#### Answer

$\sqrt{c^2-2cd+d^2+4d^2} \text{ units}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the Distance Formula to find the distance between the given points $\left(
c,c-d
\right)$ and $\left(
d,c+d
\right)$.
$\bf{\text{Solution Details:}}$
With the given points, then $x_1=
c
,$ $x_2=
d
,$ $y_1=
c-d
,$ and $y_2=
c+d
.$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
,$ then
\begin{array}{l}\require{cancel}
d=\sqrt{(c-d)^2+(c-d-(c+d))^2}
\\\\
d=\sqrt{(c-d)^2+(c-d-c-d)^2}
\\\\
d=\sqrt{(c-d)^2+(-2d)^2}
\\\\
d=\sqrt{(c-d)^2+4d^2}
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
d=\sqrt{c^2-2cd+d^2+4d^2}
.\end{array}
Hence, the distance is $
\sqrt{c^2-2cd+d^2+4d^2} \text{ units}
.$