[Geometric diagram]
Fig.9.

Let AB, Fig.9., be the given line.

From T, the station-point, draw TV parallel to AB, cutting the sight-line in V.

V is the Vanishing-point required.[Footnote 14]

[p18]
COROLLARY I.

As, if the point b is first found, V may be determined by it, so, if the point V is first found, b may be determined by it. For let AB, Fig.10., be the given line, constructed upon the paper as in Fig.8.; and let it be required to draw the line ab without using the point C'.

[Geometric diagram]
Fig.10.

Find the position of the point A in a. (ProblemI.)

[p19]
Find the vanishing-point of AB in V. (ProblemIII.)

Join aV.

Join BT, cutting aV in b.

Then ab is the line required.[Footnote 15]

COROLLARY II.

We have hitherto proceeded on the supposition that the given line was small enough, and near enough, to be actually drawn on our paper of its real size; as in the example given in AppendixI. We may, however, now deduce a construction available under all circumstances, whatever may be the distance and length of the line given.

[Geometric diagram]
Fig.11.

From Fig.8. remove, for the sake of clearness, the lines [p20] C'D', bV, and TV; and, taking the figure as here in Fig.11., draw from a, the line aR parallel to AB, cutting BT in R.

Then aR is to AB as aT is to AT.

Then — is to — as cT is to CT.

Then — is to — as TS is to TD.

That is to say, aR is the sight-magnitude of AB.[Footnote 16]

[Geometric diagram]
Fig.12.

Therefore, when the position of the point A is fixed in a, as in Fig.12., and aV is drawn to the vanishing-point; if we draw a line aR from a, parallel to AB, and make aR equal to the sight-magnitude of AB, and then join RT, the line RT will cut aV in b.

So that, in order to determine the length of ab, we need not draw the long and distant line AB, but only aR parallel to it, and of its sight-magnitude; which is a great gain, for the line AB may be two miles long, and the line aR perhaps only two inches.

[p21]
COROLLARY III.

In Fig.12., altering its proportions a little for the sake of clearness, and putting it as here in Fig.13., draw a horizontal line aR' and make aR' equal to aR.

Through the points R and b draw R'M, cutting the sight-line in M. Join TV. Now the reader will find experimentally that VM is equal to VT.[Footnote 17]

[Geometric diagram]
Fig.13.

Hence it follows that, if from the vanishing-point V we lay off on the sight-line a distance, VM, equal to VT; then draw through a a horizontal line aR', make aR' equal to the sight-magnitude of AB, and join R'M; the line R'M will cut aV in b. And this is in practice generally the most convenient way of obtaining the length of ab.

[p22]
COROLLARY IV.

Removing from the preceding figure the unnecessary lines, and retaining only R'M and aV, as in Fig.14., produce the line aR' to the other side of a, and make aX equal to aR'.

Join Xb, and produce Xb to cut the line of sight in N.

[Geometric diagram]
Fig.14.

Then as XR' is parallel to MN, and aR' is equal to aX, VN must, by similar triangles, be equal to VM (equal to VT in Fig.13.).

Therefore, on whichever side of V we measure the distance VT, so as to obtain either the point M, or the point N, if we measure the sight-magnitude aR' or aX on the opposite side of the line aV, the line joining R'M or XN will equally cut aV in b.

The points M and N are called the “Dividing-Points” of the original line AB (Fig.12.), and we resume the results of these corollaries in the following three problems.