A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonals AC and BD. 

Answer

To find the equations of the diagonals AC and BD of square ABCD given the vertices A(7, -2) and C(-1, -6), we can follow these steps:

### Step 1: Find the Midpoint O of Diagonal AC
The midpoint O of a line segment with endpoints (x1, y1) and (x2, y2) is given by the formula:
\[ O\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Here, A(7, -2) is (x1, y1) and C(-1, -6) is (x2, y2).

Calculating the coordinates of O:
\[ O\left(\frac{7 + (-1)}{2}, \frac{-2 + (-6)}{2}\right) = O\left(\frac{6}{2}, \frac{-8}{2}\right) = O(3, -4) \]

### Step 2: Find the Slope of Diagonal AC
The slope (m) of a line through two points (x1, y1) and (x2, y2) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of A and C:
\[ m_{AC} = \frac{-6 - (-2)}{-1 - 7} = \frac{-6 + 2}{-1 - 7} = \frac{-4}{-8} = \frac{1}{2} \]

### Step 3: Write the Equation of Diagonal AC
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Using point A(7, -2) and slope \(m_{AC} = \frac{1}{2}\):
\[ y - (-2) = \frac{1}{2}(x - 7) \]
Simplifying:
\[ y + 2 = \frac{1}{2}x - \frac{7}{2} \]
\[ y = \frac{1}{2}x - \frac{7}{2} - 2 \]
\[ y = \frac{1}{2}x - \frac{7}{2} - \frac{4}{2} \]
\[ y = \frac{1}{2}x - \frac{11}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 2y = x - 11 \quad \Rightarrow \quad x - 2y = 11 \]

### Step 4: Find the Slope of Diagonal BD
Since the diagonals of a square are perpendicular, the slope of diagonal BD will be the negative reciprocal of the slope of AC:
\[ m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{\frac{1}{2}} = -2 \]

### Step 5: Write the Equation of Diagonal BD
Using the midpoint O(3, -4) and slope \(m_{BD} = -2\):
\[ y - (-4) = -2(x - 3) \]
Simplifying:
\[ y + 4 = -2x + 6 \]
\[ y = -2x + 6 - 4 \]
\[ y = -2x + 2 \]
Rearranging:
\[ 2x + y = 2 \]

### Final Equations
The equations of the diagonals are:
- Diagonal AC: \(x - 2y = 11\)
- Diagonal BD: \(2x + y = 2\)