Đáp án:
$\begin{array}{l}
\left( d \right):y = a.x + b\\
a)\left( d \right) \cap Oy = \left( {0;4} \right)\\
\left( d \right) \cap Ox = \left( { - 4;0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
4 = a.0 + b\\
0 = a.\left( { - 4} \right) + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 4\\
4a = b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 4\\
a = 1
\end{array} \right.\\
Vậy\,\left( d \right):y = x + 4\\
b)A\left( {1;5} \right);B\left( { - 1;3} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
5 = a + b\\
3 = a.\left( { - 1} \right) + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b = 5\\
- a + b = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2a = 2\\
b = 5 - a
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = 5 - 1 = 4
\end{array} \right.\\
Vậy\,\left( d \right):y = x + 4
\end{array}$
