The four lines given by $12x^2 + 7xy - 12y^2 = 0$ and $12x^2 + 7xy - 12y^2 - x + 7y - 1 = 0$ will make:
Step-by-Step Solution
Key Concept: Factorize both pair equations and verify that opposite sides are parallel with equal spacing while adjacent sides are perpendicular to prove a square.
The pair of equations $12x^2 + 7xy - 12y^2 = 0$ factors as $(3x+4y)(4x-3y) = 0$, giving lines $3x+4y=0$ and $4x-3y=0$ which are perpendicular. The second equation $12x^2 + 7xy - 12y^2 + x + 7y - 1 = 0$ factors as $(3x+4y-1)(4x-3y+1) = 0$, yielding parallel lines $3x+4y-1=0$ and $4x-3y+1=0$. The distance between $3x+4y=0$ and $3x+4y-1=0$ is $\frac{1}{5}$, and the distance between $4x-3y=0$ and $4x-3y+1=0$ is also $\frac{1}{5}$. Since all four lines form a rectangle with equal perpendicular distances and the adjacent lines are perpendicular, the four lines form a square.
Correct Answer: 1,2,3,4