Question

Area of the parallelogram formed by the lines \( y = 4x, y = 4x + 1, x + y = 0 \) and \( x + y = 1 \) is

• A. \( \frac{1}{5} \)
• B. \( \frac{2}{5} \)
• C. 5
• D. 10

Hint:

For a parallelogram defined by four lines in the form \( y=m_1x+c_1, y=m_1x+c_2, y=m_2x+d_1, y=m_2x+d_2 \), the area is given by the compact formula \( \text{Area} = \frac{|(c_1-c_2)(d_1-d_2)|}{|m_1-m_2|} \). Convert the lines to this form to use it. Here \( m_1=4, m_2=-1 \), giving \( |1-0| \cdot |1-0| / |4 - (-1)| = 1/5 \).

Answer 1

The Correct Option is A

We have two pairs of parallel lines. Pair 1: \( y - 4x = 0 \) and \( y - 4x - 1 = 0 \). (Let's write as \( 4x - y = 0 \) and \( 4x - y + 1 = 0 \)) Pair 2: \( x + y = 0 \) and \( x + y - 1 = 0 \). The area of a parallelogram formed by the 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: \[ \text{Area} = \frac{|(c_1 - c_2)(d_1 - d_2)|}{|a_1b_2 - a_2b_1|} \] From our lines: \( a_1 = 4, b_1 = -1 \). \( c_1 = 0, c_2 = 1 \). \( a_2 = 1, b_2 = 1 \). \( d_1 = 0, d_2 = -1 \). Let's plug these into the formula: \[ \text{Area} = \frac{|(0 - 1)(0 - (-1))|}{|(4)(1) - (1)(-1)|} \] \[ \text{Area} = \frac{|(-1)(1)|}{|4 - (-1)|} = \frac{|-1|}{|5|} = \frac{1}{5} \] The area of the parallelogram is \( \frac{1}{5} \).