(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.

Answer

To find the equations of the diagonals AC and BD of rhombus ABCD with vertices A(1, 5) and C(-3, -1), we will follow these steps:

### Step 1: Find the slope of diagonal AC
The formula for the slope (m) between two points (x1, y1) and (x2, y2) is given by:
\[ m = \frac{y2 - y1}{x2 - x1} \]
For points A(1, 5) and C(-3, -1):
- \(x1 = 1\), \(y1 = 5\)
- \(x2 = -3\), \(y2 = -1\)

Calculating the slope:
\[ m_{AC} = \frac{-1 - 5}{-3 - 1} = \frac{-6}{-4} = \frac{3}{2} \]

### Step 2: Use point-slope form to find the equation of line AC
The point-slope form of a line is given by:
\[ y - y1 = m(x - x1) \]
Using point A(1, 5) and the slope \(m_{AC} = \frac{3}{2}\):
\[ y - 5 = \frac{3}{2}(x - 1) \]

### Step 3: Simplify the equation of line AC
Distributing the slope:
\[ y - 5 = \frac{3}{2}x - \frac{3}{2} \]
Adding 5 to both sides:
\[ y = \frac{3}{2}x - \frac{3}{2} + 5 \]
Converting 5 to a fraction:
\[ y = \frac{3}{2}x - \frac{3}{2} + \frac{10}{2} \]
Combining the constants:
\[ y = \frac{3}{2}x + \frac{7}{2} \]

### Step 4: Convert to standard form for line AC
To convert to standard form \(Ax + By + C = 0\):
\[ 2y = 3x + 7 \]
Rearranging gives:
\[ 3x - 2y + 7 = 0 \]

### Step 5: Find the slope of diagonal BD
Since diagonals of a rhombus bisect each other and are perpendicular, the slope of BD (let's call it \(m_{BD}\)) is the negative reciprocal of \(m_{AC}\):
\[ m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]

### Step 6: Find the midpoint of AC
The midpoint (M) of AC can be found using:
\[ M = \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) \]
Calculating the midpoint:
\[ M = \left(\frac{1 + (-3)}{2}, \frac{5 + (-1)}{2}\right) = \left(\frac{-2}{2}, \frac{4}{2}\right) = (-1, 2) \]

### Step 7: Use point-slope form to find the equation of line BD
Using point M(-1, 2) and slope \(m_{BD} = -\frac{2}{3}\):
\[ y - 2 = -\frac{2}{3}(x + 1) \]

### Step 8: Simplify the equation of line BD
Distributing the slope:
\[ y - 2 = -\frac{2}{3}x - \frac{2}{3} \]
Adding 2 to both sides:
\[ y = -\frac{2}{3}x - \frac{2}{3} + 2 \]
Converting 2 to a fraction:
\[ y = -\frac{2}{3}x - \frac{2}{3} + \frac{6}{3} \]
Combining the constants:
\[ y = -\frac{2}{3}x + \frac{4}{3} \]

### Step 9: Convert to standard form for line BD
To convert to standard form \(Ax + By + C = 0\):
\[ 3y = -2x + 4 \]
Rearranging gives:
\[ 2x + 3y - 4 = 0 \]

### Final Equations
The equations of the diagonals are:
- Diagonal AC: \(3x - 2y + 7 = 0\)
- Diagonal BD: \(2x + 3y - 4 = 0\)