# Simple Linear Algebra question

1. Jan 11, 2015

### Arnoldjavs3

1. The problem statement, all variables and given/known data

A line is perpendicular to the line 2x - 4y + 7 = 0 and that passes through the point P(7,2). Determine the equation of this line in Cartesian form.

2. Relevant equations

n/a
3. The attempt at a solution
Okay, so how I generally believe how to solve these problems may be wrong but i'm not sure.
1.The normal vector is (2, -4) so by using this info i found the directional vector to be (4,2)
2. After finding the directional vector, i used the dot product to find a vector that is perpendicular to (4,2) so i got (-2,4). I believe that this vector must be parallel to (2,-4) because otherwise it doesn't make sense(atleast to me)
3. After finding the normal vector to (4,2), i used this information to simulate part of the Cartesian Equation.
4. -2x + 4y = -C; What I did from here was use the specific point i was given (7,2) and implemented it in to the x and y values. I got -2(7) + 4(2) = -6
5. My cartesian equation is essentially -2x + 4y -6 = 0.

I'm getting the right answers for some questions but the wrong one for this one, can anyone help out?

2. Jan 11, 2015

### LCKurtz

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Right there, you can use (4,2) as the normal vector to your perpendicular line so try $4x+2y = C$ and make it pass through your point.

3. Jan 11, 2015

### Arnoldjavs3

Okay i see, but it looks like my fundamental understanding of the problem is incorrect, or I simply don't understand the cartesian equation.

Line 1's normal vector is -2,4
Line 2's normal vector is 4,2
Are these normal vectors collinear or perpendicular? I assumed they had to be collinear

4. Jan 11, 2015

### LCKurtz

If the lines are perpendicular their normal vectors are perpendicular. That's why their dot product is zero. Draw a picture.

5. Jan 12, 2015

### Arnoldjavs3

Yes that was what i was wondering; Thanks for your help.
I assumed that they could be either perpendicular, or the normal vectors could be collinear as well.

6. Jan 12, 2015

### LCKurtz

You don't talk about vectors being collinear. You probably mean parallel. Points can be collinear, meaning in a straight line. And, of course, if the normal vectors were parallel, the lines would be parallel, not perpendicular.

7. Jan 12, 2015

### Arnoldjavs3

I know, i was unable to communicate it properly. I know that for any pair of lines, their angles between their normal vectors will be the same. Again, thanks for the help

8. Jan 13, 2015

### HallsofIvy

Staff Emeritus
Another way of looking at this- although it is no longer "Linear Algebra" just "Calculus":
The line 2x- 4y= 7 can be rewritten as y= (1/2)x- 7/4. It has slope 1/2. A line perpendicular to it must have slope -2. So you want to find a line with slope -2, through point (7, 2). Perhaps you can use that to check your answer.