# Parallel/Perpendicular Lines

• Sep 21st 2009, 07:20 AM
BeSweeet
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....
• Sep 21st 2009, 07:24 AM
Quote:

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) .
• Sep 21st 2009, 07:25 AM
red_dog
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 $\displaystyle -\frac{2}{3}$
• Sep 21st 2009, 07:32 AM
BeSweeet
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?
• Sep 21st 2009, 02:26 PM
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?
• Sep 21st 2009, 02:35 PM
Plato
Consider the line $\displaystyle \ell: Ax+By+C=0$
Any line parallel to $\displaystyle \ell$ has this form: $\displaystyle Ax+By+D=0$.

Any line perpendicular to $\displaystyle \ell$ has this form: $\displaystyle -Bx+Ay+E=0$.
• Sep 21st 2009, 02:40 PM
BeSweeet
Quote:

Originally Posted by Plato
Consider the line $\displaystyle \ell: Ax+By+C=0$
Any line parallel to $\displaystyle \ell$ has this form: $\displaystyle Ax+By+D=0$.

Any line perpendicular to $\displaystyle \ell$ has this form: $\displaystyle -Bx+Ay+E=0$.

Well that just made things even worse (Doh).
• Sep 21st 2009, 02:56 PM
Plato
Quote:

Originally Posted by BeSweeet
Well that just made things even worse (Doh).

Why do you say that?
Example: Write lines parallel and perpendicular to $\displaystyle 3x-4y+7=0$ through the point $\displaystyle (0,1)$.

Parallel: $\displaystyle 3x-4y+D=0$
Perpendicular: $\displaystyle 4x+3y+E=0$

Now all you have to do is substitute $\displaystyle x=0~\&~y=1$ to find $\displaystyle D~\&~E$.
• Sep 21st 2009, 03:00 PM
BeSweeet
Quote:

Originally Posted by Plato
Why do you say that?
Example: Write lines parallel and perpendicular to $\displaystyle 3x-4y+7=0$ through the point $\displaystyle (0,1)$.

Parallel: $\displaystyle 3x-4y+D=0$
Perpendicular: $\displaystyle 4x+3y+E=0$

Now all you have to do is substitute $\displaystyle x=0~\&~y=1$ to find $\displaystyle D~\&~E$.

That's really confusing. Where did C go? Did you subtract it from both sides? ($\displaystyle 3x-4y=-7$).

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?
• Sep 21st 2009, 03:14 PM
Plato
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.
• Sep 21st 2009, 03:20 PM
BeSweeet
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.
• Sep 21st 2009, 03:42 PM
BeSweeet
• Sep 21st 2009, 03:51 PM
mr fantastic
Quote:

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: $\displaystyle m_{1} m_{2} = -1$.

So $\displaystyle \frac{3}{2} \, m_{2} = -1 \Rightarrow m_2 = - \frac{2}{3}$.