_{1}=mx+b

_{1}and y

_{2}=mx+b

_{2}is given by

_{2}-b

_{1}|/sqrt(m

^{2}+1).

_{1}=4, and b

_{2}=-6, you get

^{2}+1) = 10/√10 = √10.

Geometry question help please.

Tutors, sign in to answer this question.

The distance between two parallel lines y_{1}=mx+b_{1} and y_{2}=mx+b_{2} is given by

d= |b_{2}-b_{1}|/sqrt(m^{2}+1).

With m=3, b_{1}=4, and b_{2}=-6, you get

d= |-6-4|/sqrt(3^{2}+1) = 10/√10 = √10.

The line perpendicular to both of this lines has a slope of -1/3 and the length of a segment of it, which is between two those lines, gives the distance between them. Let us consider the line y=-x/3. It is perpendicular to both lines. It intercepts y=3x+4 at the point, which x-coordinate satisfies the following equation:

-x/3=3x+4; or x=-6/5=-1.2; y-coordinate is y=-1.2/-3=0.4

Analogously, it intercepts the line 3x-6 at x=1.8

-x/3=3x-6 x=9/5=1.8

y-coordinate is 1.8/-3=-0.6

Distance between two lines, 3x+4 and 3x-6, is:

d=√[(1.8-(-1.2))^{2}+(-0.6-0.4)^{2}]=√(3^{2}+(-1)^{2})=√10

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments