Thread: Parallel/Perpendicular Lines

Determine the equation of the line that passes through the point (2,1) and is:
a. Parallel to the line 3x-2y=4
b. Perpendicular to the line 3x-2y=4

First, I rearranged 3x-2y=4 to be in slope-intercept form:
y=3/2x-2

For part b, I took the slope and put it into point-slope form:
y-1=3/2(x-2)
and got the same thing as slope-intercept form.

I think I'm doing something wrong....

Originally Posted by BeSweeet

Determine the equation of the line that passes through the point (2,1) and is:
a. Parallel to the line 3x-2y=4
b. Perpendicular to the line 3x-2y=4

First, I rearranged 3x-2y=4 to be in slope-intercept form:
y=3/2x-2

For part b, I took the slope and put it into point-slope form:
y-1=3/2(x-2)
and got the same thing as slope-intercept form.

I think I'm doing something wrong....

HI

(a) m=3/2 , since it is parallel , then substitute into the equation
y-y1=m(x-x1)

(b) m=-2/3 because it is perpendicular (m1m2=-1) , then substitute into the same equation as (a) .

The point (2,1) is on the line, so the question a) doesn't make sense.

For b), the slope of the perpendicular line is

I have to determine the equation of the line. Would the equation just be in slope-intercept form once I do the thing for point-slope?

Ok, I think I got it!

For parallel lines, just find the slope, and stick that into point-slope form.

For perpendicular lines, find the slope, and find what other number it will multiply by to equal -1, then stick that into point-slope form.

I just have one [stupid] question: How about did you getting -2/3?

Consider the line
Any line parallel to has this form: .

Any line perpendicular to has this form: .

Originally Posted by Plato

Consider the line
Any line parallel to has this form: .

Any line perpendicular to has this form: .

Well that just made things even worse .

Originally Posted by BeSweeet

Well that just made things even worse .

Why do you say that?
Example: Write lines parallel and perpendicular to through the point .

Answer:
Parallel:
Perpendicular:

Now all you have to do is substitute to find .

Originally Posted by Plato

Why do you say that?
Example: Write lines parallel and perpendicular to through the point .

Answer:
Parallel:
Perpendicular:

Now all you have to do is substitute to find .

That's really confusing. Where did C go? Did you subtract it from both sides? ( ).

Isn't it easier (for parallel) to just find the slope, and stick it into point-slope form to get my equation, and for perpendicular, find the slope that is equal to -1 (when multiplied by the other number), stick it into point-slope form, and get my equation?

In your case, websites such as this seem to do more harm than good.
You need a sit-down, one-to-one, tutorial with a live instructor.
You have some fundamental misunderstandings.
Please plan to do that.

For most of my other questions, they were answered and were eventually figured out and I understood them.

I understand this problem by what I've said here, but not by what you said.

http://www.mathhelpforum.com/math-he...136-post5.html

Originally Posted by BeSweeet

Ok, I think I got it!

For parallel lines, just find the slope, and stick that into point-slope form.

For perpendicular lines, find the slope, and find what other number it will multiply by to equal -1, then stick that into point-slope form.

I just have one [stupid] question: How about did you getting -2/3?

The gradient of the line 3x-2y=4 is 3/2.

And you should know that if two lines are perpendicular, then the product of their gradients is equal to -1: .

So .