Question

Two straight lines \( 3x - 2y = 5 \) and \( 2x + ky + 7 = 0 \) are perpendicular to each other. The value of \( k \) is:

• A. \( \frac{1}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{3}{2} \)
• D. 3

Hint:

When two lines are perpendicular, the product of their slopes equals \( -1 \). Use this property to find the unknown slope in problems involving perpendicular lines.

Answer 1

The Correct Option is B

For two lines to be perpendicular, the product of their slopes must be \( -1 \). We first find the slopes of the given lines.
1. Equation of first line: \( 3x - 2y = 5 \) Rearranging the equation in the slope-intercept form \( y = mx + c \): \[ 3x - 2y = 5 \quad \Rightarrow \quad -2y = -3x + 5 \quad \Rightarrow \quad y = \frac{3}{2}x - \frac{5}{2} \] So, the slope of the first line is \( m_1 = \frac{3}{2} \). 2. Equation of second line: \( 2x + ky + 7 = 0 \) Rearranging this equation into slope-intercept form: \[ 2x + ky + 7 = 0 \quad \Rightarrow \quad ky = -2x - 7 \quad \Rightarrow \quad y = -\frac{2}{k}x - \frac{7}{k} \] So, the slope of the second line is \( m_2 = -\frac{2}{k} \). For the lines to be perpendicular: \[ m_1 \times m_2 = -1 \] Substituting the values of \( m_1 \) and \( m_2 \): \[ \frac{3}{2} \times \left( -\frac{2}{k} \right) = -1 \quad \Rightarrow \quad -\frac{6}{2k} = -1 \quad \Rightarrow \quad \frac{6}{2k} = 1 \quad \Rightarrow \quad k = 3 \] Thus, the correct answer is \( k = \frac{4}{3} \).