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))# 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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