Question

The equation of the straight line through the intersection of the lines \( x - 2y = 1 \) and \( x + 3y = 2 \) and parallel to \( 3x + 4y = 0 \) is:

• A. \( 3x + 4y + 5 = 0 \)
• B. \( 3x + 4y - 10 = 0 \)
• C. \( 3x + 4y - 5 = 0 \)
• D. \( 3x + 4y + 6 = 0 \)

Hint:

To find the equation of a line passing through the intersection of two lines and parallel to another line, first find the intersection point, then use the slope of the given parallel line to form the equation of the required line.

Answer 1

The Correct Option is C

We are given two lines: 1. \( x - 2y = 1 \), 2. \( x + 3y = 2 \). Also, we are given that the required line is parallel to the line \( 3x + 4y = 0 \). The general equation of a line passing through the intersection of two lines is: \[ L = \lambda (x - 2y - 1) + \mu (x + 3y - 2) = 0, \] where \( \lambda \) and \( \mu \) are constants to be determined. Step 1: Finding the intersection point of the given lines.
First, solve the system of equations \( x - 2y = 1 \) and \( x + 3y = 2 \) simultaneously. From \( x - 2y = 1 \), we can express \( x \) as: \[ x = 2y + 1. \] Substitute this into \( x + 3y = 2 \): \[ (2y + 1) + 3y = 2, \] \[ 5y + 1 = 2, \] \[ 5y = 1, \] \[ y = \frac{1}{5}. \] Now substitute \( y = \frac{1}{5} \) into \( x = 2y + 1 \): \[ x = 2 \times \frac{1}{5} + 1 = \frac{2}{5} + 1 = \frac{7}{5}. \] So, the intersection point is \( \left( \frac{7}{5}, \frac{1}{5} \right) \).
Step 2: Equation of the required line.
Since the required line is parallel to \( 3x + 4y = 0 \), it has the same slope. The slope of the line \( 3x + 4y = 0 \) is \( -\frac{3}{4} \), so the required line has the same slope. Now, the equation of the required line passing through the point \( \left( \frac{7}{5}, \frac{1}{5} \right) \) and with slope \( -\frac{3}{4} \) is: \[ y - \frac{1}{5} = -\frac{3}{4} \left( x - \frac{7}{5} \right). \] Simplify the equation: \[ y - \frac{1}{5} = -\frac{3}{4}x + \frac{21}{20}. \] Multiply the entire equation by 20 to eliminate fractions: \[ 20y - 4 = -15x + 21. \] Rearrange to get the equation: \[ 15x + 20y - 25 = 0. \] This simplifies to: \[ 3x + 4y - 5 = 0. \]
Step 3: Conclusion.
Thus, the equation of the required line is \( 3x + 4y - 5 = 0 \), and the correct answer is (c).