\(A\in\Delta\Rightarrow A\left(4t;3t+1\right)\)
Vì B đx với A qua I \(\Rightarrow\left\{{}\begin{matrix}x_B=2.2-4t=4-4t\\y_B=2.\dfrac{5}{2}-\left(3t+1\right)=4-3t\end{matrix}\right.\)
\(\Rightarrow B\left(4-4t;4-3t\right)\)
Có \(\overrightarrow{CA}\left(4t+2;3t-4\right)\); \(\overrightarrow{CB}\left(6-4t;-1-3t\right)\)
(Áp dụng công thức sau:
Với \(\overrightarrow{IA}\left(x;y\right)\), \(\overrightarrow{IB}\left(x';y'\right)\) => \(S_{IAB}=\dfrac{1}{2}\left|xy'-x'y\right|\))
Có \(S_{CAB}=\dfrac{1}{2}\left|\left(4t+2\right)\left(-1-3t\right)-\left(6-4t\right)\left(3t-4\right)\right|\)
\(\Leftrightarrow15=\dfrac{1}{2}\left|-44t+22\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{2}{11}\\t=\dfrac{13}{11}\end{matrix}\right.\)
\(\Rightarrow A\left(-\dfrac{8}{11};\dfrac{5}{11}\right);B\left(\dfrac{52}{11};\dfrac{50}{11}\right)\) hoặc \(A\left(\dfrac{52}{11};\dfrac{50}{11}\right);B\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)
