If, in Fig.43. or Fig.44., the lines AY and A'Y' be produced, the student will find that they meet.

Let P, Fig.45., be the point at which they meet.

From P let fall the vertical PV on the sight-line, cutting the sight-line in V.

Then the student will find experimentally that V is the vanishing-point of the line AC.[Footnote 26]

Complete the rectangle of the base AC', by drawing A'C' to V, and CC' to the vanishing-point of AA'.

Join Y'C'.

Now if YC and Y'C' be produced downwards, the student will find that they meet.

Let them be produced, and meet in P'.

Produce PV, and it will be found to pass through the point P'.

Therefore if AY (or CY), Fig.45., be any inclined line drawn in perspective by ProblemXV., and AC the relative horizontal (AC in Figs. 39, 40.), also drawn in perspective.

Through V, the vanishing-point of AV, draw the vertical PP' upwards and downwards.

Produce AY (or CY), cutting PP' in P (or P').

Then P is the vanishing-point of AY (or P' of CY).

[Geometric diagram]
Fig.45.

The student will observe that, in order to find the point P by this method, it is necessary first to draw a portion of the given inclined line by ProblemXV. Practically, it is always necessary to do so, and, therefore, I give the problem in this form.

[p54]
Theoretically, as will be shown in the analysis of the problem, the point P should be found by drawing a line from the station-point parallel to the given inclined line: but there is no practical means of drawing such a line; so that in whatever terms the problem may be given, a portion of the inclined line (AY or CY) must always be drawn in perspective before P can be found.