Key Concepts and Formulas

• Equation of a Line (Intercept Form): The equation of a line with x-intercept $a$ and y-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
• Equation of a Line (General Form): The general form of a line is $ax + by + c = 0$.
• Image of a Point in a Line: The image of a point $(x_1, y_1)$ in the line $ax + by + c = 0$ is $(\alpha, \beta)$, where $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$

Step-by-Step Solution

Step 1: Find the equation of line L

We are given that the line L has x-intercept 3 and y-intercept 1. We need to find the equation of this line in the general form.

• Using the intercept form: The equation of a line with x-intercept 3 and y-intercept 1 is: $\frac{x}{3} + \frac{y}{1} = 1$
• Converting to general form: Multiply both sides by 3 to eliminate the fraction: $x + 3y = 3$ Rearrange the terms to get the general form: $x + 3y - 3 = 0$ Thus, $a = 1$, $b = 3$, and $c = -3$.

Step 2: Set up the image formula

We are given the point $P(-1, -4)$, and we want to find its image $P'(\alpha, \beta)$ in the line $x + 3y - 3 = 0$.

• Applying the image formula: Substitute $x_1 = -1$, $y_1 = -4$, $a = 1$, $b = 3$, and $c = -3$ into the image formula: $\frac{\alpha - (-1)}{1} = \frac{\beta - (-4)}{3} = -2 \frac{(1)(-1) + (3)(-4) - 3}{1^2 + 3^2}$ $\frac{\alpha + 1}{1} = \frac{\beta + 4}{3} = -2 \frac{-1 - 12 - 3}{1 + 9}$

Step 3: Simplify the constant term

We need to simplify the right-hand side of the equation:

• Calculating the expression: $-2 \frac{-16}{10} = \frac{32}{10} = \frac{16}{5}$

Thus, the image formula becomes: $\frac{\alpha + 1}{1} = \frac{\beta + 4}{3} = \frac{16}{5}$

Step 4: Solve for $\alpha$ and $\beta$

We now have two equations to solve for $\alpha$ and $\beta$:

• Solving for $\alpha$: $\frac{\alpha + 1}{1} = \frac{16}{5}$ $\alpha + 1 = \frac{16}{5}$ $\alpha = \frac{16}{5} - 1 = \frac{16 - 5}{5} = \frac{11}{5}$

• Solving for $\beta$: $\frac{\beta + 4}{3} = \frac{16}{5}$ $\beta + 4 = \frac{48}{5}$ $\beta = \frac{48}{5} - 4 = \frac{48 - 20}{5} = \frac{28}{5}$

Therefore, the image of the point $(-1, -4)$ is $\left(\frac{11}{5}, \frac{28}{5}\right)$.

Step 5: Compare with Options

Comparing our calculated image point $\left( \frac{11}{5}, \frac{28}{5} \right)$ with the given options: (A) $\left( {{{11} \over 5},{{28} \over 5}} \right)$ (B) $\left( {{{29} \over 5},{{11} \over 5}} \right)$ (C) $\left( {{{29} \over 5},{8 \over 5}} \right)$ (D) $\left( {{8 \over 5},{{29} \over 5}} \right)$

Our result matches option (A).

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when substituting into the formula, especially with the constant term c and the -2 factor.
• Formula Confusion: Ensure you remember the image formula correctly. It's easy to mix it up with the formula for the foot of the perpendicular.
• Arithmetic Errors: Take care with fraction arithmetic and simplification, as mistakes here are common.

Summary

We determined the equation of the line L to be $x + 3y - 3 = 0$. Then, we applied the formula for finding the image of a point in a line, substituting the given point and the coefficients of the line equation. We carefully simplified the expression and solved for the coordinates of the image point. The image of the point $(-1, -4)$ in the line L is $\left( \frac{11}{5}, \frac{28}{5} \right)$.

The final answer is \boxed{\left( {{{11} \over 5},{{28} \over 5}} \right)}, which corresponds to option (A).