<s>[INST] Find the equation of two ... 
Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$.

My Attempt:
Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$.
then,
$$(a_1x+b_1y)(a_2x+b_2y)=0$$
$$(a_1a_2)x^2+(a_1b_2+b_1a_2)xy+(b_1b_2)y^2=0$$
Comparing with $x^2-6xy+8y^2=0$,
$a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$.
I got stuck at here. Please help me to continue this.
 [/INST] Note that $x^2-6xy+8y^2=(x-2y)(x-4y)$, so the lines represented by the equation $x^2-6xy+8y^2=0$ are
$$
x-2y=0 \qquad \text{and} \qquad x - 4y = 0.
$$
The line parallel to $x-2y=0$ and passing through $(2,3)$ is 
$$
(x-2) - 2(y-3) =0,
$$
which in expanded form is $x - 2y + 4 = 0$. Similarly, the other line is $x-4y+10=0$. The single equation is then
$$
(x - 2y + 4)(x - 4y + 10) = 0.
$$
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