MathematicsMedium128×since 2002Q5557The distance of the point (2,3)(2,3)(2,3) from the line 2x−3y+28=02 x-3 y+28=02x−3y+28=0, measured parallel to the line 3x−y+1=0\sqrt{3} x-y+1=03x−y+1=0, is equal toA3+423+4 \sqrt{2}3+42B636 \sqrt{3}63C4+634+6 \sqrt{3}4+63D424 \sqrt{2}42Check answerSkip