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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

nimmy
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    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. – amd Oct 31 '18 at 01:42
  • 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. – amd Oct 31 '18 at 02:22
  • Does $x:y:z = a:2s+b:c$ actually satisfy the first equation? – Blue Oct 31 '18 at 02:39
  • Sorry respected blue – nimmy Oct 31 '18 at 03:29
  • In my equation there is typo mistake I have edited now – nimmy Oct 31 '18 at 03:29
  • Respected amd thanks for reference to deal with these problems, but the problem itself is parameters a, b, c, is that method is little bit difficult, is there any other way to deal, please if you aware of those methods please share – nimmy Oct 31 '18 at 03:36

1 Answers1

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

Blue
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