# Thirteen-point conic and four-point line, are they new?

We know that Five points determine a conic and Two Points Determine a Line. Here I found a simple construct of a conic through $$7$$ points (in PS I note that how the conic through thirteen points) and a line through $$4$$ points as follows:

Let $$ABC$$ be a triangle and $$P$$ be arbitrary point in the plane, let $$A'$$, $$B'$$, $$C'$$ lie on the line through $$P$$ and parallel to $$BC$$, $$CA$$, $$AB$$ respectively such that $$AA' \parallel BB' \parallel CC'$$. I am looking for a proof that

• Seven points $$A$$, $$B$$, $$C$$, $$A'$$, $$B'$$, $$C'$$ and $$P$$ lie on a conic
• Four points: midpoints of $$AA'$$, $$BB'$$, $$CC'$$ and center of the conic are collinear? PS: Three lines: The line through midpoints of $$BC$$ and $$PA'$$, The line through midpoints of $$CA$$ and $$PB'$$ and The line through midpoints of $$AB$$ and $$PC'$$ are concurrent at the center of the conic (denote the point of concurrence is $$O$$). Then easily show that reflection of $$A, B, C, A', B', C'$$ in $$O$$ also lie on the conic. This mean the conic through $$13$$ points.

There is a systematic method to solve problems of this sort (the $$p,q$$ method) by reducing them to computations which can be done by hand. Assume the vertices are $$(0,0), (1,0), (p,q)$$. The three new points have the form $$(x+t(p-1),y+tq), (x-up,y-uq), (x+s,y)$$ with three parameters.These can be reduced to one free parameter by the parallelism condition. One then uses the condition on the rank of the matrix with rows $$(x^2 ,xy , y^2 ,x ,y, 1)$$ one for each of the seven points, which ensures that they line on a conic. Now compute the quadratic equation of the corresponding conic (with coefficients which depend on $$p,q,x,y$$). It is then routine to compute the coordinates of $$O$$ and verify the collinearity condition.