Let A(2, -3) and B(-2, 3) be vertices of a triangle ABC. If the centroid of this triangle moves on t

Q: Why do we use $(h, k)$ for the coordinates of $C$ instead of $(x, y)$?

That is a great question! We use $(h, k)$ as temporary variables to represent the coordinates of the moving point $C$. If we used $(x, y)$ from the start, it might get confused with the $x$ and $y$ in the line equation $2x + 3y = 3$. Once we find the relationship between $h$ and $k$, we replace them with $(x, y)$ to write the final equation of the locus.

Q: Why does the centroid move on a line if $C$ is just a point?

Think of it this way: the centroid $G$ is a weighted average of the vertices. Since $A$ and $B$ are fixed, the position of $G$ depends entirely on the position of $C$. If $C$ is constrained to move in a way that keeps $G$ on a specific line, then $C$ itself must be constrained to a specific path. The math we did shows that this path is also a line!

Q: Is it possible that the locus of $C$ is not a line?

In this specific problem, because the relationship between the centroid $G$ and the vertex $C$ is linear (it's just a simple average), any linear constraint on $G$ will translate into a linear constraint on $C$. If the constraint on $G$ were more complex, like a circle, the locus of $C$ would also be a circle. Here, the linearity is preserved.

Question

Let $A(2, -3)$ and $B(-2, 3)$ be vertices of a triangle $ABC$. If the centroid of this triangle moves on the line $2x + 3y = 1$, then the locus of the vertex $C$ is the line

Options:

(A) $3x - 2y = 3$

(B) $2x - 3y = 7$

(C) $3x + 2y = 5$

(D) $2x + 3y = 9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have two fixed anchors, point $A$ at $(2, -3)$ and point $B$ at $(-2, 3)$. These two points are the foundation of a triangle $ABC$.

The third vertex, $C$, is a wanderer moving across the plane. As $C$ moves, the centroid $G$ shifts its position. However, $G$ is bound to a specific path defined by the equation $2x + 3y = 3$.

Our mission is to uncover the secret path that vertex $C$ must follow to keep its centroid on that line.

### The Centroid: The Heart of the Triangle

To solve this, we must understand the relationship between the vertices and the centroid. The centroid $G$ is the geometric center of the triangle, calculated as the average of the vertices.

If we let the coordinates of $C$ be $(h, k)$, the centroid $G(x, y)$ is given by the formula:

$$G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)$$

Substituting our known values $A(2, -3)$ and $B(-2, 3)$, and our variable vertex $C(h, k)$, we get:

$$G = \left( \frac{2 + (-2) + h}{3}, \frac{-3 + 3 + k}{3} \right)$$

The constants $2$ and $-2$ cancel out, as do $-3$ and $3$. We are left with a simplified coordinate for our centroid:

$$G = \left( \frac{h}{3}, \frac{k}{3} \right)$$

### Bridging the Constraint

The problem states that the centroid $G$ must lie on the line $2x + 3y = 3$. This means the coordinates of $G$, which are $(\frac{h}{3}, \frac{k}{3})$, must satisfy this equation.

Substituting these coordinates into the line equation, we obtain:

$$2\left( \frac{h}{3} \right) + 3\left( \frac{k}{3} \right) = 3$$

To simplify this relationship, we multiply the entire equation by $3$ to clear the denominators:

$$2h + 3k = 9$$

### The Final Reveal

We have found the relationship that governs the movement of $C$. To express this as the locus of the point $(x, y)$, we replace $h$ with $x$ and $k$ with $y$.

The path of vertex $C$ is defined by the equation:

**$$2x + 3y = 9$$**

This is the line that vertex $C$ must travel along. By understanding the relationship between the centroid and the vertices, we have transformed a complex problem of motion into a clear, elegant geometric truth.

Animated Solution for Mathematics - Straight Lines: Let $A (2, - 3)$ and $B (- 2, 3)$ be vertices of a triangle $A B C$ . If the centroid of this triangle moves on the line $2 x + 3 y = 1$ , then the locus of the vertex $C$ is the line

Select Answer:

Visualized Solution

Given Vertices $A$ and $B$

Let Vertex $C$ be $(h, k)$

Centroid Formula

Substitute Coordinates into Formula

Coordinates of Centroid $G$

Centroid Path Equation

Substitute $G$ into the Path

Simplify to find Locus

Final Locus of Vertex $C$

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Centroid

Bridging the Constraint

The Final Reveal

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Visualized Solution

Given Vertices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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