4 & : 1 \\ \\ \therefore {PM \over MQ} & = {4 - 1 \over 1} \\ & = 3 \end{align*}

(a)(i)

\begin{align*} \overrightarrow{QP} & = \overrightarrow{QO} + \overrightarrow{OP} \\ & = -\overrightarrow{OQ} + \overrightarrow{OP} \\ & = - \textbf{q} + 2 \textbf{p} \end{align*}

(a)(ii)

\begin{align*} OP & : PR \\ 2 & : 1 \\ \\ \overrightarrow{OR} & = {3 \over 2} \overrightarrow{OP} \\ & = {3 \over 2} (2 \textbf{p} ) \\ & = 3 \textbf{p} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{SR} & = \overrightarrow{SO} + \overrightarrow{OR} \\ & = -\overrightarrow{OS} + \overrightarrow{OR} \\ & = -2 \overrightarrow{OQ} + \overrightarrow{OR} \\ & = -2 \textbf{q} + 3 \textbf{p} \end{align*}

(a)(iv)

\begin{align*} TQ & : PQ \\ 3 & : 1 \\ \\ \overrightarrow{ST} & = \overrightarrow{SQ} + \overrightarrow{QT} \\ & = - \overrightarrow{QS} - \overrightarrow{TQ} \\ & = - \textbf{q} - 3 \overrightarrow{PQ} \\ & = - \textbf{q} + 3 \overrightarrow{QP} \\ & = - \textbf{q} + 3 (- \textbf{q} + 2 \textbf{p} ) \\ & = - \textbf{q} - 3 \textbf{q} + 6 \textbf{p} \\ & = -4 \textbf{q} + 6 \textbf{p} \end{align*}

(b)

\begin{align*} \overrightarrow{ST} & = -4 \textbf{q} + 6 \textbf{p} \\ & = 2 (-2 \textbf{q} + 3 \textbf{p} ) \\ & = 2 \overrightarrow{SR} \\ \\ 1. & \phantom{.} S, R \text{ and } T \text{ are collinear points (i.e. lie on a straight line)} \\ 2. & \phantom{.} ST = 2SR \end{align*}

(c)

\begin{align*} { \text{Area of } \triangle SPQ \over \text{Area of } \triangle STQ} & = { {1 \over 2} \times PQ \times h \over {1 \over 2} \times TQ \times h } \\ & = {PQ \over TQ} \\ & = {1 \over 3} \\ \\ { \text{Area of } \triangle SPQ \over \text{Area of } \triangle SPT} & = {1 \over 3 - 1} \\ & = {1 \over 2} \end{align*}
\begin{align*} { \text{Area of } \triangle OPQ \over \text{Area of } \triangle SPQ} & = { {1 \over 2} \times OQ \times h \over {1 \over 2} \times SQ \times h } \\ & = {OQ \over SQ} \\ & = {1 \over 1} \\ \\ \triangle OPQ : & \phantom{.} \triangle SPQ : \triangle SPT \\ 1 \phantom{0} : & \phantom{000.} 1 \\ & \phantom{000.} 1 \phantom{00} : \phantom{00.} 2 \\ 1 \phantom{0} : & \phantom{000.} 1 \phantom{00} : \phantom{00.} 2 \\ \\ \therefore {\text{Area of } \triangle OPQ \over \text{Area of } \triangle SPT} & = {1 \over 2} \end{align*}