Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
Find the co-ordinates of its fourth vertex D, if ABCD is a square. 

Answer

To show that the points A(3, 2), B(6, -2), and C(2, -5) can be the vertices of a square and to find the coordinates of the fourth vertex D, we will follow these steps:

### Step 1: Calculate the lengths of sides AB and BC
We will use the distance formula to find the lengths of sides AB and BC.

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

**Calculate AB:**
- Points A(3, 2) and B(6, -2):
\[ AB = \sqrt{(6 - 3)^2 + (-2 - 2)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

**Calculate BC:**
- Points B(6, -2) and C(2, -5):
\[ BC = \sqrt{(2 - 6)^2 + (-5 - (-2))^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 2: Check if sides AB and BC are equal
Since \(AB = 5\) and \(BC = 5\), we can conclude that \(AB = BC\).

### Step 3: Check if AB is perpendicular to BC
To check if the lines AB and BC are perpendicular, we will calculate their slopes.

**Calculate the slope of AB:**
The slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For AB:
\[ m_{AB} = \frac{-2 - 2}{6 - 3} = \frac{-4}{3} \]

**Calculate the slope of BC:**
For BC:
\[ m_{BC} = \frac{-5 - (-2)}{2 - 6} = \frac{-3}{-4} = \frac{3}{4} \]

### Step 4: Check if the product of slopes is -1
If two lines are perpendicular, the product of their slopes should be -1:
\[ m_{AB} \times m_{BC} = \left(-\frac{4}{3}\right) \times \left(\frac{3}{4}\right) = -1 \]
Since the product is -1, AB is perpendicular to BC.

### Conclusion for vertices A, B, and C
Since \(AB = BC\) and \(AB\) is perpendicular to \(BC\), points A, B, and C can indeed be vertices of a square.

### Step 5: Find the coordinates of the fourth vertex D
To find vertex D, we will use the properties of the square. The lines AD and BC are parallel, and the lines AB and CD are also parallel.

**Equation of line CD (parallel to AB):**
Using point C(2, -5) and the slope of AB \(-\frac{4}{3}\):
\[ y - (-5) = -\frac{4}{3}(x - 2) \]
This simplifies to:
\[ y + 5 = -\frac{4}{3}x + \frac{8}{3} \]
\[ 4x + 3y = -7 \quad (1) \]

**Equation of line AD (parallel to BC):**
Using point A(3, 2) and the slope of BC \(\frac{3}{4}\):
\[ y - 2 = \frac{3}{4}(x - 3) \]
This simplifies to:
\[ 4y - 8 = 3x - 9 \]
\[ 3x - 4y = 1 \quad (2) \]

### Step 6: Solve equations (1) and (2) to find D
We will solve the system of equations:
1. \(4x + 3y = -7\)
2. \(3x - 4y = 1\)

To eliminate \(y\), we can multiply the first equation by 4 and the second by 3:
\[ 16x + 12y = -28 \quad (3) \]
\[ 9x - 12y = 3 \quad (4) \]

Now, add equations (3) and (4):
\[ 16x + 12y + 9x - 12y = -28 + 3 \]
\[ 25x = -25 \implies x = -1 \]

Substituting \(x = -1\) into equation (1):
\[ 4(-1) + 3y = -7 \]
\[ -4 + 3y = -7 \implies 3y = -3 \implies y = -1 \]

### Final Coordinates of D
Thus, the coordinates of the fourth vertex D are:
\[ D(-1, -1) \]

### Summary
The points A(3, 2), B(6, -2), and C(2, -5) can be vertices of a square, and the coordinates of the fourth vertex D are \((-1, -1)\).

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