Key Concepts and Formulas

• Centroid of a Triangle: The centroid $G$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
• Image of a Point in a Line: The image $(x', y')$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is given by: $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$
• Centroid Invariance under Reflection: The centroid of the image of a triangle under reflection is the image of the centroid of the original triangle.

Step-by-Step Solution

Step 1: Find the Centroid of the Original Triangle

The vertices of the original triangle are $A(1,3)$, $B(3,1)$, and $C(2,4)$. We want to find the coordinates of its centroid, $G$.

• Why this step? Because the centroid of the reflected triangle is simply the reflection of the original triangle's centroid. Therefore, we must first find the original centroid.

Using the centroid formula: $G = \left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right)$ $G = \left(\frac{6}{3}, \frac{8}{3}\right)$ $G = \left(2, \frac{8}{3}\right)$ Thus, the centroid of the original triangle is $G\left(2, \frac{8}{3}\right)$.

Step 2: Define the Centroid of the Reflected Triangle

Let the triangle PQR be the image of the original triangle after reflection in the line $x+2y=2$. Let the centroid of $\triangle PQR$ be $(\alpha, \beta)$.

• Why this step? This step sets up the problem. We're defining the centroid of the image triangle as $(\alpha, \beta)$, which is what we need to find.

Since reflection is an isometric transformation, the centroid of $\triangle PQR$ is the image of the centroid $G\left(2, \frac{8}{3}\right)$ in the line $x+2y-2=0$.

Step 3: Calculate the Image of the Centroid

We need to find the image $(\alpha, \beta)$ of the point $G(2, \frac{8}{3})$ in the line $x+2y-2=0$. Here, $a=1$, $b=2$, $c=-2$, $x_1=2$, and $y_1=\frac{8}{3}$.

• Why this step? This is the core calculation. We're using the formula for the reflection of a point in a line to find the coordinates of the reflected centroid.

Using the point reflection formula: $\frac{\alpha-2}{1} = \frac{\beta-\frac{8}{3}}{2} = -2\frac{(1)(2) + (2)(\frac{8}{3}) - 2}{1^2 + 2^2}$ First, calculate the constant ratio: $-2\frac{2 + \frac{16}{3} - 2}{1+4} = -2\frac{\frac{16}{3}}{5} = -2\left(\frac{16}{15}\right) = -\frac{32}{15}$ Now, solve for $\alpha$ and $\beta$:

For $\alpha$: $\frac{\alpha-2}{1} = -\frac{32}{15}$ $\alpha - 2 = -\frac{32}{15}$ $\alpha = 2 - \frac{32}{15} = \frac{30}{15} - \frac{32}{15} = -\frac{2}{15}$

For $\beta$: $\frac{\beta-\frac{8}{3}}{2} = -\frac{32}{15}$ $\beta - \frac{8}{3} = -\frac{64}{15}$ $\beta = \frac{8}{3} - \frac{64}{15} = \frac{40}{15} - \frac{64}{15} = -\frac{24}{15} = -\frac{8}{5}$

Therefore, the centroid of the reflected triangle is $(\alpha, \beta) = \left(-\frac{2}{15}, -\frac{24}{15}\right) = \left(-\frac{2}{15}, -\frac{8}{5}\right)$.

Step 4: Calculate $15(\alpha - \beta)$

We need to find the value of $15(\alpha - \beta)$.

• Why this step? This is the final calculation to answer the question.

Substitute the values of $\alpha$ and $\beta$: $15(\alpha - \beta) = 15\left(-\frac{2}{15} - \left(-\frac{24}{15}\right)\right)$ $= 15\left(-\frac{2}{15} + \frac{24}{15}\right)$ $= 15\left(\frac{22}{15}\right)$ $= 22$

Tips and Common Mistakes

• Using Centroid Property: This is the key to solving efficiently. Don't waste time finding the images of all vertices.
• Algebraic Errors: The calculations involve fractions and negative signs. Be very careful with your arithmetic.
• Image Formula: Make sure you have the correct formula for the image of a point in a line.

Summary

By using the property that the centroid of the reflected triangle is the reflection of the original centroid, we simplified the problem significantly. We found the original centroid, then used the image formula to find the reflected centroid, and finally calculated $15(\alpha - \beta)$ to get the answer 22.

The final answer is $\boxed{22}$, which corresponds to option (C).