the equation of pair of lines

Question

We know that when two lines are given, and we multiply them then we get the equation of pair of lines. My question is what is that new curve, does it have some geometrical significance? For eg. $xy=0$ represents the coordinate axis, but what does $xy=3$ represents or other such pair of lines?

Actually I was solving a question of straight line when I came across this question, the question said, "Consider a parallelogram whose sides are represented by the lines $2x + 3y = 0$; $2x + 3y - 5 = 0$; $3x - 4y = 0$ and $3x - 4y = 3$. The equation of the diagonal not passing through the origin, is", and the solution was,

$(2x + 3y)(3x - 4y) = (3x - 4y - 3)(2x + 3y - 5)$ "Diagonal 1" $(2x + 3y)(3x - 4y - 3) = (3x - 4y)(2x + 3y - 5)$ "Diagonal 2"

And so I was confused here how come we multiply two lines and equate them to find a line? And so I thought that they might represent some curve.

Answer 1

I'm going to come at this problem in a different way—a graphical way, which might give you some intuition about why this happens.

Here is a three-dimensional space, where we'll plot some functions, with $x$-axis, $y$-axis, and $z$-axis in red, green, and blue, respectively. (All images from Geogebra.)

The first function we'll plot is $2x+3y$. It looks like this:

(Actually, I've scaled everything by a factor of $1/10$, so as to get everything to fit nicely, but that won't affect the following discussion.) Now one thing you might notice is that it intersects the $x$-$y$ plane in a line, which is of course the line $2x+3y = 0$, because the $x$-$y$ plane is exactly that set of points where $z = 0$. By way of comparison, here is the plot of the function $2x+3y-5$:

Notice that this line is parallel to the first line, because it is of course the equation of the line $2x+3y = 5$. Naturally, you'll want to see $3x-4y$:

And $3x-4y-3$:

And these are, you'll have realized by now, the four lines describing your parallelogram; each of them is the intersection of the $x$-$y$ plane with another plane—a polynomial in $x$ and $y$ of degree one.

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Now, let's plot another function—the function $(2x+3y)(3x-4y)$:

It's a "saddle" function, a polynomial in $x$ and $y$ of degree two, but what I want to call attention to is its intersection with the $x$-$y$ plane, which is the pair of lines $2x+3y = 0$ and $3x-4y = 0$. This makes perfect sense, because our new function equals zero whenever either of its factors equals zero, and each of its factors corresponds to one of those lines.

You'll be unsurprised to hear that our next function is $(2x+3y-5)(3x-4y-3)$:

It looks exactly the same, except that it's been translated so that its vertex is at the point $(29/17, 9/17)$.

Here's what the two functions look like together:

You'll see that there's an arc where the two functions intersect. To see the significance of this, let's get a different perspective on this pair of functions, this time from directly overhead:

By golly, that arc from directly overhead looks exactly like one of the diagonals of the parallelogram! Can we figure out why that might be?

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To approach that, let's impose some structure on the functions we're looking at. Let $P(x, y) = 2x+3y$ and $Q(x, y) = 3x-4y$, and then the other functions are $P(x, y)-5$ and $Q(x, y)-3$. Then our two saddle functions are $P(x, y)Q(x, y)$ and $[P(x, y)-5][Q(x, y)-3]$. (To simplify terminology, let's just write $P$ and $Q$ from now on, remembering that each is a function of $x$ and $y$.)

That arc where the two saddle functions intersect (strictly speaking, the projection of that arc down onto the $x$-$y$ plane) is then just the locus of points $(x, y)$ where

$$ PQ = (P-5)(Q-3) $$

Or, equivalently, it's where

$$ R(x, y) = PQ-(P-5)(Q-3) = 0 $$

Let's expand out $R$ to get

$$ R(x, y) = PQ-(PQ-5Q-3P+15) = -5Q-3P+15 $$

One thing to observe about this is that if $P$ and $Q$ are both first-degree polynomials in $x$ and $y$, then so is $R$, which means that it graphs as a plane as well. It must therefore intersect the $x$-$y$ plane in a straight line, just as $P$ and $Q$ do.

Now what is that line? Well, since the line is characterized by the equation

$$ PQ = (P-5)(Q-3) $$

one point is given by the intersection of $Q = 0$ and $P-5 = 0$ (which leads to both sides equal to zero), and a second point is given by the intersection of $P = 0$ and $Q-3 = 0$ (ditto). But those are just two opposite vertices of the parallelogram! Therefore $R = 0$ must describe a diagonal of the parallelogram that connects those two points. And similarly,

$$ S(x, y) = P(Q-3)-(P-5)Q $$

describes another plane, whose intersection with the $x$-$y$ plane gives the other diagonal.

Here's a plot of $R$:

and a plot of $S$:

As Paul Sinclair's answer points out, this works only in this particular case; it's connected to the fact that that the expression for $R$ (and for $S$) loses its second-degree term(s). Of course, there's nothing special about $3$ and $5$ here.

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In the parlance of algebraic geometry, the set of polynomials that contain $P$ as a factor is an example of an ideal. So is the set of polynomials that contain $Q$ as a factor, or the set of polynomials that contain both $P$ and $Q$ as factors. Each of these corresponds to a set of points in the $x$-$y$ plane where those polynomials vanish (that is, are equal to zero), which are called algebraic sets (or sometimes algebraic varieties). Loosely speaking, if you multiply two ideals together, you union their corresponding algebraic sets. (I'm being a little sloppy there, but hopefully you get the idea.)

So to answer your original question, there is some significance of multiplying these functions, but only when you equate them to zero. That's why $xy = 3$ doesn't connect to either $x = 3$ or $y = 3$; the function that vanishes would be $xy-3$, but that function is irreducible; it doesn't factor into linear terms.

Answer 2

Think of a line $\ell$ as the expression $\ell_x x+\ell_y y+\ell_z$, so that the equation of the line is $\ell=0$. Then multiplication of two lines $\ell,m$ gives the equation $\ell m=0$. This is a second degree equation, and thus a conic. It's the equation of two lines, but you can think of it as the limiting case of a hyperbola, as the "bends" get sharper and sharper.

In the theory of conics, a pair of lines is called a degenerate conic.

You can also add equations. Weighted sums of lines $\ell+\lambda m=0$ define lines passing through the intersection of $\ell$ and $m$.

If $p,q$ are also lines, then weighted sums $\ell m+\lambda p q=0$ of degenerate conics $\ell m,p q$ will give you a family of non-degenerate conics that pass through the four points where $\ell m$ intersects with $p q$.

Answer 3

Let me point out something about your parallelogram solution:

• $(2x + 3y)(3x - 4y) = 6x^2 + xy - 12y^2$
• $(3x - 4y - 3)(2x + 3y - 5) = 6x^2 + xy - 12y^2 - 21x + 11y + 15$

Thus if we set them equal to each other, we can eliminate all the quadratic terms:

$$\begin{align} 6x^2 + xy - 12y^2 &= 6x^2 + xy - 12y^2 - 21x + 11y + 15\\0 &= -21x + 11y + 15\end{align}$$ leaving only a linear equation to represent the diagonal. Similarly for the Diagonal 2 equation:

• $(2x + 3y)(3x - 4y - 3) = 6x^2 + xy - 12y^2 - 6x - 9y$
• $(3x - 4y)(2x + 3y - 5) = 6x^2 + xy - 12y^2 - 15x +20y$

$$\begin{align}6x^2 + xy - 12y^2 - 6x - 9y &= 6x^2 + xy - 12y^2 - 15x +20y\\-6x - 9y &= -15x +20y\\9x-29y &= 0\end{align}$$

The fact that these cross-products reduce to lines when equated is just an algebraic artifact of both products involving lines with the same variable parts, but different constant terms. That the resulting linear equations happen to be those of the diagonals is apparently some special property of the diagonals of parallelograms that someone has discovered. There is probably some principle that it follows from, but not one that is immediately apparent to me. But regardless, it is a technique that applies strictly to the problem of finding the diagonals of a parallelogram.

It is not some general property of lines like you are assuming. It only works in this exact situation. In general an equation that has terms where the sum of the powers of the variables is $ > 1$, and such terms cannot be removed by simplifying, will NOT be the equation of a line. When the highest sum of powers of variable exponents (called the "degree" of the term) is $2$, we have a conic section: the intersection of a circular double-cone and a plane. In the special case when the plane passes through the vertex of the cone, the intersection will either be the vertex by itself, a single line tangent to the cone, or two lines. But all other planes intersect the cone in non-linear curves. $x^2, xy, y^2$ are all terms of degree $2$. The equation $xy=0$ corresponds to the special case of a plane passing through the vertex. This is why it gives two lines. The equation $xy=3$ does not correspond to such a case. Its graph is a hyperbola.

Answer 4

As pointed by Sachchidanand Prasad, the equation $xy=3$ does not represent a pair of lines.

This said, a pair of lines is... two lines, and this "curve" has exactly the signifiance of... two lines. (Well, you may see it as a degenerate hyperbola, such that the apexes come in contact while the branches straighten up.)