The routine returns true if l1 and l2 are perpendicular to each other; false if they are not; and FAIL if it is unable to determine if l1 and l2 are perpendicular to each other.

In case of FAIL, if the third optional argument is given, the condition that makes l1 and l2 perpendicular to each other is assigned to this argument.

The command with(geometry,ArePerpendicular) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{geometry}\right)\:$

$\mathrm{line}\left(\mathrm{l1}\,y=x\,\left[x\,y\right]\right)\:$

$\mathrm{line}\left(\mathrm{l2}\,y=-x\,\left[x\,y\right]\right)\:$

$\mathrm{ArePerpendicular}\left(\mathrm{l1}\,\mathrm{l2}\right)$

$\mathrm{true}$

$\mathrm{line}\left(\mathrm{l3}\,x=2\,\left[x\,y\right]\right)\:$

$\mathrm{ArePerpendicular}\left(\mathrm{l1}\,\mathrm{l3}\right)$

$\mathrm{false}$

$b≔3\:$

$\mathrm{line}\left(\mathrm{l4}\,by+ax=0\,\left[x\,y\right]\right)\:$

$\mathrm{ArePerpendicular}\left(\mathrm{l1}\,\mathrm{l4}\,'\mathrm{cond}'\right)$

ArePerpendicular:   "hint: cannot determine if -a+3 is zero"

$\mathrm{FAIL}$

$\mathrm{cond}$

$-a+3=0$

$\mathrm{assume}\left(\mathrm{cond}\right)$

$\mathrm{ArePerpendicular}\left(\mathrm{l1}\,\mathrm{l4}\,'\mathrm{cond}'\right)$

$\mathrm{true}$