Pair of Straight lines

Member
If each of the equations ax^2+2hxy+by^2+2gx+2fy+c=0 and ax^2+2hxy+by^2-2gx-2fy+c=0 represents a pair of straight lines , find the area of the parallelogram enclosed by them .

ZaidAlyafey

Well-known member
MHB Math Helper
You should start by looking at the points of intersections of lines ...

Fantini

MHB Math Helper
You should note that if these are lines, none of the terms $x^2, xy$ and $y^2$ should appear. This should tell you something about their respective coefficients. Opalg

MHB Oldtimer
Staff member
If each of the equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ and $ax^2+2hxy+by^2-2gx-2fy+c=0$ represents a pair of straight lines, find the area of the parallelogram enclosed by them.
If the conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines then it must be of the form $(lx+my+n)(px+qy+r)=0$, where (comparing coefficients) $a = lp$, $h = \frac12(lq+mp)$, $b = mq$, ..., $c = nr.$ The conic $ax^2+2hxy+by^2-2gx-2fy+c=0$ is then given by $(lx+my-n)(px+qy-r)=0.$

The answer to a question of this sort is almost sure to involve the discriminant $\Delta \mathrel{\overset{\text{def}}{=}}\begin{vmatrix}a&h \\h&b \end{vmatrix}$, so it would be worth finding this in terms of $l,m,n,p,q,r$. For a conic representing two straight lines, the discriminant is always negative, and in this case you can check that $$\Delta = \begin{vmatrix}lp & \tfrac12(lq+mp) \\ \tfrac12(lq+mp) & mq \end{vmatrix} = -\tfrac14(lq-mp)^2.$$ The line $lx+my+n = 0$ meets the lines $px+qy \pm r = 0$ at the points $\bigl(\frac{\pm mr-nq}{lq-mp},\frac{\pm lr-np}{lq-mp}\bigr)$, and the distance between those two points is $\left|\dfrac{2r\sqrt{l^2+m^2}}{lq-mp}\right|.$ So that is the length of one side of the parallelogram.

The perpendicular distance between the parallel lines $lx+my\pm n = 0$ is $\left|\dfrac{2n}{\sqrt{l^2+m^2}}\right|.$ So that is the distance between opposite sides of the parallelogram. The area $A$ of the parallelogram is therefore $$\left|\dfrac{2r\sqrt{l^2+m^2}}{lq-mp}\,\dfrac{2n}{\sqrt{l^2+m^2}}\right| = \left|\frac{2nr}{\frac12(lq-mp)}\right|.$$ In terms of the original coefficients, we can write this as $\boxed{A = \dfrac{2|c|}{\sqrt{-\Delta}}}.$

I think that there ought to be a more conceptual proof of this result, using only the original coefficients $a,h,b,g,f,c$, but I do not see one.

HallsofIvy

Well-known member
MHB Math Helper
You should note that if these are lines, none of the terms $x^2, xy$ and $y^2$ should appear. This should tell you something about their respective coefficients. I think you are misunderstanding the question. Each equation represents a pair of lines. For example $$a^2x^2- b^2y^2= (ax- by)(ax+by)= 0$$ gives the lines ax- by= 0 and ax+ by= 0.

Fantini

Thank you Halls. I was! 