When I am dealing with some geometry problem in barycentric system I come across with the following equations

$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-xy\left(\frac{1}{a}+\frac{1}{b}\right)-yz\left(\frac{1}{b}+\frac{1}{c}\right)-xz\left(\frac{1}{a}+\frac{1}{c}\right)=0, c\left(b+c\right)x-acy+a\left(a+b\right)z=0 \ $

One is the equation of curve other is equation of line in barycentric coordinates

I attempted to solve these two equations by making y as a subject and substituting in the first equation but that doesn't give any solution,

Now my question is how to get x, y, z from these two equations or how to get x: y: z from these two equations in terms of a, b, c. here a, b, c are constants such as they are the sides of the triangle.

I got one point which is common for the both equations is x: y: z as a: 2s+b: c where 2s=a+b+c How to find other point which is common for both equations.

Thanks in advance

  • 1
    $\begingroup$ I’m a bit surprised that whatever process you used to find the one solution didn’t give you a second one. Seem like somewhere along the way you would’ve been taking a square root with a choice of sign for it. It would be helpful if you showed how you found the one solution that you did. $\endgroup$ – amd Oct 31 '18 at 1:42
  • $\begingroup$ Since barycentric coordinates are homogeneous, you’re effectively trying to compute the intersection of a conic and line on the projective plane. See this question for methods. $\endgroup$ – amd Oct 31 '18 at 2:22
  • $\begingroup$ Does $x:y:z = a:2s+b:c$ actually satisfy the first equation? $\endgroup$ – Blue Oct 31 '18 at 2:39
  • $\begingroup$ Sorry respected blue $\endgroup$ – nimmy Oct 31 '18 at 3:29
  • $\begingroup$ In my equation there is typo mistake I have edited now $\endgroup$ – nimmy Oct 31 '18 at 3:29

Clearing fractions, we can write

$$\begin{align} b c x^2 + c a y^2 + a b z^2 - ( a + b ) c x y - ( b + c ) a y z - ( c + a ) b z x &= 0 \\ [4pt] ( b + c ) c x - a c y + ( a + b ) a z &= 0 \end{align} \tag{1}$$

Now, we simply eliminate $v$ to get a relation we can write as $$(c x - a z) \left(\;c^2 (-a + b + c ) x - a^2 ( a + b - c ) z \;\right) = 0 \tag{2}$$ Thus, $$cx=az \quad\to\quad y = \frac{a + 2 b + c}{a} x \tag{3a}$$ or $$c^2 (-a + b + c ) x = a^2 ( a + b - c ) z \quad\to\quad y = \frac{b (a^2 + a b + b c + c^2)}{a^2 (a + b - c)}x \tag{3b}$$ so that

$$x:y:z \;=\; a:a+2b+c:c$$ or $$x:y:z \;=\; a^2(a+b-c):b \left(\;a(a+b) + c(b+c)\;\right):c^2(-a+b+c)$$

  • $\begingroup$ Thank you very much respected blue, it was very nice approach $\endgroup$ – nimmy Oct 31 '18 at 5:13

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