The sides AB,BC,CD and DA of a quadrilateral are `x+2y=3,x=1,x-3y=4,5x+y+12=0` respectively. The angles between diagonals AC and BD is

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `30^(@)`

Answer

To find the angle between the diagonals AC and BD of the quadrilateral defined by the lines \(AB\), \(BC\), \(CD\), and \(DA\), we will follow these steps:

### Step 1: Determine the Points of Intersection
We need to find the coordinates of points A, B, C, and D by solving the equations of the lines.

1. **Finding Point B**:
- The equations for lines \(AB\) and \(BC\) are:
\[ AB: x + 2y = 3 \quad \text{(1)} \]
\[ BC: x = 1 \quad \text{(2)} \]
- Substitute \(x = 1\) into equation (1):
\[ 1 + 2y = 3 \implies 2y = 2 \implies y = 1 \]
- Thus, point \(B\) is \( (1, 1) \).

2. **Finding Point C**:
- Use \(x = 1\) in the equation for \(CD\):
\[ CD: x - 3y = 4 \quad \text{(3)} \]
- Substitute \(x = 1\) into equation (3):
\[ 1 - 3y = 4 \implies -3y = 3 \implies y = -1 \]
- Thus, point \(C\) is \( (1, -1) \).

3. **Finding Point A**:
- Solve the equations for \(AB\) and \(DA\):
\[ DA: 5x + y + 12 = 0 \quad \text{(4)} \]
- We have:
\[ x + 2y = 3 \quad \text{(1)} \]
- Multiply equation (4) by 5:
\[ 5x + 5y + 60 = 0 \implies 5x + y = -12 \quad \text{(5)} \]
- Now, subtract equation (1) multiplied by 5 from equation (5):
\[ 5x + y - (5x + 10y) = -12 - 15 \]
\[ -9y = -27 \implies y = 3 \]
- Substitute \(y = 3\) back into equation (1):
\[ x + 2(3) = 3 \implies x + 6 = 3 \implies x = -3 \]
- Thus, point \(A\) is \( (-3, 3) \).

4. **Finding Point D**:
- Solve equations for \(DA\) and \(CD\):
\[ CD: x - 3y = 4 \quad \text{(3)} \]
- Substitute \(5x + y + 12 = 0\) into equation (3):
\[ 5x + y = -12 \implies y = -5x - 12 \]
- Substitute into equation (3):
\[ x - 3(-5x - 12) = 4 \implies x + 15x + 36 = 4 \implies 16x = -32 \implies x = -2 \]
- Substitute \(x = -2\) back into \(y = -5x - 12\):
\[ y = -5(-2) - 12 = 10 - 12 = -2 \]
- Thus, point \(D\) is \( (-2, -2) \).

### Step 2: Calculate the Slopes of Diagonals AC and BD
1. **Slope of AC**:
- Points \(A(-3, 3)\) and \(C(1, -1)\):
\[ \text{slope of AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1 \]

2. **Slope of BD**:
- Points \(B(1, 1)\) and \(D(-2, -2)\):
\[ \text{slope of BD} = \frac{-2 - 1}{-2 - 1} = \frac{-3}{-3} = 1 \]

### Step 3: Determine the Angle Between the Diagonals
- The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
- Here, \( m_1 = -1 \) and \( m_2 = 1 \):
\[ \tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{0} \right| \text{ (undefined)} \]
- Since the tangent is undefined, this indicates that the angle is \(90^\circ\).

### Conclusion
The angle between the diagonals AC and BD is \(90^\circ\).