To do this we simply reverse the foregoing problem. Thus let P be the given perspective point. From point of sight S draw a line through P till it cuts AB at m. From distance D draw another line through P till it cuts the base at n. From m drop perpendicular, and then with centre m and radius mn describe arc, and where it cuts that perpendicular is the required point P·. We often have to make use of this problem.
[ LII]
How to put a Given Line into Perspective
This is simply a question of putting two points into perspective, instead of one, or like doing the previous problem twice over, for the two points represent the two extremities of the line. Thus we have to find the perspective of A and B, namely a·b·. Join those points, and we have the line required.
Fig. 109.
If one end touches the base, as at A (Fig. 110), then we have
but to find one point, namely b. We also find the perspective of the angle mAB, namely the shaded triangle mAb. Note also that the perspective triangle equals the geometrical triangle.