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Question
If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed point
Options
(2, 2/3)
(2/3, 2)
(−2, 2/3)
none of these
Solution
(2/3, 2)
Given:
a + b + c = 0
Substituting c = − a − b in 3ax + by + 2c = 0, we get:
\[3ax + by - 2a - 2b = 0\]
\[ \Rightarrow a\left( 3x - 2 \right) + b\left( y - 2 \right) = 0\]
\[ \Rightarrow \left( 3x - 2 \right) + \frac{b}{a}\left( y - 2 \right) = 0\]
This line is of the form
\[L_1 + \lambda L_2 = 0\], which passes through the intersection of the lines \[L_1 \text { and } L_2\] i.e.
\[3x - 2 = 0 \text { and } y - 2 = 0\].
Solving \[3x - 2 = 0 \text { and } y - 2 = 0\],we get:
\[x = \frac{2}{3}, y = 2\]
Hence, the required fixed point is \[\left( \frac{2}{3}, 2 \right)\].
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