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For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

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.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.