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

Question

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

Accepted Answer

Since $3x+4y=0 \implies y=-\dfrac34x$ is one factor, substitute into $6x^2-xy+4cy^2=0$:
$$6x^2-x\left(-\frac34x\right)+4c\left(\frac34x\right)^2=0 \implies 6x^2+\frac34x^2+4c\cdot\frac9{16}x^2=0$$
$$\left(6+\frac34+\frac{9c}4\right)x^2=0 \implies 6+\frac34+\frac{9c}4=0 \implies \frac{9c}4=-\frac{27}4 \implies c=-3$$

Solutions · 1