To find the value of \(|\tan \theta|\) where \(\theta\) is the angle between the diagonals \(PR\) and \(QS\) of the 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:
- \(PQ: x + 2y - 3 = 0\)
- \(QR: x - 1 = 0\)
- \(RS: x - 3y - 4 = 0\)
- \(SP: 5x + y + 12 = 0\)
### Step 2: Find the intersection points to determine the vertices of the quadrilateral
We need to find the points \(P\), \(Q\), \(R\), and \(S\) by solving the equations of the lines.
#### Finding Point Q:
From \(QR: x - 1 = 0\), we have \(x = 1\).
Substituting \(x = 1\) into \(PQ: x + 2y - 3 = 0\):
\[
1 + 2y - 3 = 0 \implies 2y = 2 \implies y = 1
\]
Thus, \(Q(1, 1)\).
#### Finding Point R:
Substituting \(x = 1\) into \(RS: x - 3y - 4 = 0\):
\[
1 - 3y - 4 = 0 \implies -3y = 3 \implies y = -1
\]
Thus, \(R(1, -1)\).
#### Finding Point P:
Now, we will find \(P\) by solving \(PQ\) and \(SP\). We already have \(PQ: x + 2y - 3 = 0\) and \(SP: 5x + y + 12 = 0\).
From \(PQ: x + 2y - 3 = 0\), we can express \(y\) in terms of \(x\):
\[
2y = 3 - x \implies y = \frac{3 - x}{2}
\]
Substituting this into \(SP\):
\[
5x + \frac{3 - x}{2} + 12 = 0
\]
Multiplying through by 2 to eliminate the fraction:
\[
10x + 3 - x + 24 = 0 \implies 9x + 27 = 0 \implies x = -3
\]
Substituting \(x = -3\) back into \(PQ\) to find \(y\):
\[
-3 + 2y - 3 = 0 \implies 2y = 6 \implies y = 3
\]
Thus, \(P(-3, 3)\).
#### Finding Point S:
Now, we will find \(S\) by solving \(RS\) and \(SP\):
Using \(RS: x - 3y - 4 = 0\):
\[
x = 3y + 4
\]
Substituting into \(SP: 5x + y + 12 = 0\):
\[
5(3y + 4) + y + 12 = 0 \implies 15y + 20 + y + 12 = 0 \implies 16y + 32 = 0 \implies y = -2
\]
Substituting \(y = -2\) back into \(RS\):
\[
x - 3(-2) - 4 = 0 \implies x + 6 - 4 = 0 \implies x = -2
\]
Thus, \(S(-2, -2)\).
### Step 3: Find the slopes of the diagonals PR and QS
Now we have the points:
- \(P(-3, 3)\)
- \(Q(1, 1)\)
- \(R(1, -1)\)
- \(S(-2, -2)\)
#### Slope of diagonal PR:
\[
m_{PR} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1
\]
#### Slope of diagonal QS:
\[
m_{QS} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 1}{-2 - 1} = \frac{-3}{-3} = 1
\]
### Step 4: Calculate \(|\tan \theta|\)
The formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\]
Substituting \(m_{PR} = -1\) and \(m_{QS} = 1\):
\[
\tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{1 - 1} \right|
\]
Since the denominator becomes zero, this means that the lines are perpendicular, and thus \(\theta = 90^\circ\).
Therefore, \(|\tan \theta| = \infty\).
### Final Answer
The value of \(|\tan \theta|\) is \(\infty\).
