Equation of pair of straight lines perpendicular to another pair of lines Find the equation of the pair of straight lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3$ 
The solution to the above problem is as such : 
The required equation is $4(x-2)(y-1)$
$ xy-x-2y+2$
However I couldn’t understand how we can arrive at this solution ? Could you please explain how it’s done ? Thanks ! 
 A: Note that the equation of the lines represented by the equation $$4xy + 2x + 6y +3=0$$ are given by $2x +3=0$ and $2y +1=0$.
Therefore you can find the equation of the lines perpendicular to these two lines and passing through $(2,1) $.
A: If you find where the lines intersect, then you can transform the equation by translating, rotating, and then translating again. A pair of intersecting lines is a degenerate hyperbola, and there are several ways to find the center of a hyperbola from its general conic equation, such as the one discussed in this question. Applying that formula produces the point $\left(-\frac32,-\frac12\right)$.  
Now, to transform the equation, first translate this point to the origin by making the substitutions $x\mapsto x-\frac32$, $y\mapsto y-\frac12$.  Simplifying the resulting equation gives $4xy=0$. This looks plausible: if the pair of lines intersect at the origin, their equation shouldn’t have any $x$ or $y$ terms. We can drop the constant factor of $4$ at this point to make things simpler. (It should now be obvious that the original lines parallel the coordinate axes, but we don’t really need to know what the individual lines are.) Now, rotate by making the substitutions $x\mapsto-y$ and $y\mapsto x$. This changes the equation to $-xy=0$, but again, we can drop the factor of $-1$. Finally, translate the origin to the new intersection point via $x\mapsto x-2$, $y\mapsto y-1$ to get $(x-2)(y-1) = 0$, which expands into $xy-x-2y+2=0$.
