Đáp án:
\(a = - \frac{1}{2}\) hoặc \(a = - \frac{9}{2}\)
Giải thích các bước giải:
$\begin{array}{l}
A\left( {2;3} \right) \in d \Leftrightarrow 3 = 2a + b \Rightarrow b = 3 - 2a \Rightarrow y = ax + 3 - 2a\\
Cho\,y = 0 \Rightarrow 0 = ax + 3 - 2a \Leftrightarrow x = \frac{{2a - 3}}{a}\\
x > 0 \Leftrightarrow \frac{{2a - 3}}{a} > 0 \Leftrightarrow \left[ \begin{array}{l}
a > \frac{3}{2}\\
a < 0
\end{array} \right.\,\,\left( 1 \right)\\
\Rightarrow d\,cat\,Ox\,tai\,B\left( {\frac{{2a - 3}}{a};0} \right)\\
Cho\,x = 0 \Rightarrow y = a.0 + 3 - 2a \Leftrightarrow y = 3 - 2a\\
y > 0 \Leftrightarrow 3 - 2a > 0 \Leftrightarrow a < \frac{3}{2}\,\,\left( 2 \right)\\
\Rightarrow d\,cat\,Oy\,tai\,C\left( {0;3 - 2a} \right)\\
Tu\,\left( 1 \right)\,va\,\left( 2 \right) \Rightarrow a < 0\\
{S_{OBC}} = 16 \Leftrightarrow \frac{1}{2}OB.OC = 16 \Leftrightarrow \frac{1}{2}.\frac{{2a - 3}}{a}.\left( {3 - 2a} \right) = 16\\
\Leftrightarrow \left( {2a - 3} \right)\left( {3 - 2a} \right) = 32a \Leftrightarrow - 4{a^2} + 12a - 9 = 32a\\
\Leftrightarrow 4{a^2} + 20a + 9 = 0 \Leftrightarrow \left[ \begin{array}{l}
a = - \frac{1}{2}\\
a = - \frac{9}{2}
\end{array} \right.\left( {TM} \right)
\end{array}$
