What is the distance between parallel lines whose equations are y=-x+2 and y=-x+8?

1 Answer
Alan P.
Jun 16, 2017

Distance: color(magenta)(6/sqrt(2))6√2 units

Explanation:

{: ("at "x=0,y=-x+2,rarr,y=2), (,y=-x+8,rarr,y=8), ("at "y=2,y=-x+2,rarr,x=0), (,y=-x+8,rarr,x=6) :}

Giving us the points
color(white)("XXX")(x,y) in {(0,2),(0,8),(6,2)}

The vertical distance between the two lines is the vertical distance between (0,2) and (0,8), namely 6 units.

The horizontal distance between the two lines is the horizontal distance between (0,2) and (6,2), namely 6 units (again).

Consider the triangle formed by these 3 points.
The length of the hypotenuse (based on the Pythagorean Theorem) is 6sqrt(2) units.

The area of the triangle using the horizontal vertical sides is "Area"_triangle=1/2xx6xx6=36/2 sq.units.

But we can also get this area using the perpendicular distance from the hypotenuse (let's call this distance d).
Note that d is the (perpendicular) distance between the two lines.
#"Area"_triangle = 1/2 * 6sqrt(2) * d " sq.units

Combining our two equations for the area gives us
color(white)("XXX")36/2=(6sqrt(2)d)/2

color(white)("XXX")rarr d=6/sqrt(2)

enter image source here

Related questions
See all questions in Equations of Parallel Lines
Impact of this question
5752 views around the world
You can reuse this answer
Creative Commons License