\[{d_1} \cap {d_2} = A \to \left\{ \begin{array}{l}
x + 2y - 5 = 0 \\
4x + 13y - 10 = 0 \\
\end{array} \right. \to A\left( {9; - 2} \right)\]
Gọi E là điểm đối xứng với C qua d1
\[{\begin{array}{*{20}{c}}
{CE \bot {d_1}\, \to C \in E:2\left( {x - 4} \right) - 1\left( {y - 3} \right) = 0} \\
{ \to CE:2x - y - 5 = 0} \\
{CE \cap {d_1} = I \to \left\{ {\begin{array}{*{20}{c}}
{2x - y - 5 = 0} \\
{x + 2y - 5 = 0} \\
\end{array}} \right. \to I\left( {3;1} \right) \to E\left( {2; - 1} \right)} \\
\end{array}}\]
\[{\left\{ {\begin{array}{*{20}{c}}
{E\left( {2; - 1} \right)} \\
{A\left( {9; - 2} \right)} \\
\end{array}} \right. \to AB:1\left( {x - 2} \right) + 7\left( {y + 1} \right) = 0}\]
\[AB. x + 7y + 5 = 0 \to B\left( { - 5 - 7b;b} \right)\]
\[\begin{array}{*{20}{c}}
{M\left( {\frac{{{x_b} + {x_c}}}{2};\frac{{{y_b} + {y_c}}}{2}} \right),M \in {d_2} \to 4\left( {\frac{{ - 1 - 7b}}{2}} \right) + 13\left( {\frac{{b + 3}}{2}} \right) - 10 = 0} \\
{ \to b = 1 \to B\left( { - 12;1} \right)} \\
\end{array}\]
“Bài dự thi event box toán 10”
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
