The co-ordinates of the third vertex of an equilateral triangle whose two vertices are at (3,4) and (-2,3) are

Options

A. (1,1) or (1,-1)
B. `((1+sqrt(3))/2,(7-5sqrt(3))/2)` or `((1-sqrt(3))/2,(7+5sqrt(3))/2)`
C. `(-sqrt(3),sqrt(3))` or `(sqrt(3),sqrt(3))`
D. None of these

Answer

To find the coordinates of the third vertex of an equilateral triangle given two vertices at (3, 4) and (-2, 3), we can follow these steps:

### Step 1: Identify the vertices
Let the vertices of the triangle be A(3, 4) and B(-2, 3). We need to find the coordinates of the third vertex C(x, y).

### Step 2: Calculate the distance between A and B
The distance between points A and B can be calculated using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and B:
\[ AB = \sqrt{((-2) - 3)^2 + (3 - 4)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \]

### Step 3: Set up the equations for distances
Since triangle ABC is equilateral, the distances AC and BC must also equal AB:
1. \( AC = \sqrt{(x - 3)^2 + (y - 4)^2} = \sqrt{26} \)
2. \( BC = \sqrt{(x + 2)^2 + (y - 3)^2} = \sqrt{26} \)

### Step 4: Square both equations to eliminate the square roots
Squaring both sides of the equations gives us:
1. \( (x - 3)^2 + (y - 4)^2 = 26 \) (Equation 1)
2. \( (x + 2)^2 + (y - 3)^2 = 26 \) (Equation 2)

### Step 5: Expand both equations
Expanding Equation 1:
\[ (x - 3)^2 + (y - 4)^2 = 26 \implies x^2 - 6x + 9 + y^2 - 8y + 16 = 26 \implies x^2 + y^2 - 6x - 8y + 25 = 26 \implies x^2 + y^2 - 6x - 8y - 1 = 0 \]
(Equation 1 simplified)

Expanding Equation 2:
\[ (x + 2)^2 + (y - 3)^2 = 26 \implies x^2 + 4x + 4 + y^2 - 6y + 9 = 26 \implies x^2 + y^2 + 4x - 6y - 13 = 0 \]
(Equation 2 simplified)

### Step 6: Subtract the two equations
Now, we can subtract Equation 1 from Equation 2:
\[ (x^2 + y^2 + 4x - 6y - 13) - (x^2 + y^2 - 6x - 8y - 1) = 0 \]
This simplifies to:
\[ 10x + 2y - 12 = 0 \implies 5x + y = 6 \quad \text{(Equation 3)} \]

### Step 7: Substitute Equation 3 into one of the original distance equations
We can express \( y \) in terms of \( x \) using Equation 3:
\[ y = 6 - 5x \]

Substituting this into Equation 1:
\[ (x - 3)^2 + ((6 - 5x) - 4)^2 = 26 \]
This simplifies to:
\[ (x - 3)^2 + (2 - 5x)^2 = 26 \]

### Step 8: Expand and simplify
Expanding:
\[ (x - 3)^2 + (2 - 5x)^2 = 26 \]
\[ (x^2 - 6x + 9) + (25x^2 - 20x + 4) = 26 \]
\[ 26x^2 - 26x - 13 = 0 \]
Dividing by 13:
\[ 2x^2 - 2x - 1 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2} \]

### Step 10: Find corresponding y values
Substituting \( x \) back into Equation 3:
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 Answer
Thus, the coordinates of the third vertex C are:
1. \( \left(\frac{1 + \sqrt{3}}{2}, \frac{7 - 5\sqrt{3}}{2}\right) \)
2. \( \left(\frac{1 - \sqrt{3}}{2}, \frac{7 + 5\sqrt{3}}{2}\right) \)