What is the area of the parallelogram enclosed by a pair of straight lines?

In summary, the equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ and $ax^2+2hxy+by^2-2gx-2fy+c=0$ represent a pair of straight lines. The area of the parallelogram enclosed by these lines can be found by using the discriminant of the conic and the distance between the lines. This can be written as $A = \dfrac{2|c|}{\sqrt{-\Delta}}$, where $\Delta$ is the discriminant. There may be a more conceptual proof of this result using the original coefficients, but it is not yet known.
  • #1
Suvadip
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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
 
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  • #2
You should start by looking at the points of intersections of lines ...
 
  • #3
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. :)
 
  • #4
suvadip said:
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.
 
  • #5
Fantini said:
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.
 
  • #6
Thank you Halls. I was! :)
 

FAQ: What is the area of the parallelogram enclosed by a pair of straight lines?

What is the equation of a pair of straight lines?

The general equation of a pair of straight lines can be written as ax² + 2hxy + by² + 2gx + 2fy + c = 0, where a, b, and h are not all zero.

How many solutions can a pair of straight lines have?

A pair of straight lines can have a maximum of two solutions.

What is the condition for a pair of straight lines to be perpendicular?

The condition for a pair of straight lines to be perpendicular is that the product of their slopes must be -1.

Can a pair of straight lines intersect at more than one point?

No, a pair of straight lines can only intersect at one point. If they intersect at more than one point, then they are the same line.

How can we determine the type of intersection between two pairs of straight lines?

The type of intersection between two pairs of straight lines can be determined by calculating the discriminant of their general equation. If the discriminant is positive, the lines intersect at two distinct points. If it is zero, they intersect at one point. And if it is negative, they do not intersect.

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