Are the two straight lines for two linear simultaneous equations with C1=C2=0 and a1b2-a2b1=0, coincident? Do they pass through the origin?

Answer 1

We need to make some assumptions here about the variables you've listed--somehow, they've got to be inserted into simultaneous equations. Based on the two equations given, it's reasonable to claim we're working with a general system of equations that looks like this:

a₁·x + b₁·y = C₁

a₂·x + b₂·y = C₂

...which is how all of your "2-variables-in-2-equations" systems of equations look (after you replace all the a_i, b_i and C_i guys with plain old numbers).

The first expression given in the question tells us that the C_i's are 0, so now we have

a₁·x + b₁·y = 0

a₂·x + b₂·y = 0.

Part 1: "Do they pass through the origin?"

Whatever else happens, we're already in a position to answer part of the question: Do the lines pass through the origin? Well, "passing through the origin" means the lines contain the point (0,0), which is a handy way to say that the ordered pair (0,0) satisfies the equations. That is, if you set x equal to 0 and y equal to zero (as in the point (0,0)), the equation is true. Let's try it:

a₁·0 + b₁·0 = 0 ⌈ Remember: we haven't declared a₁, a₂, b₁ or b₂ to be any particular

a₂·0 + b₂·0 = 0 ⌊ numbers; we're asking if these equations are true no matter what they are.

No matter what a₁, a₂, b₁ and b₂ are, multiplying them by zero gets us zero, so our equations become

0 + 0 = 0

0 + 0 = 0,

which is true. That means the equations are true when x=0 and y=0, so the origin satisfies the equations. That is, yes, the lines pass through the origin.

Part 2: Are the lines coincident?

There are three ways things can go when you have two expressions representing lines in the plane. Thinking of them as two lines,

1. they can cross at a point
2. they can be two distinct lines parallel to each other, and so never cross
3. they can be coincident, which just means they are the same line,

Since we've just shown that both of the lines contain the origin, we know that they share a point in common--so we're not in the second case (i.e., the equations do not represent a pair of distinct parallel lines). What is left to check? We need to know if the lines have the same slope. Now, since you've made it this far, I'm first going to tell you that the phrasing of the question suggests that your teacher taught you a trick to find this out. The rule is that in any system of equations

a₁·x + b₁·y = C₁

a₂·x + b₂·y = C₂,

the two lines represented by the equations have the same slope if (and only if!) a₁·b₂ – a₂·b₁ = 0. Phrased this way, it's a nifty consequence from Linear Algebra that can be stated in terms of something called the "determinant of the coefficient matrix"--but there's no reason you need to understand that to play this game in Algebra I. So, even though that may sound familiar, I'm going to continue as if you didn't know the rule just yet.

Let's, for the sake of clarity, put the equations in slope-intercept form--that is, let's solve for y. After all, if we want to know if their slopes are the same, this tells us right away! Here we go:

a₁·x + b₁·y = 0 → b₁·y = -a₁·x → y = -(a₁/b₁)·x

a₂·x + b₂·y = 0 → b₂·y = -a₂·x → y = -(a₂/b₂)·x

So the slope of the first line is -a₁/b₁, and the slope of the second line is -a₂/b₂. Let's suppose they are equal, then manipulate the equation a little:

-a₁/b₁ = -a₂/b₂

⇒ divide each side by -1 ⇒

a₁/b₁ = a₂/b₂

⇒ multiply each side by b₁ ... ⇒

a₁ = b₁·a₂/b₂

⇒ ...then by b₂ ... ⇒

a₁·b₂ = b₁·a₂, or a₁·b₂ = a₂·b₁ (just switching the order of b₁ and a₂ there).

So... the slopes being the same is equivalent to having a₁·b₂ = a₂·b₁. And if we subtract a₂·b₁ from each side, we see that this is all equivalent to

a₁·b₂ – a₂·b₁= 0,

which is the equation from the "nifty trick" I mentioned above.

Where were we going with all this? Well, we know that the lines from the question have a point in common (the origin), and we wanted to know if they had the same slope. We just showed that they have the same slope if (and only if) a₁·b₂ – a₂·b₁= 0... . In the statement of the question were told that a₁·b₂ – a₂·b₁= 0, so we now know that the lines have the same slope! It follows that they are, in fact, the same line. That is (in the language of the question), the lines are coincident.