Question

If one of the lines given by $6x^2-xy+4c{y}^2=0$ is $3x+4y=0$, then $c$ equals

• A. $-3$
• B. $-1$
• C. $3$
• D. $1$

Answer 1

The correct option is A

$-3$

Explanation for the correct option

Step 1: information required for the solution

Since one of the equations of the line is $3x+4y=0$. Let the other line's equation be $y=mx+a$.

Then the product of the equation of the first line and third line will be $0$.

Step 2: Calculation for the value of $\mathit{c}$.

The product of the first and third line,

$\begin{array}{rcl}\therefore \left(3x+4y\right)\left(y-mx+a\right)& =& 0\\ & \Rightarrow & 3xy-3m{x}^2+3xa+4y^2-4mxy+4ay=0\\ & \Rightarrow & -3m{x}^2+4y^2+\left(3-4m\right)xy+\left(3x+4y\right)a=0\end{array}$

Now, suppose that $c=0$ then the above equation will be,

$-3m{x}^2+\left(3-4m\right)xy+4y^2=0$

Compare this equation with $6x^2-xy+4c{y}^2=0$

To find the value of $m$ we have,

$\begin{array}{rcl}\therefore \frac{3-4m}{-1}& =& \frac{-3m}{6}\\ & \Rightarrow & 18-24m=3m\\ & \Rightarrow & 18=27m\\ & \Rightarrow & m=\frac{2}{3}\end{array}$

Also, $\frac{4}{4c}=\begin{array}{r}\frac{3-4m}{-1}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{r}\frac{-3m}{6}\end{array}$

From this we have,

$\begin{array}{rcl}\therefore \frac{1}{c}& =& \frac{-3}{6}\left(\frac{2}{3}\right)\\ & \Rightarrow & \frac{1}{c}=\frac{-2}{6}\\ & \Rightarrow & c=-3\end{array}$

Hence, the correct option is (A).