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# The base BC of triangle ABC is bisected at (p,q) and the equation of sides AB and aAC are px+qy+1 and qx+py=1.Then the equation of median through A is

Ajay
209 Points
4 years ago
$Given\quad Equations\quad of\quad side\quad AB\quad and\quad AC\quad as\\ px+qy\quad =\quad 1\quad and\quad qx+py\quad =1\quad \\ Also\quad Given\quad median\quad of\quad BC\quad is\quad D(p,q).\\ Solving\quad for\quad equations\quad will\quad give\quad point\quad A\quad as\quad follows\\ (\quad \frac { 1 }{ p+q } ,\quad \frac { 1 }{ p+q } )\\ Now\quad we\quad have\quad to\quad find\quad the\quad equation\quad ofline\quad AD\quad which\quad can\quad be\quad easily\quad found\\ using\quad formula\\ y-{ y }_{ 1 }\quad =\quad \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } (x-{ x }_{ 1 })$
Abhinav Joshi
24 Points
4 years ago
U can also use the formula for finding bisector of two given lines, because A is being bisected here, put point (P,Q) in the two equations obtained, the one which satisfies, is the answer.So, when I solved , in both it is not satisfying, so check ur question once again