Simple Linear Algebra question

[Arnoldjavs3]

Homework Statement

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.

Homework 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 into 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?

 

[LCKurtz]

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Homework Statement

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.

Homework 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)

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.

 

[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
Because ((4)(2) + (-4)(2)) = 0; I'm confused about this.

 

[LCKurtz]

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
Because ((4)(2) + (-4)(2)) = 0; I'm confused about this.

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

 

[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.

 

[LCKurtz]

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.

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.

 

[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

 

[HallsofIvy]

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.