Đáp án:
Bài 1:
1 điểm nằm trên trục tung có x=0
$\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
y = \left( {m + 4} \right).x + 11 = 11\\
y = x + {m^2} + 2 = {m^2} + 2
\end{array} \right.\\
\Rightarrow \left( {0;11} \right) \equiv \left( {0;{m^2} + 2} \right)\\
\Rightarrow 11 = {m^2} + 2\\
\Rightarrow {m^2} = 9\\
\Rightarrow \left[ \begin{array}{l}
m = 3\\
m = - 3
\end{array} \right.\\
B2:\\
+ )Theo\,Pytago\\
A{B^2} = A{H^2} + H{B^2} = {3^2} + {4^2} = 25\\
\Rightarrow AB = 5\left( {cm} \right)\\
+ )\Delta ABH \sim \Delta CAH\left( {g - g} \right)\\
\Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{BH}}{{AH}} = \dfrac{{AH}}{{CH}} = \dfrac{4}{3}\\
\left\{ \begin{array}{l}
AC = \dfrac{{15}}{4} = 3,75\left( {cm} \right)\\
CH = \dfrac{9}{4} = 2,25\left( {cm} \right)
\end{array} \right.\\
\Rightarrow BC = BH + CH = 6,25\left( {cm} \right)\\
\Rightarrow AM = \dfrac{{BC}}{2} = 3,125\left( {cm} \right)
\end{array}$
