Question

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

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

Answer 1

The correct option is B

$(\frac{2}{3},2)$

Given :

$a+b+c=0$

Substituting $c=-a-b$ in $3ax+by+2c=0,$ we get :

$3ax+by-2a-2b=0$

$\Rightarrow a(3x-2)+b(y-2)=0$

$\Rightarrow (3x-2)+\frac{b}{a}(y-2)=0$

This line is of the form $L_1+\lambda L_2=0,$ which passes through the intersection of the lines $L_1$ and $L_2.$

i.e. $3x-2$ and $y-2=0$

Solving $3x-2=0$ and $y-2=0,$ we get:

$x=\frac{2}{3},y=2$

Hence, the required fixed point is $(\frac{2}{3},2).$