Question

The area of the quadrilateral formed by the lines x+2y+3=0, 2x+4y+9=0, x-2y+3=0 and 3x-6y+11=0 is

• A. \( \frac{5}{12} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{7}{12} \)

Hint:

Recognize pairs of parallel lines by checking if their slope is the same. The area of a parallelogram formed by the lines \(y=m_1x+c_1\), \(y=m_1x+c_2\), \(y=m_2x+d_1\), \(y=m_2x+d_2\) is \(|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}|\). The determinant formula used in the solution is a more general version.

Answer 1

The Correct Option is B

Let's analyze the given lines.
L1: \(x+2y+3=0\). L2: \(2x+4y+9=0 \implies x+2y+4.5=0\). These lines are parallel.
L3: \(x-2y+3=0\). L4: \(3x-6y+11=0 \implies x-2y+11/3=0\). These lines are also parallel.
The quadrilateral formed is a parallelogram.
The area of a parallelogram formed by lines \(a_1x+b_1y+c_1=0\), \(a_1x+b_1y+c_2=0\), \(a_2x+b_2y+d_1=0\), and \(a_2x+b_2y+d_2=0\) is given by the formula:
Area = \( |\frac{(c_1-c_2)(d_1-d_2)}{a_1b_2 - a_2b_1}| \).
Here, \(c_1=3, c_2=4.5\) for the first pair of lines. And \(d_1=3, d_2=11/3\) for the second pair.
The coefficients are \(a_1=1, b_1=2\) and \(a_2=1, b_2=-2\).
Area = \( |\frac{(3 - 4.5)(3 - 11/3)}{(1)(-2) - (1)(2)}| \).
Numerator: \( (-1.5) \cdot (9/3 - 11/3) = (-3/2) \cdot (-2/3) = 1 \).
Denominator: \( -2 - 2 = -4 \).
Area = \( |\frac{1}{-4}| = \frac{1}{4} \).