Question

Two families of lines are given by $ax+by+c=0$ and $4a^2+9b^2-c^2-12ab=0$. Then the line common to both the families is

• A. a line passing through (-1,2) and (2,3)
• B. a line passing through (3,2) and (2,3)
• C. a line passing through (-3,-2) and (-2,-3)
• D. a line passing through (2,-3) and (-2,3)

Hint:

When a family of lines $ax+by+c=0$ is constrained by a quadratic relation in $a,b,c$, try to factor the relation into two linear constraints. Each linear constraint will define a fixed point (a "concurrent point") that a subset of the family of lines passes through.

Answer 1

The Correct Option is D

Step 1: Factorize the condition on the parameters a, b, and c.
The given condition is $4a^2+9b^2-c^2-12ab=0$. Rearranging the terms, we can identify a perfect square: \[ (4a^2 - 12ab + 9b^2) - c^2 = 0. \] \[ (2a - 3b)^2 - c^2 = 0. \] This is a difference of squares, which factors as: \[ ((2a-3b) - c)((2a-3b) + c) = 0. \]

Step 2: Identify the two linear relationships between a, b, and c.
This factorization implies that for any line in the family, its parameters must satisfy one of two conditions: Case 1: $2a - 3b - c = 0$. Case 2: $2a - 3b + c = 0$.

Step 3: Find the fixed point associated with Case 1.
The family of lines is defined by $ax+by+c=0$. From Case 1, we can substitute $c = 2a-3b$. \[ ax + by + (2a-3b) = 0. \] Group the terms with parameters $a$ and $b$: \[ a(x+2) + b(y-3) = 0. \] For this to be true for all possible values of $a$ and $b$, the coefficients must be zero. This means the line must pass through a fixed point. $x+2=0 \implies x=-2$. $y-3=0 \implies y=3$. So, one fixed point is $P_1(-2, 3)$.

Step 4: Find the fixed point associated with Case 2.
From Case 2, we substitute $c = -2a+3b$. \[ ax + by + (-2a+3b) = 0. \] Group the terms with parameters $a$ and $b$: \[ a(x-2) + b(y+3) = 0. \] For this to be true for all $a$ and $b$, the coefficients must be zero. $x-2=0 \implies x=2$. $y+3=0 \implies y=-3$. So, the second fixed point is $P_2(2, -3)$.

Step 5: Identify the common line.
The family of lines consists of all lines that pass through either $P_1(-2,3)$ or $P_2(2,-3)$. The question asks for the "line common to both families," which refers to the line that passes through both of these fixed points. This is the line passing through $(2,-3)$ and $(-2,3)$, which matches option (D).