When we are given an equation of a pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$, the homogeneous part of the equation i.e., $ax^2+2hxy+by^2$ gives the angle between the two straight lines constituting the pair. The other terms influence the point of intersection of the pair. When the other terms are zero or the equation consists of only the homogeneous part, it represents a pair of straight lines having the origin as their point of intersection.

On a graph paper, on varying the non-homogeneous part of the entire equation we actually translate the system without rotating it. The new set of lines remain parallel to the original set.

Is it possible to achieve or obtain the equation of pair of straight lines which are perpendicular to another pair of straight lines, under the condition, the new pair of straight lines passes through the origin (for the sake of simplicity, or you may consider an arbitrary point $(p,q)$)? How are the homogeneous part of the original equation and the new equation related? In this case, how would the behaviour of non-homogeneous terms affect our result?

  • $\begingroup$ Seems like all you need to do is rotate 90 degrees about the intersection point. The effect of this on the quadratic part is fairly simple, but gets a bit ugly for the rest of the equation. $\endgroup$ – amd Oct 20 '19 at 18:23
  • $\begingroup$ @amd, Thanks for your comment. Is it not possible to express the rest of the equation as a simple operation? I can think only of this - just break the joint equation, find slopes of new lines based on existing lines, then find the constant terms by substitution, then again multiply these to get the joint equation. I think this is a very long process. Is there any alternative to this? $\endgroup$ – user14250 Oct 21 '19 at 3:04
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    $\begingroup$ Express the conic equation in matrix form and apply the rotation to that. $\endgroup$ – amd Oct 21 '19 at 8:19

You can transform the quadratic equation directly: just apply a 90° rotation to it. In the simple case that the lines intersect at the origin, the resulting equation is $bx^2-2hxy+ay^2=0$.

In the general case, the intersection point is $(x_0,y_0)=\left({bg-fh\over h^2-ab},{af-gh\over h^2-ab}\right)$, so just translate the above equation: $b(x-x_0)^2-2h(x-x_0)(y-y_0)+a(y-y_0)^2=0$. Expand and rearrange as desired.


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