Đáp án:
$\begin{array}{l}
1){x_A} = - 1\\
\Rightarrow {y_A} = \dfrac{1}{2}x_A^2 = \dfrac{1}{2}\\
{x_B} = 2\\
\Rightarrow {y_B} = \dfrac{1}{2}x_B^2 = 2\\
\Rightarrow A\left( { - 1;\dfrac{1}{2}} \right);B\left( {2;2} \right)\\
2)\left( d \right):y = a.x + b\\
Do:A,B \in \left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{2} = - a + b\\
2 = 2a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = \dfrac{1}{2}\\
b = 1
\end{array} \right.\\
\Rightarrow \left( d \right):y = \dfrac{1}{2}x + 1\\
hay:\left( d \right):x - 2y + 2 = 0\\
\Rightarrow {d_{O - d}} = \dfrac{{\left| {0 - 2.0 + 2} \right|}}{{\sqrt {1 + {{\left( { - 2} \right)}^2}} }} = \dfrac{2}{{\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{5}
\end{array}$
