Đáp án:
a) Gọi đường thẳng đi qua O và vuông góc với Δ là y=a.x
$\begin{array}{l}
\Leftrightarrow a.\left( { - 2} \right) = - 1\\
\Leftrightarrow a = \dfrac{1}{2}\\
\Leftrightarrow y = \dfrac{1}{2}.x\\
Xet: - 2x + 1 = \dfrac{1}{2}.x\\
\Leftrightarrow 2x + \dfrac{1}{2}x = 1\\
\Leftrightarrow \dfrac{5}{2}.x = 1\\
\Leftrightarrow x = \dfrac{2}{5}\\
\Leftrightarrow y = \dfrac{1}{2}.\dfrac{2}{5} = \dfrac{1}{5}\\
\Leftrightarrow H\left( {\dfrac{2}{5};\dfrac{1}{5}} \right)\\
\Leftrightarrow OH \bot \Delta \\
{d_{O - \Delta }} = OH = \sqrt {{{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}} = \dfrac{{\sqrt 5 }}{5}
\end{array}$
b) Gọi đt đi qua M và vuông góc với Δ là $y = a.x + b$
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
a.\left( { - 2} \right) = - 1\\
- 3 = a.\left( { - 1} \right) + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \dfrac{1}{2}\\
b = a - 3 = - \dfrac{5}{2}
\end{array} \right.\\
\Leftrightarrow y = \dfrac{1}{2}.x - \dfrac{5}{2}\\
Xet: - 2x + 1 = \dfrac{1}{2}.x - \dfrac{5}{2}\\
\Leftrightarrow 2x + \dfrac{1}{2}x = 1 + \dfrac{5}{2}\\
\Leftrightarrow \dfrac{5}{2}.x = \dfrac{7}{2}\\
\Leftrightarrow x = \dfrac{7}{5}\\
\Leftrightarrow y = - 2x + 1 = - 2.\dfrac{7}{5} + 1 = \dfrac{{ - 9}}{5}\\
\Leftrightarrow H\left( {\dfrac{7}{5};\dfrac{{ - 9}}{5}} \right)\\
\Leftrightarrow MH \bot \Delta \\
\Leftrightarrow {d_{M - \Delta }} = MH = \sqrt {{{\left( { - 1 - \dfrac{7}{5}} \right)}^2} + {{\left( { - 3 + \dfrac{9}{5}} \right)}^2}} = \dfrac{{6\sqrt 5 }}{5}
\end{array}$
