The diagonals AC and BD of a rhombus intersect at (5, 6). If A (3, 2) then equation of diagonal BD is

Options

A. y - x = 1
B. 2 y - x = 17
C. y - 2x + 4 = 0
D. 2y + x = 17

Answer

To find the equation of diagonal BD of a rhombus where the diagonals intersect at point (5, 6) and point A is given as (3, 2), we can follow these steps:

### Step 1: Identify the coordinates of the intersection point and point A
The intersection point of the diagonals AC and BD is given as O(5, 6). Point A is given as A(3, 2).

### Step 2: Find the slope of line AO
To find the slope (m1) of line AO, we use the slope formula:
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of points A(3, 2) and O(5, 6):
\[ m_1 = \frac{6 - 2}{5 - 3} = \frac{4}{2} = 2 \]

### Step 3: Determine the slope of line BD
Since the diagonals of a rhombus are perpendicular to each other, the slope of line BD (m2) can be found using the relationship:
\[ m_1 \cdot m_2 = -1 \]
Substituting m1:
\[ 2 \cdot m_2 = -1 \implies m_2 = -\frac{1}{2} \]

### Step 4: Use point-slope form to find the equation of line BD
We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Here, we will use point O(5, 6) and the slope m2 = -\frac{1}{2}:
\[ y - 6 = -\frac{1}{2}(x - 5) \]

### Step 5: Simplify the equation
Expanding the equation:
\[ y - 6 = -\frac{1}{2}x + \frac{5}{2} \]
Adding 6 to both sides:
\[ y = -\frac{1}{2}x + \frac{5}{2} + 6 \]
Converting 6 to a fraction:
\[ y = -\frac{1}{2}x + \frac{5}{2} + \frac{12}{2} = -\frac{1}{2}x + \frac{17}{2} \]

### Step 6: Rearranging to standard form
To express the equation in standard form (Ax + By + C = 0):
\[ \frac{1}{2}x + y - \frac{17}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ x + 2y - 17 = 0 \]

### Final Answer
The equation of diagonal BD is:
\[ x + 2y - 17 = 0 \]