Question

Find the distance between the pair of parallel lines, y = -2x+1 & y = -2x+16.
Choose one answer.
a. square root of 45, or appx 6.7
b. 15
c. square root of 54, or appx 7.3
d. square root of 50, or appx 7.1

Answer 1

Step #1: determine an equation of a perpendicular line to these 2 lines.

Step #2: solve 2 linear systems. The first is between the new line and the first of the parallel lines. The second is between the new line and the second of the parallel lines.

Step #3: use distance formula on the two points determined from step #2.

Answer 2

That didn't help much. Idk how to do this

Answer 3

To start off with step #1, you'll know the slope of the perpendicular since it is the negative reciprocal of the given lines.

Answer 4

lol can u just give me the answer?

Answer 5

We have to show you HOW to get the answer at this website and guide you through the process. That process is done with your efforts. This website isn't an "ask questions, get answers with no explanations and learn nothing" website. You won't learn anything and you'll just get stuck always asking questions for every problem you have.

Plus, students and tutors get in trouble for just giving answers. Just giving answers will not help anyway.

Answer 6

Show how u get the answer thats how I learn

Answer 7

that'll be fine

Answer 8

Idk how to do this

Answer 9

np. So, for the perpendicular line slope, it will be the negative reciprocal of -2 which is +1/2 or 1/2 because (-2)(1/2) = -1 and we got -2 from the original equations that are in the form y= mx + b.

For the equation of the perpendicular line, we can use the origin, which is point (0, 0), for convenience.

The point-slope form of the equation is:\[y - y _{1} = m(x - x _{1})\]where (x1, y1) is (0, 0) and "m" = slope =1/2, so we get
y - 0 = (1/2)(x - 0) which is y = x/2 end of step #1

Answer 10

Ok I understaNd that part

Answer 11

Step #2:
part a) solve the linear system: y = x/2 and y = -2x + 1.
substituting: x/2 = -2x + 1, so x = -4x + 2 x = 2/5 so y = 1/5

part b) solve the linear system: y = x/2 and y = -2x + 16.
susbstituting: x/2 = -2x + 16, so x = -4x + 32 x = 32/5 so y = 16/5

end of step #2

Answer 12

Ok

Answer 13

step #3:

distance formula: d = sqrt[(x1 - x2)^2 + (y1 - y2)^2]

d = sqrt[(32/5 - 2/5)^2 + (16/5 - 1/5)^2]

at this point, it's just simplification. Still need help?

Answer 14

Yeah. Im not good at math at all

Answer 15

np. d = sqrt[(30/5)^2 + (15/5)^2] = sqrt[(6)^2 + (3)^2]
=sqrt(36 + 9) = sqrt(45)

and that's it! Yes, it's a lot of steps, but they are all logical and if you break it down this way, you'll be ok. The hardest part is conceptualizing the perpendicular line as the shortest distance between the two lines.

Answer 16

Thank you. I actually understand it. Ur better than my geometry teacher

Answer 17

math can be tough or you can be tough on math. I try to explain because students eventuright away like that better. Gives them independence.

Answer 18

And you're quite welcome.