ective 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.

[Illustration: Fig. 108.]

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.

[Illustration: Fig. 109.]

[Illustration: Fig. 110.]

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.

[Illustration: Fig. 111.]

When the line required is parallel to the base line of the picture, then
the perspective of it is also parallel to that base (see Rule 3).

LIII

TO FIND THE LENGTH OF A GIVEN PERSPECTIVE LINE

A perspective line _AB_ being given, find its actual length and the
angle at which it is placed.

This is simply the reverse of the previous problem. Let _AB_ be the
given line. From distance _D_ through _A_ draw _DC_, and from _S_, point
of sight, through _A_ draw _SO_. Drop _OP_ at right angles to base,
making it equal to _OC_. Join _PB_, and line _PB_ is the actual length
of _AB_.

This problem is useful in finding the position of any given line or
point on the perspective plane.

[Illustration: Fig. 112.]

LIV

TO FIND THESE POINTS WHEN THE DISTANCE-POINT IS INACCESSIBLE

[Illustration: Fig. 113.]

If the distance-point is a long way out of the picture, then the same
result can be obtained by using the half distance and half base, as
already shown.

From _a_, half of _mP_', draw quadrant _ab_, from _b_ (half base), draw
line from _b_ to half Dist., which intersects _Sm_ at _P_, precisely the
same point as would be obtained by using the whole distance.

LV

HOW TO PUT A GIVEN TRIANGLE OR OTHER RECTILINEAL FIGURE INTO PERSPECTIVE

Here we simply put three points into perspective to obtain the giv