\begin{align*} \text{Let coordinates} & \text{ of } D \text{ be } (a, b) \\ \\ \text{Mid-point of } CD & \text{ is } \left({1 + a \over 2}, {4 + b \over 2}\right) \\ \\ \text{Comparing } & \text{with } M(2.5, 2.5), \end{align*} \begin{align*} {1 + a \over 2} & = 2.5 &&& {4 + b \over 2} & = 2.5 \\ 1 + a & = 2(2.5) &&& 4 + b & = 2(2.5) \\ 1 + a & = 5 &&& 4 + b & = 5 \\ a & = 5 - 1 &&& b & = 5 - 4 \\ a & = 4 &&& b & = 1 \end{align*}
$$ \therefore D(4, 1) $$

\begin{align*} y & = mx + c \\ y & = -2x + 2 \\ \\ \text{Let } & y = 0, \\ 0 & = -2x + 2 \\ 2x & = 2 \\ x & = 1 \phantom{000000} [x \text{-intercept}] \end{align*}

\begin{align*} \tan \theta & = {1 \over 2} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \theta & = \tan^{-1} \left(1 \over 2\right) \\ \theta & = 26.565^\circ \\ \theta & \approx 26.6^\circ \end{align*}

\begin{align*} \tan \alpha & = {2 \over 1} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \alpha & = \tan^{-1} \left(2 \over 1\right) \\ \alpha & = 63.435^\circ \\ \\ \text{Obtuse angle} & = 180^\circ - 63.435^\circ \\ & = 116.565^\circ \\ & \approx 116.6^\circ \end{align*}

\begin{align*} \text{Gradient of } EF & = {3 - 2 \over 2.5 - 1} \\ & = {2 \over 3} \\ \\ \text{Gradient of } FG & = {4 - 3 \over 4 - 2.5} \\ & = {2 \over 3} \\ \\ \therefore E, F \text{ and } G & \text{ lie on the same line} \end{align*}

\begin{align*} \text{Gradient of } EF & = {3 - 2 \over 2.5 - 1} \\ & = {2 \over 3} \\ \\ y & = mx + c \\ y & = {2 \over 3}x + c \\ \\ \text{Using } & E(1, 2), \\ 2 & = {2 \over 3}(1) + c \\ 2 & = {2 \over 3} + c \\ {4 \over 3} & = c \\ \\ \text{Eqn of } EF: & \phantom{0} y = {2 \over 3}x + {4 \over 3} \\ \\ \text{Let } & x = 4, \\ y & = {2 \over 3}(4) + {4 \over 3} \\ y & = 4 \\ \\ \therefore G(4, 4) \text{ lies on line } & \text{passing through } E \text{ and } F \end{align*}

\begin{align*} \underbrace{ \overrightarrow{AB} }_{\text{Moving from } A \text{ to } B} & = {2 \choose 1} \phantom{000000} [2 \text{ units in }x \text{-direction, } 1 \text{ unit in } y \text{-direction}] \\ \\ \overrightarrow{AC} & = 3 \overrightarrow{AB} \\ & = 3 {2 \choose 1} \\ & = {6 \choose 3} \\ \\ x \text{-coordinate of } C & = -2 + 6 = 4 \\ \\ y \text{-coordinate of } C & = 1 + 3 = 4 \\ \\ \therefore & \phantom{.} C(4, 4) \end{align*}

\begin{align*} x \text{-coordinate of } C & = -2 + 6 = 4 \\ \\ y \text{-coordinate of } C & = 1 + 3 = 4 \\ \\ \therefore & \phantom{.} C(4, 4) \end{align*}

\begin{align*} \left( {1 + 5 \over 2}, {4 + 2 \over 2} \right) & = (3, 3) \\ \\ \text{Mid-point of } AB & \phantom{.} (3, 3) \\ \\ \text{Gradient of } AB & = {4 - 2 \over 1 - 5} \\ & = -{1 \over 2} \\ \\ m_1 \times m_2 & = - 1 \\ \text{Gradient of perpendicular bisector} \times -{1 \over 2} & = -1 \\ \text{Gradient of perpendicular bisector} & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using midpoint } (3, 3), & \text{ let } x = 3 \text{ and } y = 3, \\ 3 & = 2(3) + c \\ 3 & = 6 + c \\ 3 - 6 & = c \\ -3 & = c \\ \\ \therefore \text{Equation of perpendicular bisector: } & y = 2x - 3 \end{align*}