# Math Help - Parallel/Perpendicular Lines

1. ## 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....

2. 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) .

3. 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 $-\frac{2}{3}$

4. 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?

5. 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?

6. Consider the line $\ell: Ax+By+C=0$
Any line parallel to $\ell$ has this form: $Ax+By+D=0$.

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

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

Any line perpendicular to $\ell$ has this form: $-Bx+Ay+E=0$.
Well that just made things even worse .

8. Originally Posted by BeSweeet
Well that just made things even worse .
Why do you say that?
Example: Write lines parallel and perpendicular to $3x-4y+7=0$ through the point $(0,1)$.

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

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

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

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

Now all you have to do is substitute $x=0~\&~y=1$ to find $D~\&~E$.
That's really confusing. Where did C go? Did you subtract it from both sides? ( $3x-4y=-7$).

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?

10. 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.

11. 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.

12. 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: $m_{1} m_{2} = -1$.

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