The interpretation that you mean a perpendicular distance of 20 fits with there being only two possible locations for the translated line segment. The vector $\langle x_2-x_1,y_2-y_1\rangle$ is along the given line; the vector $\langle y_2-y_1,-x_2+x_1\rangle$ is orthogonal (perpendicular) to the given line (in one particular direction—$\langle -y_2+y_1,x_2-x_1\rangle$ would be in the other direction); and the vector $\begin{align} \frac{20\langle y_2-y_1,-x_2+x_1\rangle}{||\langle y_2-y_1,-x_2+x_1\rangle||} &=\frac{20}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}}\langle y_2-y_1,-x_2+x_1\rangle\\ &=\left\langle\frac{20(y_2-y_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}},\frac{20(-x_2+x_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}}\right\rangle \end{align}$ is orthogonal to the given line and has length 20. Translate your two given point by this vector and you'll have two points on the line you want: $(x_1,y_1)\to\left(x_1+\frac{20(y_2-y_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}},y_1+\frac{20(-x_2+x_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}}\right)$ $(x_2,y_2)\to\left(x_2+\frac{20(y_2-y_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}},y_2+\frac{20(-x_2+x_1)}{\sqrt{(y_2-y_1)^2+(-x_2+x_1)^2}}\right)$