The distance between the lines represented by 16 x^2+9 y^2+48 x-24 x y-36 y+35=0 is ...... units

Step-by-step solution

Given the equation 16x^2 + 9y^2 + 48x - 24xy - 36y + 35 = 0 . The quadratic terms 16x^2 - 24xy + 9y^2 factor as (4x - 3y)^2 , leading to the rewritten form (4x - 3y)^2 + 48x - 36y + 35 = 0 . Observing that 48x - 36y = 12(4x - 3y) , the equation simplifies to (4x - 3y)^2 + 12(4x - 3y) + 35 = 0 . Substituting Z = 4x - 3y gives the quadratic equation Z^2 + 12Z + 35 = 0 , which factors as (Z + 5)(Z + 7) = 0 , providing Z = -5 or Z = -7 . Replacing Z yields two linear equations in standard form: 4x - 3y + 5 = 0 and 4x -