Key Concepts and Formulas

• Distance Between Parallel Lines: The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

• Absolute Value Equations: The equation $|x| = a$ (where $a > 0$) has two solutions: $x = a$ and $x = -a$.

Step-by-Step Solution

Step 1: Analyze the Given Lines and Prepare for Distance Calculation

We are given the line $L_0: 2x - y + 3 = 0$, and two other lines $L_1: 4x - 2y + \alpha = 0$ and $L_2: 6x - 3y + \beta = 0$. We need to find the values of $\alpha$ and $\beta$ such that the distances from $L_0$ to $L_1$ and $L_0$ to $L_2$ are $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ respectively. To use the distance formula, the coefficients of $x$ and $y$ must be identical.

Step 2: Scale the Equation of $L_0$ for Comparison with $L_1$

To compare $L_0$ and $L_1$, we multiply the equation of $L_0$ by 2: $2(2x - y + 3) = 0 \implies 4x - 2y + 6 = 0$ Now we can compare $4x - 2y + 6 = 0$ with $4x - 2y + \alpha = 0$. The distance between these lines is given as $\frac{1}{\sqrt{5}}$.

Step 3: Calculate Possible Values of $\alpha$

Using the distance formula, the distance between $4x - 2y + 6 = 0$ and $4x - 2y + \alpha = 0$ is: $d = \frac{|\alpha - 6|}{\sqrt{4^2 + (-2)^2}} = \frac{|\alpha - 6|}{\sqrt{16 + 4}} = \frac{|\alpha - 6|}{\sqrt{20}} = \frac{|\alpha - 6|}{2\sqrt{5}}$ We are given that this distance is $\frac{1}{\sqrt{5}}$, so: $\frac{|\alpha - 6|}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$ Multiplying both sides by $2\sqrt{5}$ gives: $|\alpha - 6| = 2$ This gives two possible cases:

• $\alpha - 6 = 2 \implies \alpha = 8$
• $\alpha - 6 = -2 \implies \alpha = 4$

Step 4: Scale the Equation of $L_0$ for Comparison with $L_2$

To compare $L_0$ and $L_2$, we multiply the equation of $L_0$ by 3: $3(2x - y + 3) = 0 \implies 6x - 3y + 9 = 0$ Now we can compare $6x - 3y + 9 = 0$ with $6x - 3y + \beta = 0$. The distance between these lines is given as $\frac{2}{\sqrt{5}}$.

Step 5: Calculate Possible Values of $\beta$

Using the distance formula, the distance between $6x - 3y + 9 = 0$ and $6x - 3y + \beta = 0$ is: $d = \frac{|\beta - 9|}{\sqrt{6^2 + (-3)^2}} = \frac{|\beta - 9|}{\sqrt{36 + 9}} = \frac{|\beta - 9|}{\sqrt{45}} = \frac{|\beta - 9|}{3\sqrt{5}}$ We are given that this distance is $\frac{2}{\sqrt{5}}$, so: $\frac{|\beta - 9|}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$ Multiplying both sides by $3\sqrt{5}$ gives: $|\beta - 9| = 6$ This gives two possible cases:

• $\beta - 9 = 6 \implies \beta = 15$
• $\beta - 9 = -6 \implies \beta = 3$

Step 6: Calculate the Sum of All Possible Values of $\alpha$ and $\beta$

The possible values of $\alpha$ are 8 and 4, and the possible values of $\beta$ are 15 and 3. The sum of all possible values is: $8 + 4 + 15 + 3 = 30$

Common Mistakes & Tips

• Incorrect Scaling: Ensure you multiply the entire equation by the scaling factor, not just the $x$ and $y$ terms.
• Missing Absolute Value Cases: Remember to consider both positive and negative cases when solving absolute value equations.

Summary

We scaled the equation of the line $2x - y + 3 = 0$ to match the coefficients of $x$ and $y$ in the equations $4x - 2y + \alpha = 0$ and $6x - 3y + \beta = 0$. We then used the distance formula to find the possible values of $\alpha$ and $\beta$, and summed them. The sum of all possible values of $\alpha$ and $\beta$ is 30.

The final answer is \boxed{30}.