Question

Show that the distance between the parallel lines y=mx+u and y=mx+v is

Answer 1

\[\frac{ \left| u-v \right| }{ \sqrt{1+m ^{2}} }\]

Answer 2

@ganeshie8

Answer 3

ok

Answer 4

now how do u prove that it equals that equation I inserted in the beginning

Answer 5

You want to use the distance formula, I am guessing

Answer 6

yes

Answer 7

\[
d^2 = (x_2-x_1)^2+((mx_2+u) - (mx_2+v))^2
\]

Answer 8

can you show me how you got that equation?

Answer 9

\[
d^2 = (x_2-x_1)^2+(y_2-y_1)^2=(x_2-x_1)^2+((mx_2+u) - (mx_1+v))^2
\]

Answer 10

does it have to be d squared?

Answer 11

No, I just didn't want to add a square root just yet

Answer 12

What class is this anyway?

Answer 13

math analysis honors

Answer 14

ok now what?

Answer 15

I guess for now you want to expand things a bit.

Answer 16

I'm curious if you know how to minimize a function.

Answer 17

no im confused for what to do next

Answer 18

You can draw a perpendicular to the two lines through the origin, which has slope negative reciprocal of m .

Answer 19

\[
d^2 = (x_2-x_1)^2+(m(x_2-x_1)+u -v)^2
\]

Answer 20

then get the coordinates of the intersection points of this perpendicular and the two lines, and then use distance formula

Answer 21

ok I get how you got that formula but im still lost on how to continue it

Answer 22

So find the intersection points for the following two system of equations
y = -x/m
y = mx+u

and

y = -x/m
y = mx + v

Answer 23

so for the first system we have
-x/m = mx + u
-x = m^2 x + um
-um = m^2 x + x
-um = x( m^2 + 1)
-um/(m^2+1) = x

Answer 24

and y = -x/m, so plugging in gives , y = u/(m^2 + 1)
so first intersection point is
(-um/(m^2+1) , u/(m^2 + 1))

I will let you find the other intersection point.

Then use distance formula on the two points

Answer 25

@wio so whats after that

Answer 26

Well, I am guessing you don't know calculus, so you probably want to use a geometric method.