The number of possible distinct straight lines passing through \((2,3)\) and forming a triangle with co-ordinate axes whose area is 12 sq. units are,

Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\)
Since line passes through the point \((2,3)\), we get \(\frac{2}{a}+\frac{3}{b}=1\)
\(\therefore \quad 2 b+3 a=a b\)
As per the given condition, Area of the triangle \(=12\) sq. units
\(\begin{array}{ll}
\therefore & \frac{1}{2}|a b|=12 \\
\therefore & a b= \pm 24 \\
& \text {Case I : } \\
& \text {ab }=24 \\
\therefore \quad & 2 b+3 a=24...(ii) \\
\therefore \quad & 2 \cdot \frac{24}{a}{ }^2+3 a=24 ...[from (i)]\\
\therefore \quad & 16=8 a ...[from (ii)]\\
\therefore \quad & a^2-8 a+16=0 \\
\therefore \quad & a=4 \\
\therefore \quad & b=6
\end{array}\)
\(\begin{array}{ll}
& \text { Case II : } \\
& a b=-24 \\
\therefore \quad & 2\left(\frac{-24}{a}\right)+3 a=-24 \\
\therefore \quad & a^2+8 a-16=0 \\
\therefore \quad & a=\frac{-8 \pm \sqrt{64+64}}{2}=-4 \pm 4 \sqrt{2}
\end{array}\)
...[from (i) and (ii)]
\(\therefore \quad\) b will also have 2 values.
\(\therefore \quad\) The required number of lines is 3.