1998 AHSME Problem 25

$Let\quad point\quad A,\quad B,\quad C,\quad D\quad be\quad the\quad (0,2),\quad (4,0),\quad (7,3)\quad and\quad (m,n)\quad respectively.\\ \\ Step\quad 1:\quad To\quad find\quad the\quad reflection\quad line\\ Since\quad this\quad is\quad a\quad folding\quad problem,\quad the\quad points\quad can\quad be\quad treated\quad as\quad reflection\quad along\quad the\quad perpendicular\\ bisector\quad of\quad AB.\\ \\ By\quad using\quad the\quad point\quad A\quad and\quad B,\quad the\quad equation\quad of\quad perpendicular\quad bisector\quad of\quad AB\quad is\\ 2x-y-3=0\\ \\ Step\quad 2:\quad To\quad find\quad point\quad D\\ \\ Since\quad CD\quad is\quad parallel\quad to\quad AB,\quad we\quad have\\ -\frac { 1 }{ 2 } =\frac { 3-n }{ 7-m } \quad \cdots \cdots (1)\\ Since\quad the\quad mid-point\quad of\quad AB\quad passes\quad through\quad the\quad perpendicular\quad bisector\quad of\quad AB,\quad we\quad have\\ \frac { 2\left( 7+m \right) }{ 2 } -\frac { 3+n }{ 2 } -3=0\quad \cdots \cdots (2)\\ By\quad solving\quad (1)\quad and\quad (2),\quad we\quad have\quad (m,n)=(\frac { 3 }{ 5 } ,\frac { 31 }{ 5 } )\\$