Question

Hi, I need some help with these, I am really tired b/c of my surgery, an explanation would be enough, thank you ever so much.


1) Given the following line and point not on the line, find the distance between them.

y = 7x + 5

(-8,-1)

2) Find the distance between the lines.

y=32x+32

y=32x−5

3) Find the distance between the lines.

y=3x+10

y=3x−20

4) Given the following line and point not on the line, find the distance between them.

y=−23x+2

(-2,-1)

Answer 1

I will not do them all for you, they are all virtually the same.
1) Given the following line and point not on the line, find the distance between them.

y = 7x + 5

(-8,-1)
==============================
find a line perpendicular to that one through that point
slope = m' = -1/m = -1/7
y = -x/7 + b
-1 = 8/7 + b
b = -15/7
so the line we want is y = -x/7 - 15/7
or 7 y = -x -15
now where does that hit the first line?
7 y = - 1 x - 15
y = 7 x + 5 multiply this by 7

7 y = - x - 15
7 y = 49 x + 35
-----------------------subtract
0 = -50 x - 50
x = -1
then y = 7x+5 = -2 so that point is at (-1 , -2)
NOW
I suspect you can find the distance between
(-1,-2) and (-8,-1)
sqrt (7^2+1^1) = sqrt 50 = 5 sqrt 2

Answer 2

by the way lines in #2 and #3 are parallel, so pick any old point on one of them

Answer 3

thank you so
much

Answer 4

You are welcome.