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(C) The equation of the line $L$ with $x$-intercept $3$ and $y$-intercept $1$ is given by $\frac{x}{3} + \frac{y}{1} = 1$,which simplifies to $x + 3y

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$\left(\frac{11}{5}, \frac{28}{5}\right)$

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(C) The equation of the line $L$ with $x$-intercept $3$ and $y$-intercept $1$ is given by $\frac{x}{3} + \frac{y}{1} = 1$,which simplifies to $x + 3y - 3 = 0$.To find the image $(x', y')$ of the point $(x_0, y_0) = (-1, -4)$ in the line $ax + by + c = 0$,we use the formula $\frac{x' - x_0}{a} = \frac{y' - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.Substituting the values: $\frac{x' - (-1)}{1} = \frac{y' - (-4)}{3} = -2 \frac{1(-1) + 3(-4) - 3}{1^2 + 3^2}$.$\frac{x' + 1}{1} = \frac{y' + 4}{3} = -2 \frac{-1 - 12 - 3}{1 + 9} = -2 \frac{-16}{10} = \frac{32}{10} = \frac{16}{5}$.For $x'$: $x' + 1 = \frac{16}{5} \Rightarrow x' = \frac{16}{5} - 1 = \frac{11}{5}$.For $y'$: $\frac{y' + 4}{3} = \frac{16}{5}$ $\Rightarrow y' + 4 = \frac{48}{5}$ $\Rightarrow y' = \frac{48}{5} - 4 = \frac{28}{5}$.Thus,the image is $\left(\frac{11}{5}, \frac{28}{5}\right)$.

Let $L$ denote the line in the $xy$-plane with $x$ and $y$ intercepts as $3$ and $1$ respectively. Then the image of the point $(-1, -4)$ in this line is

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