Question

If one of the lines given by \( 6x^2 - xy + 4cy^2 = 0 \) is \( 3x + 4y = 0 \), then \( c \) equals

• A. \( -3 \)
• B. \( 1 \)
• C. \( 3 \)
• D. \( -1 \)

Hint:

To solve problems like this, substitute the line equation into the given pair of straight lines equation and simplify the resulting expression.

Answer 1

The Correct Option is A

The equation representing the pair of straight lines is: \[ 6x^2 - xy + 4cy^2 = 0. \] Given that one of the lines is \( 3x + 4y = 0 \), we can express \( y \) as \( y = -\frac{3}{4}x \). Substituting this into the given equation: \[ 6x^2 - x\left(-\frac{3}{4}x\right) + 4c\left(-\frac{3}{4}x\right)^2 = 0. \] Simplifying the expression: \[ 6x^2 + \frac{3}{4}x^2 + 4c\left(\frac{9}{16}x^2\right) = 0. \] Dividing by \( x^2 \) (assuming \( x eq 0 \)): \[ 6 + \frac{3}{4} + \frac{36c}{16} = 0. \] To eliminate fractions, multiply the entire equation by 16: \[ 96 + 12 + 36c = 0. \] Combine the constant terms: \[ 108 + 36c = 0. \] Solve for \( c \): \[ 36c = -108 \quad \Rightarrow \quad c = -3. \] Final Answer: \[ \boxed{-3} \]