If `a+b+c=0` , then the family of lines `3a x+b y+2c=0` pass through fixed point `a. (2,2/3) b. (2/3,2) c. (-2,2/3) d`. none of these

Explanation

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) + b/a (y - 2) = 0 `

This line is of the form` L_1 + λL_2 = 0`, which passes through the intersection of the lines `L_1 `and `L_2`,

i.e. `3x – 2 = 0` and `y – 2 = 0` .

Solving `3x – 2 = 0` and` y – 2 = 0`, we get: `x = 2/3 , y = 2`
Hence, the required fixed point is `(2/3,2)`