Let the equations of the sides PQ, QR, RS and SP of a quadrilateral PQRS are `x+2y-3=0, x-1=0, x-3y-4=0` and `5x+y+12=0` respectively. If `theta` is the angle between the diagonals PR and QS, then the value of `|tan theta|` is equal to

Options

A. 2
B. `-2`
C. 1
D. Not defined

Answer

To find the value of \(|\tan \theta|\), where \(\theta\) is the angle between the diagonals \(PR\) and \(QS\) of quadrilateral \(PQRS\), we will follow these steps:

### Step 1: Identify the equations of the sides
The equations of the sides of the quadrilateral are given as:
1. \(PQ: x + 2y - 3 = 0\)
2. \(QR: x - 1 = 0\)
3. \(RS: x - 3y - 4 = 0\)
4. \(SP: 5x + y + 12 = 0\)

### Step 2: Find the intersection points (vertices of the quadrilateral)
We will find the intersection points of the lines to get the coordinates of the vertices \(P\), \(Q\), \(R\), and \(S\).

**Finding point \(Q\):**
- From \(QR: x - 1 = 0\), we have \(x = 1\).
- Substitute \(x = 1\) into \(PQ: x + 2y - 3 = 0\):
\[ 1 + 2y - 3 = 0 \implies 2y = 2 \implies y = 1 \]
Thus, \(Q(1, 1)\).

**Finding point \(R\):**
- From \(QR: x - 1 = 0\), we have \(x = 1\).
- Substitute \(x = 1\) into \(RS: x - 3y - 4 = 0\):
\[ 1 - 3y - 4 = 0 \implies -3y = 3 \implies y = -1 \]
Thus, \(R(1, -1)\).

**Finding point \(P\):**
- From \(SP: 5x + y + 12 = 0\) and \(PQ: x + 2y - 3 = 0\).
- Substitute \(y\) from \(PQ\) into \(SP\):
\[ y = \frac{3 - x}{2} \]
Substitute into \(SP\):
\[ 5x + \frac{3 - x}{2} + 12 = 0 \]
Multiply through by 2 to eliminate the fraction:
\[ 10x + 3 - x + 24 = 0 \implies 9x + 27 = 0 \implies x = -3 \]
Substitute \(x = -3\) back to find \(y\):
\[ y = \frac{3 - (-3)}{2} = \frac{6}{2} = 3 \]
Thus, \(P(-3, 3)\).

**Finding point \(S\):**
- From \(SP: 5x + y + 12 = 0\) and \(RS: x - 3y - 4 = 0\).
- Substitute \(y\) from \(RS\) into \(SP\):
\[ y = \frac{x - 4}{3} \]
Substitute into \(SP\):
\[ 5x + \frac{x - 4}{3} + 12 = 0 \]
Multiply through by 3:
\[ 15x + x - 4 + 36 = 0 \implies 16x + 32 = 0 \implies x = -2 \]
Substitute \(x = -2\) back to find \(y\):
\[ y = \frac{-2 - 4}{3} = -2 \]
Thus, \(S(-2, -2)\).

### Step 3: Calculate the slopes of the diagonals \(PR\) and \(QS\)
**Slope of \(PR\):**
\[ \text{slope of } PR = \frac{y_R - y_P}{x_R - x_P} = \frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1 \]

**Slope of \(QS\):**
\[ \text{slope of } QS = \frac{y_S - y_Q}{x_S - x_Q} = \frac{-2 - 1}{-2 - 1} = \frac{-3}{-3} = 1 \]

### Step 4: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \(m_1 = -1\) and \(m_2 = 1\):
\[ \tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{1 - 1} \right| \]
Since the denominator is zero, \(\tan \theta\) is undefined, which indicates that the diagonals are perpendicular.

### Final Result
Thus, \(|\tan \theta|\) is undefined, which corresponds to the angle being \(90^\circ\).