Question

Give a vector parametric equation for the line through the point (1,−4) that is perpendicular to the line (2t−1,2t−2)

Answer 1

If I remember correctly, we are looking for two parametric equations whose coefficient in front of t is 2 and where x = 1 and y = -4 is a point at some value of t. Wouldn't (2t + 1, 2t - 4) be our answer?

Answer 2

How would you put that into a vector parametric equation?

Answer 3

x = 2t + 1 and y = 2t - 4. <x , y> = <2t+1, 2t-4>

Answer 4

How did you figure out that x= 2t+1 and y=2t-4?

Answer 5

Well, I knew that the coefficient in front of t would be 2, then, I knew that there needed to be a value of t which would yield the point (1,-4). Thus, I let the constant for x equal 1, since 2(0) + 1 = 1 and the constant for y to be 4, since 2(0) + 4 = 4

Answer 6

How did you know that t=2 since it is a line perpendicular to it should the slope be its inverse?

Answer 7

Oh you are correct. I was thinking parallel. Then our parametric equations would be <x, y> = < t/2 + 1, t/2 - 4> right?

Answer 8

That does not seem to be right and it says it is equivalent to the equation <2t+1,2t-4>
so I am not sure what impact the number in front of t makes

Answer 9

It should be the negative inverse, so <x, y> = <-t/2 + 1, -t/2 -4>. The coefficient of t sort of acts like the slope of a non parameterized equation.

Answer 10

I can not figure this one out but thank you so much for all of your help.

Answer 11

Ok, my apologies for the confusion. Good luck!