2

If $6a^2-3b^2-c^2+7ab-ac+4bc=0$, then the family of lines $ax+by+c=0$ is concurrent at

  1. $(-2,-3)$
  2. $(3,-1)$
  3. $(2,3)$
  4. $(-3,1)$

Multiple answers are possible.

I am not able to group terms of the expression $6a^2-3b^2-c^2+7ab-ac+4bc=0$ so as to get expression in the form $\lambda_1L_1+\lambda_2L_2=0$, where $L_1$ and $L_2$ denote two lines out of the given family. Please help.

Tejas
  • 2,082
  • 3
    A first observation is that $6a^2-3b^2-c^2+7ab-ac+4bc=(2a+3 b-c)(3a-b+c)$. This could be helpful. – Dennis Gulko Dec 08 '13 at 15:14
  • @DennisGulko Indeed, your factorization gives immediately the points $(-2,-3)$ and $(3,-1)$ as points on lines $ax+by+c=0.$ – coffeemath Dec 08 '13 at 15:20
  • 1
    @coffeemath I think 'concurrent' (used by OP) would be the wrong word to use. All we know is that the lines must pass through one of those 2 points. Not all lines need to pass through $(-2,-3)$. – Calvin Lin Dec 08 '13 at 15:24
  • @Calvin Yes I agree, each line must pass through at least one of the points, not necessarily through either particular point. (+1 on comment) – coffeemath Dec 08 '13 at 16:33

0 Answers0