# SOLUTION: how can i find the cordinates of foot of the perpendicular in co-ordinate geometry?

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 Click here to see ALL problems on Length-and-distance Question 37332: how can i find the cordinates of foot of the perpendicular in co-ordinate geometry? Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website! 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 LET US DO GENERAL AND SPECIFIC CASE WITH AN EXAMPLE. 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 AX+BY+C=0 SINCE Q IS THE FOOT OF PERPENDICULAR FROM P TO LINE L, IT LIES ON LINE L.HENCE AH+BK+C=0.......................OR.........H+K=1 ....................I PQ IS PERPENDICULAR TO L...SO PRODUCT OF SLOPE OF PQ AND SLOPE OF LINE L =-1 SLOPE OF PQ =(K-2)/(H-1)............OR..............(K-Y1)/(H-X1) SLOPE OF LINE L =-1/1=-1..................OR...............-A/B HENCE {(K-2)/(H-1)}*(-1)=-1...........OR.....{(K-Y1)/(H-X1)}{-A/B}=-1 (K-2)/(H-1)=1...............OR.........(K-Y1)/(H-X1)=B/A (K-2)=(H-1).................OR.........A(K-Y1)=B(H-X1) K-2=H-1.......................OR.....AK-AY1=BH-BX1 H-K=-1...............OR...........BH-AK=BX1-AY1.....................II NOW SOLVE EQNS.I AND II TO GET H,K....FOR THE NUMERICAL EXAMPLE WE GET H+K+H-K=1-1=0 2H=0 H=0 0+K=1 K=1 HENCE THE FOOT OF PERPENDICULAR FROM P(1,2) TO THE LINE L ...X+Y=1 IS Q.(0,1)