Q.Show that $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ represents a pair of parallel lines.

The triangle formed by $x^2-4y^2=0$ and $x=a$ has area $\dfrac{1}{2}a^2$.

Factor $x^2-4y^2=(x-2y)(x+2y)=0$, so the two lines are $x=2y$ and $x=-2y$, both passing through the origin $(0,0)$.

The line $x=a$ meets $x=2y$ at $y=\dfrac{a}{2}$, giving the point $\left(a,\dfrac{a}{2}\right)$, and meets $x=-2y$ at $y=-\dfrac{a}{2}$, giving $\left(a,-\dfrac{a}{2}\right)$.

So the triangle has vertices $(0,0)$, $\left(a,\dfrac{a}{2}\right)$, $\left(a,-\dfrac{a}{2}\right)$. Its base is the vertical segment on $x=a$ of length $\dfrac{a}{2}-\left(-\dfrac{a}{2}\right)=a$, and the height (distance from the origin to the line $x=a$) is $a$.

Area $=\dfrac12\times\text{base}\times\text{height}=\dfrac12\times a\times a=\dfrac12 a^2$.

Writing $6x^2-xy+4cy^2$ as the product $(3x+4y)(ax+by)$ and matching coefficients gives $c=-3$.

Since $3x+4y=0$ is one of the two lines represented by $6x^2-xy+4cy^2=0$, we can write:

$6x^2-xy+4cy^2 = (3x+4y)(ax+by)$

for some constants $a,b$ (up to an overall scalar, but let's match directly). Expanding the right side:

$(3x+4y)(ax+by) = 3a\,x^2 + (3b+4a)xy + 4b\,y^2$

Matching the $x^2$ coefficient: $3a=6 \implies a=2$.

Matching the $y^2$ coefficient: $4b=4c \implies b=c$.

Matching the $xy$ coefficient: $3b+4a=-1 \implies 3c+8=-1 \implies 3c=-9 \implies c=-3$.