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

Jun 16, 2017

Distance: $\textcolor{m a \ge n t a}{\frac{6}{\sqrt{2}}}$ 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
$\textcolor{w h i t e}{\text{XXX}} \left(x , y\right) \in \left\{\begin{matrix}0 & 2 \\ 0 & 8 \\ 6 & 2\end{matrix}\right\}$

The vertical distance between the two lines is the vertical distance between $\left(0 , 2\right) \mathmr{and} \left(0 , 8\right)$, namely $6$ units.

The horizontal distance between the two lines is the horizontal distance between $\left(0 , 2\right) \mathmr{and} \left(6 , 2\right)$, namely $6$ units (again).

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

The area of the triangle using the horizontal vertical sides is ${\text{Area}}_{\triangle} = \frac{1}{2} \times 6 \times 6 = \frac{36}{2}$ 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
$\textcolor{w h i t e}{\text{XXX}} \frac{36}{2} = \frac{6 \sqrt{2} d}{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow d = \frac{6}{\sqrt{2}}$