Question

If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed point
(a) (2, 2/3)
(b) (2/3, 2)
(c) (−2, 2/3)
(d) none of these

Answer 1

(b) $\left(\frac{2}{3},2\right)$
Given:
a + b + c = 0

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

$$\begin{gathered} 3ax+by-2a-2b=0 \\ \Rightarrow a\left(3x-2\right)+b\left(y-2\right)=0 \\ \Rightarrow \left(3x-2\right)+\frac{b}{a}\left(y-2\right)=0 \end{gathered}$$

This line is of the form $L_1+\lambda L_2=0$, which passes through the intersection of the lines $L_1\mathrm{and}{L}_2$, i.e. $3x-2=0\mathrm{and}y-2=0$.

Solving $3x-2=0\mathrm{and}y-2=0$, we get:

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

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