If $6a^2 - 3b^2 - c^2 + 7ab - ac + 4bc = 0$, then the family of lines $ax + by + c = 0$ is concurrent at ordered pairs $(A, B)$ and $(C, D)$. $A > 0$ then the value of $A - B - C - D$ is ______.
Step-by-Step Solution
Key Concept: The constraint equation factors into two linear conditions in $a, b, c$, each defining a family of lines concurrent at a fixed point.
We need to find values of $a, b, c$ satisfying $6a^2 - 3b^2 - c^2 + 7ab - ac + 4bc = 0$. This quadratic form can be factored as $(2a + 3b - c)(3a - b + c) = 0$. This gives two linear relations: $2a + 3b - c = 0$ or $3a - b + c = 0$. From the first: $c = 2a + 3b$, giving line $ax + by + 2a + 3b = 0$ or $a(x+2) + b(y+3) = 0$, concurrent at $(-2, -3)$. From the second: $c = -3a + b$, giving line $ax + by - 3a + b = 0$ or $a(x-3) + b(y+1) = 0$, concurrent at $(3, -1)$. The two points of concurrency are $(-2, -3)$ and $(3, -1)$. With $A = 3 > 0$, we have $A - B - C - D = 3 - (-1) - (-2) - (-3) = 3 + 1 + 2 + 3 = 9$.
Correct Answer: 9