Question

y+2=1.52(x-1), give a vector parallel to this line, and a vector perpendicular to this line

Answer 1

so you have the equation of a line:
y+2=1.52(x-1) we see that the slope of this line is 1.52 You may rewrite the equation of the line as follows:
y=1.52x-1.52-2
y=1.52x-3.52
Agree?
So if we want a vector that is parallel to this line, it must have the same slope.
It has to have slope 1.52
The way we do this very easily without confusion let's do it in 2 steps:
STEP 1) find the equation of the line that is parallel to y+2=1.52(x-1)
Answer: y=1.52x why? because the slope is the same as your original equation so they are parallel
Step 2) Make a vector from the parallel line you found (by the way they are many you could have y=1.52x + "any number" but i kept it simple and made that "any number" 0)
How? Well, we simply cut off a segment from the line and we get a vector
How about the vector that goes from (0,0) to (1, 1.52)
So your vector is v=<1, 1.52>

Do you understand? Before I explain a vector that is perpendicular.

Answer 2

thank you. but how can i use the line equation to illustrate? i know the vector line equation is l(t)= point a+t(point b - point a). i found two point which are point a(1,-2) and point b is (0,-3.52) would you mind help me on this point?

Answer 3

When you have two points, in order to create a vector where one point is the head and the other is the tail is as follows:
Point A=(a1, a2) and Point B=(b1,b2)
Vector AB=<b1-a1,b2-a2>
Notice we use soft brackets parenthesis for points and sharp brackets < > for vectors.

Answer 4

A is the initial point B is the final point

Answer 5

how can we use that line(t) which i think it's vector to show the parallel line?

Answer 6

I'm trying to draw it so that you can easily see it :-)

Answer 7

ok!