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.
On
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 $$ (x - 2y + 4)(x - 4y + 10) = 0. $$
We may set $a_1=a_2=1$ to have $$b_1+b_2=-6,\quad b_1b_2=8$$ giving $\{b_1,b_2\}=\{-2,-4\}$.
So, we get $$x^2-6xy+8y^2=(x-2y)(x-4y)$$ Hence, the equation we want is $$((x-2)-2(y-3))((x-2)-4(y-3))=0,$$ i.e. $$(x-2y+4)(x-4y+10)=0$$