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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)
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