If the centroid and a vertex of an equilateral triangle are (2,3) and (4,3) respectively, then the other two vertices are

Options

A. `(1,2+-sqrt(3))`
B. `(1,3+-sqrt(3))`
C. `(2,2+-sqrt(3))`
D. `(2,3+-sqrt(3))`

Answer

To find the other two vertices of the equilateral triangle given the centroid and one vertex, we can follow these steps:

### Step 1: Identify the Given Points
We have the centroid \( G(2, 3) \) and one vertex \( A(4, 3) \).

### Step 2: Use the Centroid Formula
The centroid \( G \) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the known values:
\[ G(2, 3) = \left(\frac{4 + x_2 + x_3}{3}, \frac{3 + y_2 + y_3}{3}\right) \]

### Step 3: Set Up the Equations
From the x-coordinates:
\[ 2 = \frac{4 + x_2 + x_3}{3} \implies 4 + x_2 + x_3 = 6 \implies x_2 + x_3 = 2 \quad \text{(1)} \]
From the y-coordinates:
\[ 3 = \frac{3 + y_2 + y_3}{3} \implies 3 + y_2 + y_3 = 9 \implies y_2 + y_3 = 6 \quad \text{(2)} \]

### Step 4: Determine the Properties of the Equilateral Triangle
In an equilateral triangle, the median also serves as the altitude. Since \( A(4, 3) \) and the centroid \( G(2, 3) \) lie on the same horizontal line \( y = 3 \), the other two vertices \( B \) and \( C \) must have the same x-coordinate.

Let’s denote the x-coordinate of points \( B \) and \( C \) as \( x \). Thus, we can write:
\[ x_2 = x \quad \text{and} \quad x_3 = x \]
Substituting into equation (1):
\[ x + x = 2 \implies 2x = 2 \implies x = 1 \]

### Step 5: Find the y-coordinates of B and C
Now substituting \( x_2 = 1 \) and \( x_3 = 1 \) into equation (2):
\[ y_2 + y_3 = 6 \]
Let’s denote \( y_2 = y \) and \( y_3 = 6 - y \).

### Step 6: Use the Distance Formula
The distance \( AB \) must equal \( AC \) since it is an equilateral triangle. The distance \( AB \) is given by:
\[ AB = \sqrt{(4 - 1)^2 + (3 - y)^2} = \sqrt{9 + (3 - y)^2} \]
The distance \( AC \) is given by:
\[ AC = \sqrt{(4 - 1)^2 + (3 - (6 - y))^2} = \sqrt{9 + (y - 3)^2} \]
Setting these distances equal:
\[ \sqrt{9 + (3 - y)^2} = \sqrt{9 + (y - 3)^2} \]
Squaring both sides and simplifying:
\[ 9 + (3 - y)^2 = 9 + (y - 3)^2 \]
This simplifies to:
\[ (3 - y)^2 = (y - 3)^2 \]
Since both sides are equal, we can conclude that:
\[ 3 - y = y - 3 \quad \text{or} \quad 3 - y = -(y - 3) \]

### Step 7: Solve the Equations
1. From \( 3 - y = y - 3 \):
\[ 3 + 3 = 2y \implies 6 = 2y \implies y = 3 \]
(This is not valid since it would mean B and C coincide with A)

2. From \( 3 - y = -(y - 3) \):
\[ 3 - y = -y + 3 \implies 3 = 3 \quad \text{(always true)} \]
So we need to find the values of \( y \) such that \( y_2 \) and \( y_3 \) are \( 3 + \sqrt{3} \) and \( 3 - \sqrt{3} \).

### Step 8: Final Coordinates
Thus, the coordinates of vertices \( B \) and \( C \) are:
\[ B(1, 3 + \sqrt{3}), \quad C(1, 3 - \sqrt{3}) \]

### Conclusion
The other two vertices of the equilateral triangle are:
\[ (1, 3 + \sqrt{3}) \quad \text{and} \quad (1, 3 - \sqrt{3}) \]