1

I have points $A, B, C, K$, with $K = -A$. I want to define $M$ as the intersection of $BC$ and $AK$. Representing the points as complex numbers in the plane, I've set up the equations:

$m = b + \lambda_1(c - b)$

$m = \lambda_2a$

And now I'm completely confused. The equations are simply algebraic reformulations of the lines. The intersection of the lines is unique. I should be able to solve this system for $m$ in terms of $a, b, c$, no? I don't think that's possible. Is there something I'm missing?

turmur
  • 23
  • Presumably you're working in 2 dimensions (otherwise 2 lines need not intersect). For given points $a, b, c$, Set up the equation in each coordinate, which gives you 2 equations and 2 unknowns ($\lambda_1, \lambda_2$). – Calvin Lin Jan 24 '22 at 00:12
  • Calvin, I think you misunderstood my question. Clearly I can solve the above system for $\lambda_1$ and $\lambda_2$, but I want to solve it for $m$ in terms of $a, b, c$. I'm sorry if this is a very stupid question still. – turmur Jan 24 '22 at 00:17
  • After you found $ \lambda_1$, you have $ m = b + \lambda_1(c-b)$, don't you? And yes, this will equal to $ \lambda_2 a$ . $\quad$ (So, I'm not quite sure what you're asking for then.) – Calvin Lin Jan 24 '22 at 00:23
  • 1
    I think I figured it out. I was confused as to how I was supposed to extract the coordinates of the variables, so I was only getting $\lambda_1, \lambda_2$ in terms of $m$. It seems like I can just take the conjugate of both equations to do this, which is obvious in retrospect. Thank you for your comment, as it helped, and sorry once again. – turmur Jan 24 '22 at 00:26
  • @turmur What you describe is equivalent to writing both equations in the form $,\alpha z + \bar \alpha \bar z + \beta = 0,$ then eliminating $,\bar z,$ between the two equations. – dxiv Jan 24 '22 at 03:21

0 Answers0