# Straight line in a plane

1. Jun 19, 2015

### vktsn0303

The title does not say much. But my actual question is as follows.
Let us suppose a line ax+by=0.
This means A.B = 0 with A and B being vectors, where A = [a,b] and B = [x,y].
Therefore, A is perpendicular to B.

Now my question is if A is also perpendicular to line ax+by=0 (I'm not sure if this makes sense).
I read this statement in a book and I'm not able to understand how this came to be.

2. Jun 19, 2015

### Staff: Mentor

The equation ax + by = 0 is being interpreted as the dot product of A and B, as you defined them above.
The line ax + by = 0 can be written as y = (-a/b)x, where b ≠ 0. The slope of this line is m = -a/b. The slope of the vector A = <a, b> is b/a. Since -a/b and b/a are negative reciprocals of each other, the vector A and the line ax + by = 0 are perpendicular.

3. Jun 19, 2015

### RyanH42

You know the dot product rule If $\vec{K}=(x_1,y_1)$ and $\vec{L}=(x_2,y_2)$, the dot product will be $x_1x_2+y_1y_2$.Now Do the same thing A and B.This is the answer why $\vec{A}.\vec{B}=0$.Lets come to your question.
Do you now write line like a vector? Translate line into a vector form then make a dot product of A.If you do that you will also understand why A is perpandicular to line.

4. Jun 19, 2015

### vktsn0303

Thanks Mark44, I had forgotten the negative reciprocal rule of slopes for perpendicular lines. Everything makes sense now.

5. Jun 19, 2015

### vktsn0303

Is it OK to say that A is perpendicular to ax+by=0 because A.B=0? Another statement from the same book conveying this meaning.

6. Jun 19, 2015

### RyanH42

Yes thats true.

7. Jun 19, 2015

### vktsn0303

8. Jun 19, 2015

### vktsn0303

I think the above quoted message is misleading here.
My question would have been as follows:
Is it OK to say that A is perpendicular to ax+by=0 because A is perpendicular to B?

9. Jun 19, 2015

### RyanH42

A is perpendicular to B means A.B=0 you are asking same question again.

10. Jun 19, 2015

### vktsn0303

How is it right to say that A is perpendicular to ax+by=0 just because A is perpendicular to B?

11. Jun 19, 2015

### RyanH42

If A.B=0, (A=(a,b) and B=(x,y)) we get ax+by=0.Now this means ax=-yb and then y=-ax/b this is a line equation If you write it like a vector you will have to write (x,y) so (x,-ax/b) then If you make dot product rule you get ax+b(-ax/b)=0 so A and line is perpandicular.How can we conclude this idea A.B=0 If you didnt know A.B=0 or you didnt conclude thats not zero you cannot tell me that A is perpandicular to K.Let give an example.A=(3,4) and B=(x;y) now we want to sure that 3x+4y=0 is perpandicular to A.then you need to make 3x+4y=0 like a vector.3x=-4y y=-3x/4 then write it like a vector lets call it K vector.K=(x,-3x/4) now we want to know A.K=? Lets make it , 3x+4(-3x/4)=0 we get zero.So it means A.K=0

If I was made 3x+4y=1 which it means A and B are not perpandicular.Lets see what will happen now.Again we want to know A.K=?. A=(3,4) and K=(x,1-3x/4) If you make dot product you can easily see A.K≠0 means they are not perpandicular.So If A and B are not perpandicular then line and A will be not perpandicular
I hope this helps

12. Jun 19, 2015

### vktsn0303

Thank you RyanH42.

13. Jun 19, 2015

### Staff: Mentor

I explained this in post #2.

14. Jun 19, 2015

### vktsn0303

Post #2 helped me understand the negative reciprocal rule for perpendicularity.
Post #11 helped me understand the negative reciprocal rule for perpendicularity being applicable, in context of post #10, only after A.B=0 being valid.

The combination of both posts helped me understand everything.
Thanks to both of you.