Help with a relatively simple linear algebra proof

[paulrb]

Homework Statement

Show that the lines given by the equation ax + by + c = 0 and bx - ay + d = 0 (where a, b, c, d are in R) are perpendicular by finding a vector in the direction of each line and showing that these vectors are orthogonal. (Hint: Watch out for the cases in which a or b equals zero.)

Homework Equations

Vectors a and b are orthogonal if the dot product of a and b are 0.

The Attempt at a Solution

I have spent a long time trying to figure out this problem but I don't even know how to start. I don't know how to create a vector in the same direction as a line unless I know 2 points on the line or the slope of the line. In this case I don't know either because the equation is so general. That is, unless choosing my own values of a, b, and c is acceptable (Is it?).

I cannot use many techniques to prove this since not much has been introduced in the course. Only basic vector properties, the dot product, and projections have been introduced so far.

 

[Dick]

Choosing a, b and c is not acceptable. Choosing a value of x or y is. Since those equations are supposed to be true for all x and y. Pick x=0 and x=1. Or y=0 and y=1. Now you have two points.

 

[paulrb]

Thank you, I'm not sure why I didn't think of it like that before...

 

[paulrb]

Sorry, I didn't have time to actually do the problem before. Now that I'm looking at it again, I'm still confused. How do I relate the first equation to the second equation?

For example, if I choose x=0 y=0 for the first equation, what do I do with that? I get c = 0, but that doesn't tell me anything about the second equation. That I can do x=1 y=1 and get c = -(a+b) giving me the vector [1,1] for the first equation but again I don't see what that tells me. I can't just plug in he same values of x and y for the second equation, because obviously that will just give me the same vector.

Sorry if that doesn't make sense...I feel like I'm missing something obvious, but I don't know what it is.

 

[Hitman2-2]

Start with the line ax + by + c = 0. To get a vector in the direction of this line, you need two points. I think this is where you are getting confused.

Take the line y = 3x + 2 as an example. How do you get a point on this line? Well x is the independent variable so you can choose any number. Choose x=0 for simplicity. Then you get that y = 2. So a point on this line is (0,2)

Going back to ax + by + c = 0, you cannot pick x=0 and y=0 at the same time. The value of y depends on what you choose for x (or vice versa). So if you choose x=0, find the y that satisfies the equation a*0 + by + c = 0.

 

[paulrb]

Ok...put the points for y won't be numbers, but variables.

if x = 0, y = -(c/b) giving point (0, -c/b)
if x = 1, y = (-c-a)/b giving point (1, -c/b - a/b)

Now I have a vector: [1, -a/b]

using y = 0 and y = 1 for the second equation will give me [a/b, 1]

These are orthogonal! Thank you!
Now all that is left is the hint part of the problem. b cannot equal 0, or it is invalid. Is there anything special I have to add to account for that?

 

[Hitman2-2]

go back to the equations for the lines ax + by + c = 0 and bx - ay + d = 0. If you substitute b=0 into both equations, what are the lines that do get?

 

[paulrb]

I worked on it a bit more and figured everything out. Thanks for your help.