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)

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