Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD. 

Answer

To determine whether the points A(3, 2), B(6, -2), and C(2, -5) can be the vertices of a square, we need to check two conditions:
1. The lengths of the sides AB, BC, and AC must be equal.
2. The slopes of adjacent sides must be negative reciprocals (indicating they are perpendicular).

### Step 1: Calculate the lengths of sides AB, BC, and AC

**Length of AB:**
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(6 - 3)^2 + (-2 - 2)^2} \]
\[ = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

**Length of BC:**
\[ BC = \sqrt{(2 - 6)^2 + (-5 + 2)^2} = \sqrt{(-4)^2 + (-3)^2} \]
\[ = \sqrt{16 + 9} = \sqrt{25} = 5 \]

**Length of AC:**
\[ AC = \sqrt{(2 - 3)^2 + (-5 - 2)^2} = \sqrt{(-1)^2 + (-7)^2} \]
\[ = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

### Step 2: Check if AB and BC are perpendicular

**Slope of AB:**
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 2}{6 - 3} = \frac{-4}{3} \]

**Slope of BC:**
\[ \text{slope of } BC = \frac{-5 + 2}{2 - 6} = \frac{-3}{-4} = \frac{3}{4} \]

**Product of slopes:**
\[ \text{slope of } AB \times \text{slope of } BC = \left(-\frac{4}{3}\right) \times \left(\frac{3}{4}\right) = -1 \]

Since the product of the slopes is -1, AB is perpendicular to BC.

### Conclusion for Step 1:
Since AB = BC and they are perpendicular, A, B, and C can be vertices of a square.

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### Step 3: Find the equation of side AD of the square

Since AD is parallel to BC, it will have the same slope as BC.

**Slope of AD:**
\[ \text{slope of } AD = \frac{3}{4} \]

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting point A(3, 2):
\[ y - 2 = \frac{3}{4}(x - 3) \]
Multiplying through by 4 to eliminate the fraction:
\[ 4(y - 2) = 3(x - 3) \]
\[ 4y - 8 = 3x - 9 \]
Rearranging gives:
\[ 3x - 4y + 1 = 0 \]

### Step 4: Find the equation of diagonal BD

**Slope of AC:**
\[ \text{slope of } AC = \frac{-5 - 2}{2 - 3} = \frac{-7}{-1} = 7 \]

Since diagonals are perpendicular, the slope of BD will be the negative reciprocal of the slope of AC:
\[ \text{slope of } BD = -\frac{1}{7} \]

Using point-slope form with point B(6, -2):
\[ y - (-2) = -\frac{1}{7}(x - 6) \]
\[ y + 2 = -\frac{1}{7}x + \frac{6}{7} \]
Multiplying through by 7 to eliminate the fraction:
\[ 7y + 14 = -x + 6 \]
Rearranging gives:
\[ x + 7y - 8 = 0 \]

### Final Equations
- Equation of side AD: \(3x - 4y + 1 = 0\)
- Equation of diagonal BD: \(x + 7y - 8 = 0\)

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