Đáp án:
$\begin{array}{l}
a)d//Ox:y = 0\\
\Rightarrow \left\{ \begin{array}{l}
m + 2 = 0\\
- m + 8 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m = - 2\\
m \ne 8
\end{array} \right.\\
\Rightarrow m = - 2\\
Vậy\,m = - 2\\
b)d//Oy\\
Do:Oy:x = 0\\
\Rightarrow \left\{ \begin{array}{l}
m - 3 = 0\\
- m + 8 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m = 3\\
m \ne 8
\end{array} \right.\\
Vậy\,m = 3
\end{array}$
c) Gọi điểm đt luôn đi qua với mọi m là M(x;y)
$\begin{array}{l}
\Rightarrow \left( {m + 2} \right).x + \left( {m - 3} \right).y - m + 8 = 0\left( {\forall m} \right)\\
\Rightarrow m.x + 2x + m.y - 3y - m + 8 = 0\forall m\\
\Rightarrow \left( {x + y - 1} \right).m = 3y - 2x - 8\forall m\\
\Rightarrow \left\{ \begin{array}{l}
x + y - 1 = 0\\
3y - 2x - 8 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + y = 1\\
2x - 3y = - 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 2
\end{array} \right.\\
\Rightarrow M\left( { - 1;2} \right) \equiv A
\end{array}$
Vậy d luôn đi qua điểm A ( -1 ; 2 )
