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i have two triangles. Say a , b , c and p, q, r and the projection of the abc to pqr

a - > p
b - > q
c - > r

here known point values are a b c p q and r unknown.

$\overline{PR}=\overline{AC}$ and $\overline{QR}=\overline{BC}$

from these information how can we find r coordinates. please suggest me.

David K
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Suri
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  • Please add some punctuation. Especially in the sentence that tried to tell us which quantities are known and which are unknown. – bubba Aug 31 '14 at 12:31
  • And a projection of a triangle into another, what does it mean? In LaTeL you can represent the distance between two points by \overline{AB}. I'll edit that part. – PenasRaul Aug 31 '14 at 12:52
  • Projection implies that the coordinates (a and b) are projected by using delaunay triangulation and r is not projected because it is outside the quadrilateral need to be adjusted. sorry for my english – Suri Sep 01 '14 at 04:39

1 Answers1

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Given known coordinates for $A,$ $B,$ and $C,$ you can compute the lengths $\overline{AC}$ and $\overline{BC}$ by the Pythagorean Theorem. Given known coordinates for $P$ and $Q,$ and given the lengths $\overline{PR}$ and $\overline{QR},$ apply the Pythagorean Theorem to the unknown coordinates of $R$ and the known coordinates of $P$ and $Q,$ and set the results equal to the known lengths $\overline{PR}$ and $\overline{QR}.$ The resulting system of quadratic equations has two solutions. Which one of those is $R$ depends on whether your projection preserves the orientation of the triangle or reverses the orientation of the triangle. (In informal terms, does it merely move, rotate, and/or stretch the triangle, or does it flip the triangle over?)

The "system of quadratic equations" in the previous paragraph will consist of two equations, each of which is the equation of a circle. Solving the system is the same as finding the intersections of two circles. Other examples of this problem are worked out in detail here and here, and there is a more general solutiion here.

Community
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David K
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  • Hi thanks for the reply these are my triangle values A(0,0),,C(8.7,-5),B(0,10) and P(0,0),Q(13,15) R(x,y) I computed the two equation PQ and PR equal to AB and AC (x^2) + (y^2) = 100, (x^2) + (y^2)-26x,30y-294 = 0. Please suggest me how can i solve this – Suri Dec 29 '14 at 07:11
  • The last equation looks almost like something you could use, except it has an unexpected comma (was that supposed to be a subtraction sign?) and the constant term is very different than I would expect. Make sure you are setting $\overline{QR}$ (not $\overline{PQ}$) equal to $\overline{BC}$ (not $\overline{AB}$). You already have enough data to compute $\overline{PQ}$ and $\overline{AB}$ directly and to find out that they are not equal; this is of no use to you. The rest of the solution is shown in many other places so I have given references. – David K Dec 29 '14 at 19:43