Đáp án:
$\begin{array}{l}
a)\overrightarrow {AC} = \left( {1;2} \right)\\
BH \bot AC\\
\Rightarrow VTCP:BH = \overrightarrow {AC} = \left( {1;2} \right)\\
\Rightarrow VTPT:\overrightarrow u = \left( {2; - 1} \right)\\
\Rightarrow PTTS:\left\{ \begin{array}{l}
x = 2t + 4\\
y = - t + 5
\end{array} \right.\\
PTTQ:x + 2y + d = 0\\
Do:B\left( {4;5} \right)\\
\Rightarrow 4 + 2.5 + d = 0\\
\Rightarrow d = - 14\\
\Rightarrow PTTQ:x + 2y - 14 = 0\\
Vậy:\left\{ \begin{array}{l}
BH:\left\{ \begin{array}{l}
x = 2t + 4\\
y = - t + 5
\end{array} \right.\\
BH:x + 2y - 14 = 0
\end{array} \right.\\
b)M\left( {\dfrac{{4 + 1}}{2};\dfrac{{5 + 1}}{2}} \right) \Rightarrow M\left( {\dfrac{5}{2};3} \right)\\
\Rightarrow \overrightarrow {AM} = \left( {\dfrac{5}{2};4} \right)\\
\Rightarrow VTPT\,AM = \overrightarrow {AM} = \left( {\dfrac{5}{2};4} \right)\\
\Rightarrow VTCP = \overrightarrow u = \left( {4; - \dfrac{5}{2}} \right)\\
\Rightarrow PTTS:\left\{ \begin{array}{l}
x = \dfrac{5}{2}t\\
y = 4t - 1
\end{array} \right.\\
PTTQ:4x - \dfrac{5}{2}y + c = 0\\
A\left( {0; - 1} \right)\\
\Rightarrow - \dfrac{5}{2}.\left( { - 1} \right) + c = 0\\
\Rightarrow c = - \dfrac{5}{2}\\
\Rightarrow PTTQ:4x - \dfrac{5}{2}y - \dfrac{5}{2} = 0\\
Vậy\,AM:\left\{ \begin{array}{l}
PTTS:\left\{ \begin{array}{l}
x = \dfrac{5}{2}t\\
y = 4t - 1
\end{array} \right.\\
PTTQ:4x - \dfrac{5}{2}y - \dfrac{5}{2} = 0
\end{array} \right.
\end{array}$
