The equations of two sides AB and AC of a triangle ABC are $$4x + y = 14$$ and $$3x - 2y = 5$$, respectively. The point $$\left(2, -\frac{4}{3}\right)$$ divides the third side BC internally in the ratio $$2 : 1$$. The equation of the side BC is

Let B=(b₁,b₂) on 4x+y=14 and C=(c₁,c₂) on 3x-2y=5. Point (2,-4/3) divides BC in 2:1: (2c₁+b₁)/3=2, (2c₂+b₂)/3=-4/3. So 2c₁+b₁=6, 2c₂+b₂=-4. Also 4b₁+b₂=14, 3c₁-2c₂=5. From 1st: b₁=6-2c₁, b₂=-4-2c₂. Sub in 4th: 4(6-2c₁)+(-4-2c₂)=14 → 20-8c₁-2c₂=14 → 8c₁+2c₂=6 → 4c₁+c₂=3. Also 3c₁-2c₂=5 → c₂=(3c₁-5)/2. 4c₁+(3c₁-5)/2=3 → 11c₁/2=11/2 → c₁=1, c₂=-1. b₁=4, b₂=-2.

Line BC through (4,-2) and (1,-1): slope = (-1+2)/(1-4) = -1/3. y+2 = -1/3(x-4) → 3y+6=-x+4 → x+3y+2=0.

Option (1): x+3y+2=0.