Geometry questions

Let the slopes of the perspective lines passing through origin O(0,0) be m₁, m₂, m₃ and let y=m₁x be the line joining the corresponding generic vertices A of perspective triangles; similarly y=m₂x joins generic vertices B and y=m₃x joins generic vertices C. (A generic vertex is A, B or C followed by a subscripted index. So, ABC is the generic triangle and AᵣBᵣCᵣ where r=1, 2, etc., identify a particular triangle with the same perspectivity centre O.)

We can define the vertices of three perspective triangles A₁B₁C₁, A₂B₂C₂, A₃B₃C₃:

A₁(a₁,m₁a₁), B₁(b₁,m₂b₁), C₁(c₁,m₃c₁),

A₂(a₂,m₁a₂), B₂(b₂,m₂b₂), C₂(c₂,m₃c₂),

A₃(a₃,m₁a₃), B₃(b₃,m₂b₃), C₃(c₃,m₃c₃).

Triangles A₁B₁C₁ and A₂B₂C₂ determine points on the perspectivity axis, and therefore establish the equation of this axis. The sides A₁B₁ and A₂B₂ (extended as necessary) intersect at some point P; similarly A₁C₁ and A₂C₂ intersect at Q, and B₁C₁ and B₂C₂ intersect at R. The line PQR is the axis. To be in perspective with these triangles, triangle A₃B₃C₃ must be constrained so that A₃B₃ lies on the line A₁P, A₃C₃ on the line A₁Q, and B₃C₃ on the line B₁R.

ago by Top Rated User (786k points)