The area of the quadrilateral formed by the lines $2\mathrm{x}+3\mathrm{y}+5=0$ , $3\mathrm{x}-2\mathrm{y}+4=0$ ,   $4x+6y-4=0$and $6\mathrm{x}-4\mathrm{y}+3=0$  is ( in square units )

detailed solution

Correct option is C

Rewrite the given lines as    2x+3y+5=0  …… (1)    3x−2y+4=0  …… (2)   2x+3y−2=0 …… (3)   3x−2y+32=0…… (4)Equations (1) and (3) are parallel lines, equations (2) and (4) are parallel linesEquations (1) and (2) are perpendicular lines, and equations (2) and (3) are perpendicular linesThe figure formed by the lines is rectangle. The area of the rectangle formed is the product of the distances between parallel linesThe distance between two parallel lines ax+by+c1=0  and  ax+by+c2=0 is  c2−c1a2+b2The distance between two parallel lines (1) and (3) is  5+222+32=713The distance between other two parallel lines (2) and (4) is  32−422+32=5213Therefore, the area is 713⋅5213=3526