An equilateral triangle has one vertex at (3, 4) and another at (-2, 3). Find the coordinates of the third vertex.

Answer

To find the coordinates of the third vertex of the equilateral triangle with vertices at \( A(3, 4) \) and \( B(-2, 3) \), we can follow these steps:

### Step 1: Calculate the Length of Side AB
First, we need to calculate the distance \( AB \) using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points \( A(3, 4) \) and \( B(-2, 3) \):
\[ AB = \sqrt{((-2) - 3)^2 + (3 - 4)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \]

### Step 2: Use the Properties of Equilateral Triangle
In an equilateral triangle, all sides are equal. Therefore, the lengths \( AC \) and \( BC \) must also equal \( AB \). We denote the third vertex as \( C(x, y) \).

### Step 3: Set Up the Equations for Distances AC and BC
We can set up the equations for the distances \( AC \) and \( BC \):
\[ AC = \sqrt{(x - 3)^2 + (y - 4)^2} = \sqrt{26} \]
Squaring both sides gives:
\[ (x - 3)^2 + (y - 4)^2 = 26 \quad \text{(1)} \]

For \( BC \):
\[ BC = \sqrt{(x + 2)^2 + (y - 3)^2} = \sqrt{26} \]
Squaring both sides gives:
\[ (x + 2)^2 + (y - 3)^2 = 26 \quad \text{(2)} \]

### Step 4: Expand the Equations
Expanding both equations:
1. From equation (1):
\[ (x - 3)^2 + (y - 4)^2 = 26 \implies x^2 - 6x + 9 + y^2 - 8y + 16 = 26 \]
This simplifies to:
\[ x^2 + y^2 - 6x - 8y - 1 = 0 \quad \text{(3)} \]

2. From equation (2):
\[ (x + 2)^2 + (y - 3)^2 = 26 \implies x^2 + 4x + 4 + y^2 - 6y + 9 = 26 \]
This simplifies to:
\[ x^2 + y^2 + 4x - 6y - 13 = 0 \quad \text{(4)} \]

### Step 5: Subtract Equations (3) and (4)
Now, we can subtract equation (3) from equation (4):
\[ (x^2 + y^2 + 4x - 6y - 13) - (x^2 + y^2 - 6x - 8y - 1) = 0 \]
This simplifies to:
\[ 10x + 2y - 12 = 0 \]
Rearranging gives:
\[ 5x + y = 6 \quad \text{(5)} \]

### Step 6: Substitute Equation (5) into (3)
Now, we can substitute \( y = 6 - 5x \) from equation (5) into equation (3):
\[ x^2 + (6 - 5x)^2 - 6x - 8(6 - 5x) - 1 = 0 \]
Expanding this:
\[ x^2 + (36 - 60x + 25x^2) - 6x - 48 + 40x - 1 = 0 \]
Combining like terms:
\[ 26x^2 - 26x - 13 = 0 \]
Dividing through by 13 gives:
\[ 2x^2 - 2x - 1 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2} \]

### Step 8: Find Corresponding y Values
Substituting \( x \) back into equation (5):
1. For \( x = \frac{1 + \sqrt{3}}{2} \):
\[ y = 6 - 5\left(\frac{1 + \sqrt{3}}{2}\right) = 6 - \frac{5 + 5\sqrt{3}}{2} = \frac{12 - 5 - 5\sqrt{3}}{2} = \frac{7 - 5\sqrt{3}}{2} \]

2. For \( x = \frac{1 - \sqrt{3}}{2} \):
\[ y = 6 - 5\left(\frac{1 - \sqrt{3}}{2}\right) = 6 - \frac{5 - 5\sqrt{3}}{2} = \frac{12 - 5 + 5\sqrt{3}}{2} = \frac{7 + 5\sqrt{3}}{2} \]

### Final Coordinates of the Third Vertex
Thus, the coordinates of the third vertex \( C \) can be:
1. \( C\left(\frac{1 + \sqrt{3}}{2}, \frac{7 - 5\sqrt{3}}{2}\right) \)
2. \( C\left(\frac{1 - \sqrt{3}}{2}, \frac{7 + 5\sqrt{3}}{2}\right) \)