Question

If $a+b+c=0$, then the family of lines $3ax+by+2c=0$ pass through fixed point:

  • A $\left(2, \frac{2}{3}\right)$
  • B $\left(\frac{2}{3}, 2\right)$
  • C $\left(-2, \frac{2}{3}\right)$
  • D none of these

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Answer

:

The correct option is B: $$ \left(\frac{2}{3}, 2\right) $$

Given: $$ a + b + c = 0 $$

First, substitute $c = -a - b$ in the equation: $$ 3ax + by + 2c = 0 $$ This results in: $$ 3ax + by + 2(-a - b) = 0 \ 3ax + by - 2a - 2b = 0 $$ Rewriting it, we obtain: $$ a(3x - 2) + b(y - 2) = 0 $$ Considering the form $L_1 + \lambda L_2 = 0$, this equation represents a line passing through the intersection of the lines $L_1$ and $L_2$, where: $$ L_1: 3x - 2 = 0 $$ $$ L_2: y - 2 = 0 $$

Next, solve the system of equations:

  1. $3x - 2 = 0$
  2. $y - 2 = 0$

Solving for $x$ and $y$: $$ 3x - 2 = 0 \ x = \frac{2}{3} $$ $$ y - 2 = 0 \ y = 2 $$

Hence, the intersection point (and thus the fixed point) is: $$ \left(\frac{2}{3}, 2\right) $$


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