Let the image of the point $A(2,4)$ in the line mirror $D E$ be $C(\alpha, \beta)$. Then, $A C$ is perpendicular to $D E$.


$\therefore$ The coordinate of point $B$ are $\left(\frac{\alpha+2}{2}, \frac{\beta+4}{2}\right)$.
Since, point $B$ lies on the line $2 x+3 y-6=0$
$$
\begin{aligned}
& \therefore \quad 2\left(\frac{\alpha+2}{2}\right)+3\left(\frac{\beta+4}{2}\right)-6=0 \\
& \Rightarrow \quad 2 \alpha+4+3 \beta+12-12=0 \\
& \Rightarrow \quad 2 \alpha+3 \beta+4=0 \\
&
\end{aligned}
$$
Since $\quad A C \perp D E$
$\therefore$ Slope of $A C \times$ slope of $D E=-1$
$$
\begin{aligned}
& \Rightarrow \quad \frac{\beta-4}{\alpha-2} \times \frac{-2}{3}=-1 \\
& \Rightarrow \quad 2 \beta-8=3 \alpha-6 \\
& \Rightarrow \quad 3 \alpha-2 \beta+2=0 \\
&
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get $\alpha=\frac{-14}{13}$ and $\beta=\frac{-8}{13}$
$\therefore$ Image of point $(2,4)$ is $\left(\frac{-14}{13}, \frac{-8}{13}\right)$.