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Two projective triangles are in perspectivity from a point $O$ when their three corresponding-vertex projective lines $P{P}^{\prime }$, $Q{Q}^{\prime }$, and $R{R}^{\prime }$ all pass through $O$.

Desargues's theorem: perspectivity from a point implies perspectivity from a line. More explicitly, if $P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }$ are concurrent, then the three corresponding-side intersectionsare collinear. The usual statement assumes nondegenerate projective triangles and well-defined, distinct corresponding-side intersections. The projective line containing $X,Y,Z$ is the axis of perspectivity from a line. All intersections are projective: parallel lines in an affine chart meet on its projective line at infinity.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.