Ana.stasia
Junior Member
- Joined
- Sep 28, 2020
- Messages
- 118
Looks complicated. Do you have to use that formula?The question is
Determine the equation of the line that is parallel to the lines 3x-2y-1 = 0 and 3x-2y-13 = 0 and is equally distant from both lines.
I have written the formula we are supposed to use along the way.
I also know everything so far I did is correct. But what do I do now, with only p???
View attachment 25055
Looks complicated. Do you have to use that formula?
I don't understand how you got the expressions for p1 and p2.
If you are free to use any method, here's what I would do:
1. Convert the 2 equations into slope/intercept form.
2. The slopes of parallel lines are the same - this gives you the slope
3. The y-intercept you need is midway between the other 2 intercepts (you may need to prove it, not too difficult).
I hate formulas. The formula that you are supposed to use is for what purpose? Where did you get the angle you are using? What is the definition of p?
In equation 1, if y = 0, x = 1/3, and the angle is 0. If x = 0, y = -1/2 and the angle is - pi/2 radians.
[MATH]\dfrac{1}{3} * cos(0) + 0 * sin(0) - p = 0 \implies \dfrac{1}{3} - p = 0 \implies p = \dfrac{1}{3}.[/MATH]
And (1/3, 0) is clearly distant from the origin by 1/3
[MATH]0 * cos (- \pi/2) - \dfrac{1}{2} * sin(- \pi /2) - p = 0 \implies - \dfrac{1}{2} * (-1) - p \implies p = 1/2.[/MATH]
And (0, -1/2) is clearly distant from the origin by 1/2.
Where did these square roots of 13 even come from? And why?
Can you do the problem now?
The problem is almost trivial, with no need to use a formula! They really force you to use specific formulas for it??Determine the equation of the line that is parallel to the lines 3x-2y-1 = 0 and 3x-2y-13 = 0 and is equally distant from both lines.