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

Given: 

a + b + c = 0 

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

3ax + by – 2a – 2b = 0 

⇒ a (3x – 2) + b (y – 2) = 0 

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

This line is of the form L₁ + λL₂ = 0, 

which passes through the intersection of the lines L₁ and L₂, i.e. 3x – 2 = 0 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 \(\Big(\frac{2}{3},2\Big)\)