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For i = 1, 2, and 3, let $a_ix + b_iy + c_i = 0$ be three equations of 3 (non-special cased) straight lines. From which, the co-ordinates of the vertices can be found. Using these co-ordinates, via the shoe-lace formula, we can found (A), the area of the triangle thus formed.

Is there any document that provides a direct (fully simplified) formula to find A via the a’s, b’s and c’s?

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    $\begingroup$ related : this $\endgroup$ – mathlove Jun 30 '15 at 12:01
  • $\begingroup$ @mathlove Great. thanks for the link. $\endgroup$ – Mick Jun 30 '15 at 13:04
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Put $a_1b_2-a_2b_1=:d_3$, and define $d_1$, $d_2$ similarly. If $d_3\ne0$ the map $$T:\quad (x,y)\mapsto\left\{\eqalign{u&:=a_1x+b_1 y\cr v&:=a_2x+b_2 y\cr}\right.\tag{1}$$ maps the $(x,y)$-plane bijectively onto the $(u,v)$-plane. It's inverse computes to $$T^{-1}:\quad (u,v)\mapsto\left\{\eqalign{x&={1\over d_3}(b_2u-b_1v)\cr y&={1\over d_3}(-a_2u+a_1v)\cr}\right.\ .\tag{2}$$ The triangle $\triangle$ in question is mapped onto a triangle $\triangle'$, two of whose sides are given by $$g_1':\quad u=-c_1,\qquad g_2':\quad v=-c_2\ ,$$ which implies $$A_3'=g_1'\wedge g_2'=(-c_1,-c_2)\ .$$ In order to obtain the $(u,v)$-equation of the third side $g_3'$ of $\triangle'$ we have to plug $x$ and $y$ from $(2)$ into the third equation $a_3x+b_3y+c_3=0$. After multiplication with $d_3$ this results in $$a_3(b_2u-b_1v)+b_3(-a_2u+a_1v)+c_3d_3=0\ ,$$ or $$g_3':\quad d_1 u+d_2 v=c_3 d_3\ .$$ In this way we obtain $$A_1'=g_2'\wedge g_3'=\left( {1\over d_1}(c_2d_2+c_3d_3),-c_2\right),\quad A_2'=g_1'\wedge g_3'=\left(-c_1, {1\over d_2}(c_1d_1+c_3d_3)\right)\ . $$ Now the side $A_1'A_3'$ of $\triangle'$ is horizontal and has length $$|A_1'\>A_3'|= \left|{1\over d_1}(c_2d_2+c_3d_3)+c_1\right|={|\Delta|\over|d_1|}\ ,$$ where $\Delta:=c_1d_1+c_2d_2+c_3d_3$. Similarly the side $A_2'A_3'$ of $\triangle'$ is vertical and has length $$|A_2'\>A_3'|= {|\Delta|\over|d_2|}\ .$$ It follows that the area of $\triangle'$ comes to $${\rm area}(\triangle')={1\over2}{\Delta^2\over |d_1d_2|}\ .$$ In order to obtain ${\rm area}(\triangle)$ we have to multiply this by $\bigl|\det T^{-1}\bigr|={\displaystyle{1\over |d_3|}}$. The final result then is $${\rm area}(\triangle)={1\over2}{\Delta^2\over |d_1d_2d_3|}\ ,$$ as given here.

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