Let a line intersect the co-ordinate axes in points \(A\) and \(B\) such that the area of the triangle \(O A B\) is 12 sq. units. If the line passes through the point \((2,3)\), then the equation of the line is

Area of \(\triangle A O B=12\) sq. units
\(\begin{array}{ll}
& \frac{1}{2} a b=12 \\
\therefore \quad & a b=24 \\
\therefore \quad & b=\frac{24}{a}...(i)
\end{array}\)
Equation of line in point slope form is
\(\begin{aligned}
& y-a=\frac{-a}{b}(x-0) \\
& y-a=\frac{-a x}{\frac{24}{a}}
\end{aligned}\)
...[from (i)]
\(y-\mathrm{a}=\frac{-\mathrm{a}^2 x}{24}...(ii)\)
Since line passes through \((2,3)\)
\(\begin{aligned}
& \therefore \quad 3-a=\frac{-a^2(2)}{24} \\
& \quad \Rightarrow 3-a=\frac{-a^2}{12} \\
& \\
& \quad \Rightarrow a^2-12 a+36=0 \\
& \\
& \quad \Rightarrow(a-6)^2=0 \\
& \quad \Rightarrow a=6 \\
& \therefore \quad \\
& \therefore=\frac{24}{6}=4...[from(i)]
\end{aligned}\)
Required equation of line is
\(\begin{aligned}
& y-6=\frac{-36 x}{24}...[from(ii)] \\
& 2 y-12=-3 x \\
& \Rightarrow 3 x+2 y-12=0 \\
& \Rightarrow 3 x+2 y=12
\end{aligned}\)
...[from (ii)]