JEE Main 202429 Jan, Evening shiftMathematicsMediumCoordinate Geometry478× since 2002The 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 toAOption A. 3+423+4 \sqrt{2}3+42BOption B. 636 \sqrt{3}63COption C. 4+634+6 \sqrt{3}4+63DOption D. 424 \sqrt{2}42CheckSkip · show solution
