Let the triangle PQR be the image of the triangle with vertices (1, 3), (3,1) and (2,4) in the line

Q: Wait, why can't I just reflect the three vertices and then find the centroid? Isn't that safer?

While it is technically 'safer' in terms of logic, it is a disaster for your time management in a JEE Advanced exam. Reflecting three points involves solving the reflection formula three times, which triples your chances of making a silly arithmetic error. In competitive exams, elegance is not just about beauty; it is about efficiency and error reduction. Trust the property of the centroid!

Q: Is the centroid of the reflected triangle always the reflection of the original centroid?

Yes, absolutely. Reflection is a linear transformation. The centroid is the arithmetic mean of the vertices. Because the transformation is linear, the image of the mean is the mean of the images. This holds true for any affine transformation, not just reflection. It is a powerful tool to keep in your arsenal.

Q: I always mess up the signs in the reflection formula. How do I remember it?

Think of the formula $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$ as a vector projection. The term $-2(ax_1 + by_1 + c)$ accounts for the fact that you are moving from the point to the line, and then doubling that distance to get to the other side. Always write your line in the form $ax + by + c = 0$ before you start, and the signs will take care of themselves.

Question

Let the triangle PQR be the image of the triangle with vertices (1, 3), $(3,1)$ and $(2,4)$ in the line $x+2y=2$ If the centroid of $	riangle PQR$ is the point $(\alpha,\beta)$, then $15(\alpha-\beta)$ is equal to:

Options:

(A) 19

(B) 24

(C) 21

(D) 22

Sigma Percentile · Accepted Answer

### The Art of the Shortcut: Mastering Geometric Reflections

Welcome, future engineer. Today, we are going to dissect a problem that separates the 'brute-force' students from the 'strategic' thinkers.

When you see a problem asking for the reflection of a triangle, your instinct might be to panic. You might think you have to find the reflection of point $A$, then point $B$, then point $C$, and then calculate the centroid of the new triangle.

Stop right there. If you do that, you are playing the game on 'Hard Mode' when 'Easy Mode' is staring you right in the face. Let us embark on this journey of geometric elegance.

### The Centroid Shortcut

The core of this problem lies in a beautiful property of geometry: the centroid is a linear combination of the vertices. In the world of coordinate geometry, reflection is an isometry—a transformation that preserves distance and orientation.

Because of this, the centroid of the image triangle is simply the reflection of the centroid of the original triangle. We do not need to reflect three points; we only need to reflect one.

Let us calculate the original centroid, $G_1$, of the triangle with vertices $A(1, 3)$, $B(3, 1)$, and $C(2, 4)$. The formula is:

$$G_1 = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$$

Substituting our values, we get:

$$G_1 = \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) = \left( 2, \frac{8}{3} \right)$$

Take a breath. We have reduced a complex triangle problem to a single point reflection problem. This is the JEE mindset: simplify, simplify, simplify.

### The Reflection Mechanics

Now, we must reflect the point $G_1(2, 8/3)$ across the line $x + 2y - 2 = 0$. We use the standard reflection formula for a point $(x_1, y_1)$ across the line $ax + by + c = 0$:

$$\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Here, our line is $1x + 2y - 2 = 0$, so $a=1$, $b=2$, and $c=-2$. Our point is $x_1=2$ and $y_1=8/3$.

Let us plug these into the formula. The constant ratio on the right side becomes:

$$\frac{-2(1(2) + 2(8/3) - 2)}{1^2 + 2^2}$$

Notice how the $2$ and $-2$ cancel out inside the parenthesis? That is the universe rewarding your patience. We are left with:

$$\frac{-2(16/3)}{5} = -\frac{32}{15}$$

This is our constant ratio.

### The Final Calculation

Now, we solve for $\alpha$ and $\beta$ individually. For $\alpha$, we have:

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

For $\beta$, we have:

$$\frac{\beta - 8/3}{2} = -\frac{32}{15} \Rightarrow \beta - \frac{8}{3} = -\frac{64}{15}$$

Solving for $\beta$:

$$\beta = \frac{8}{3} - \frac{64}{15} = \frac{40 - 64}{15} = -\frac{24}{15}$$

We are almost there. The question asks for $15(\alpha - \beta)$. Substituting our values:

$$15\left(-\frac{2}{15} - \left(-\frac{24}{15}\right)\right) = 15\left(\frac{-2 + 24}{15}\right)$$

The $15$s cancel out, leaving us with $-2 + 24$, which equals **22**.

Look at that! The complexity dissolves into a clean, integer answer. This is the beauty of mathematics. When you approach problems with the right tools and a calm mind, the path clears itself.

Animated Solution for Mathematics - Straight Lines: Let the triangle PQR be the image of the triangle with vertices (1, 3), $(3, 1)$ and $(2, 4)$ in the line $x + 2 y = 2$ If the centroid of $△ P QR$ is the point $(α, β)$ , then $15 (α - β)$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Setup

The Centroid Property

Calculating Original Centroid $G_{1}$

Substituting Vertices for $G_{1}$

Simplifying $G_{1}$ Coordinates

The Image Formula

Setting up the Equation

Computing the Constant Ratio

Finding the x-coordinate $α$

Finding the y-coordinate $β$

Final Expression Setup

The Final Result

The Sigma Insight: Distance of a Point from a Line

The Art of the Shortcut

The Centroid Shortcut

The Reflection Mechanics

The Final Calculation

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