document.write( "Question 37332: how can i find the cordinates of foot of the perpendicular in co-ordinate geometry? \n" ); document.write( "
Algebra.Com's Answer #22991 by venugopalramana(3286)\"\" \"About 
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how can i find the cordinates of foot of the perpendicular in co-ordinate geometry?IS IT PLANE C.G...OR 3D C.G?..LET ME TELL YOU FIRST FOR PLANE C.G.IF YOU NEED FOR 3D PLEASE COME BACK AND I SHALL EXPLAIN
\n" ); document.write( "LET US DO GENERAL AND SPECIFIC CASE WITH AN EXAMPLE.
\n" ); document.write( "LET US FIND THE FOOT OF THE PERPENDICULAR Q- SAY (H,K) FROM A POINT P (1,2)-SAY IN GENERAL (X1,Y1) ON TO A LINE L GIVEN BY EQN.X+Y=1..SAY IN GENERAL
\n" ); document.write( "AX+BY+C=0
\n" ); document.write( "SINCE Q IS THE FOOT OF PERPENDICULAR FROM P TO LINE L, IT LIES ON LINE L.HENCE
\n" ); document.write( "AH+BK+C=0.......................OR.........H+K=1 ....................I
\n" ); document.write( "PQ IS PERPENDICULAR TO L...SO PRODUCT OF SLOPE OF PQ AND SLOPE OF LINE L =-1
\n" ); document.write( "SLOPE OF PQ =(K-2)/(H-1)............OR..............(K-Y1)/(H-X1)
\n" ); document.write( "SLOPE OF LINE L =-1/1=-1..................OR...............-A/B
\n" ); document.write( "HENCE
\n" ); document.write( "{(K-2)/(H-1)}*(-1)=-1...........OR.....{(K-Y1)/(H-X1)}{-A/B}=-1
\n" ); document.write( "(K-2)/(H-1)=1...............OR.........(K-Y1)/(H-X1)=B/A
\n" ); document.write( "(K-2)=(H-1).................OR.........A(K-Y1)=B(H-X1)
\n" ); document.write( "K-2=H-1.......................OR.....AK-AY1=BH-BX1
\n" ); document.write( "H-K=-1...............OR...........BH-AK=BX1-AY1.....................II
\n" ); document.write( "NOW SOLVE EQNS.I AND II TO GET H,K....FOR THE NUMERICAL EXAMPLE WE GET
\n" ); document.write( "H+K+H-K=1-1=0
\n" ); document.write( "2H=0
\n" ); document.write( "H=0
\n" ); document.write( "0+K=1
\n" ); document.write( "K=1
\n" ); document.write( "HENCE THE FOOT OF PERPENDICULAR FROM P(1,2) TO THE LINE L ...X+Y=1
\n" ); document.write( "IS Q.(0,1)
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