If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0$ then c is equal to?

Question

If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0,$ then $c$ is equal to?

I have tried solving it by putting $x = \frac{-4y}{3}$ inside the given equation, as $3x + 4y$ is a solution of $6x^2 – xy + 4cy^2 = 0$, but I am getting two variables, $c$ and $y$. Please tell me how to solve it.

Answer 1

Assume that equation of other lines $x-my = 0$, so the given equation of pair of straight lines can be written as $$2(x-my)(3x + 4y) = 0$$ Now,compare it with original equation you would get a value of $m = 3/2$ and consequently of $c$ which is $-3$

Answer 2

$(3x+4y)(ax+by)=3ax^2+(4a+3b)xy+4by^2=6x^2 – xy + 4cy^2 = 0\implies$

$3a=6, 4a+3b=-1, 4b=4c\implies $

$a=2, b=-3, c=-3.$

Answer 3

Your idea is good; plugging in $x=-\tfrac43y$ yields $$0=\frac{96}{9}y^2+\frac43y^2+4cy^2=\frac49\left(36+12c\right)y^2,$$ which must hold for all $y$. Then $36+12c=0$ so $c=-3$.

Alternatively, you could note that the point $(x,y)=(4,-3)$ is on the line, and hence on the curve, so $$6\cdot4^2-4\cdot(-3)+4\cdot c\cdot(-3)^2=0.$$ This shows that $24+3+9c=0$ so $c=-3$.