We have hitherto been examining the conditions of horizontal and vertical lines only, or of curves enclosed in rectangles.

We must, in conclusion, investigate the perspective of inclined lines, beginning with a single one given in position. For the sake of completeness of system, I give in Appendix II. Article III.

Fig. 39.

Fig. 40.

The development of this problem from the second. But, in practice, the position of an inclined line may be most conveniently defined by considering it as the diagonal of a rectangle, as A B in Fig. 39., and I shall therefore, though at some sacrifice of system, examine it here under that condition.

If the sides of the rectangle A c and a d are given, the slope of the line a b is determined; and then its position will depend on that of the rectangle. If, as in Fig. 39., the rectangle is parallel to the picture plane, the line A B must be so also. If, as in Fig. 40., the rectangle is inclined to the picture plane, the line

A B will be so also. So that, to fix the position of a b, the line A c must be given in position and magnitude, and the height a d.If these are given, and it is only required to draw the single line a b in perspective, the construction is entirely simple; thus: Draw the line A c by Problem I.

Let a c, Fig. 41., be the line so drawn. From a and c raise the vertical lines a d, c b. Make a d equal to the sight-magnitude of A D. From d draw d b to the vanishing-point of a c, cutting b c in b.

Join a b. Then a b is the inclined line required.

Fig. 41.

Fig. 42.

If the line is inclined in the opposite direction, as D c in Fig. 42., we have only to join d c instead of a b in Fig. 41., and d c will be the line required.

I shall hereafter call the line A c, when used to define the position of an inclined line A B (Fig. 40.), the relative horizontal of the line A b.

## Observation

In general, inclined lines are most needed for gable roofs, in which, when the conditions are properly stated, the vertical height of the gable, x y, Fig. 43., is given, and the base line, A c, in position. When these are given, draw A c; raise vertical A D; make A D equal to sight-magnitude of x Y; complete the perspective-rectangle a d b c; join a b and d c (as by dotted lines in figure); and through the intersection of the dotted lines draw vertical x Y, cutting d b in y.

Fig. 43.

Join a y, c y; and these lines are the sides of the gable. If the length of the roof A a' is also given, draw in perspective the complete parallelopiped a' d' b c, and from Y draw Y y' to the vanishing-point of A a',

Fig. 44.

cutting d b' in y'.Join a' y, and you have the slope of the farther side of the roof.

The construction above the eye is as in Fig. 44.; the roof is reversed in direction merely to familiarise the student with the different aspects of its lines.

## III. Analysis Of Problem XV

We proceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distances B c' and A c (Fig. 6. page 222.), as well as the direct distances T D and T d' are unequal.

In Fig. 6., here repeated (Fig. 76.), produce C' B downwards, and make c' E equal to c A.

Join A E.

Then, by the second Corollary of Problem II., A E is a horizontal line.

Draw T v parallel to A E, cutting the sight-line in V.

.'. v is the vanishing-point of A E.

Complete the constructions of Problem II. and its second Corollary.

Then by Problem II. a b is the line A B drawn in perspective; and by its Corollary a e is the line A E drawn in perspective.

Fig. 76.

From v erect perpendicular v p, and produce a b to cut it in P.

Join T p, and from e draw e f parallel to A E, and cutting A T in f

Now in triangles e b T and A E T, as e b is parallel to E B and e f to A E; - c b: e f:: E B: A E.

But T v is also parallel to A E and P v to e b.

Therefore also in the triangles a p v and a V T, e b: e f:: P v: v T.

Therefore PV:vt::e b:A E.

And, by construction, angle T V P = angle A E B.

Therefore the triangles T v P, a e b, are similar; and T P is parallel to A B.

Now the construction in this problem is entirely general for any inclined line A B, and a horizontal line A E in the same vertical plane with it.

So that if we find the vanishing-point of A E in v, and from v erect a vertical v p, and from T draw T p parallel to A B, cutting V P in P, P will be the vanishing-point of A B, and (by the same proof as that given at page 226.) of all lines parallel to it.

Next, to find the dividing-point of the inclined line.

Fig. 77.

I remove some unnecessary lines from the last figure and repeat it here, Fig. 77., adding the measuring-line a M, that the student may observe its position with respect to the other lines before I remove any more of them.