Đáp án:
\(\begin{array}{l}
a)\,m = - \dfrac{1}{2}\\
b)\,A\left( {2m + 1;0} \right);B\left( {0; - 2m - 1} \right)\\
OH = \dfrac{{\left| {2m + 1} \right|}}{{\sqrt 2 }}\\
m = \dfrac{{ - \sqrt 2 \pm 1}}{2}\\
c)\,y = - x
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\,Thay\,x = 0;y = 0\,\,ta\,duoc\\
0 = 0 - 2m - 1 \Rightarrow m = - \dfrac{1}{2}\\
b)\,A\left( {{x_A};0} \right) \in Ox \Rightarrow 0 = {x_A} - 2m - 1 \Rightarrow {x_A} = 2m + 1\\
\Rightarrow A\left( {2m + 1;0} \right)\\
B\left( {0;{y_B}} \right) \in Oy \Rightarrow {y_B} = - 2m - 1 \Rightarrow B\left( {0; - 2m - 1} \right)\\
\Rightarrow OA = OB = \left| {2m + 1} \right|\\
\dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} = \dfrac{2}{{{{\left( {2m + 1} \right)}^2}}}\\
\Rightarrow OH = \dfrac{{\left| {2m + 1} \right|}}{{\sqrt 2 }}\\
OH = \dfrac{{\sqrt 2 }}{2} \Rightarrow \left| {2m + 1} \right| = \sqrt 2 \\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{{\sqrt 2 - 1}}{2}\\
m = \dfrac{{ - \sqrt 2 - 1}}{2}
\end{array} \right.\\
c)\,Trung\,diem\,I\,cua\,doan\,AB:\\
\left\{ \begin{array}{l}
{x_I} = \dfrac{{2m + 1}}{2}\\
{y_I} = \dfrac{{ - 2m - 1}}{2}
\end{array} \right.\\
\Rightarrow {y_I} = - {x_I}\\
Qui\,tich\,trung\,diem\,I\,la\,duong\,thang:\,y = - x
\end{array}\)
