Gọi B(5-2b;b) thuộc BD
M(0;0) là TĐ của BC => C(2b-5;-b)
Ta có:
\(\eqalign{
& \overrightarrow {KC} = \left( {2b - 11; - b - 2} \right) \cr
& \overrightarrow {M'B} = \left( {3 - 2b;b - 4} \right) \cr
& \overrightarrow {KC} .\overrightarrow {M'B} = 0 \cr
& \Leftrightarrow \left( {2b - 11} \right)\left( {3 - 2b} \right) + \left( { - b - 2} \right)\left( {b - 4} \right) = 0 \cr
& \Leftrightarrow 6b - 4{b^2} - 33 + 22b - {b^2} + 4b - 2b + 8 = 0 \cr
& \Leftrightarrow - 5{b^2} + 30b - 25 = 0 \cr
& \Leftrightarrow \left[ \matrix{
b = 5 \hfill \cr
b = 1 \hfill \cr} \right. \Rightarrow \left[ \matrix{
B\left( { - 5;5} \right)\,\,\left( {loai} \right) \hfill \cr
B\left( {3;1} \right)\,\,\left( {tm} \right) \hfill \cr} \right. \cr} \)
\( \Rightarrow C\left( { - 3; - 1} \right)\)
\(\eqalign{
& B\left( {3;1} \right)\,;C\left( { - 3; - 1} \right) \cr
& PT\,\,AB:\,\,{{x - 3} \over {2 - 3}} = {{y - 1} \over {4 - 1}} \cr
& \Leftrightarrow 3\left( {x - 3} \right) = - y + 1 \Leftrightarrow 3x + y - 10 = 0 \cr
& PT\,\,AC:\,\,{{x + 3} \over {6 + 3}} = {{y + 1} \over {2 + 1}} \cr
& \Leftrightarrow x + 3 = 3\left( {y + 1} \right) \Leftrightarrow x - 3y = 0 \cr
& A = AB \cap AC \Rightarrow A\left( {3;1} \right) \cr} \)
