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Pair of Straight lines

suvadip

Member
Feb 21, 2013
69
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 .

Please help
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You should start by looking at the points of intersections of lines ...
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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
Feb 7, 2012
2,703
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
Jan 29, 2012
1,151
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 [tex]a^2x^2- b^2y^2= (ax- by)(ax+by)= 0[/tex] gives the lines ax- by= 0 and ax+ by= 0.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Thank you Halls. I was! :)