# SOLUTION: Hi, not sure if this is appropriate topic, anyway here goes for the point A (3,-1) and the line with equation x-y=1, find: a The equation of the perpendicular to the line thr

Algebra ->  Algebra  -> Length-and-distance -> SOLUTION: Hi, not sure if this is appropriate topic, anyway here goes for the point A (3,-1) and the line with equation x-y=1, find: a The equation of the perpendicular to the line thr      Log On

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 Geometry: Length, distance, coordinates, metric length Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Length-and-distance Question 35483: Hi, not sure if this is appropriate topic, anyway here goes for the point A (3,-1) and the line with equation x-y=1, find: a The equation of the perpendicular to the line through A b The foot of the perpendicular from A to the line c The distance of A from the line Hope someone can help, thank youAnswer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Hi, not sure if this is appropriate topic, anyway here goes for the point A (3,-1) and the line with equation x-y=1....SAY LINE L find: a The equation of the perpendicular to the line through A EQN.OF LINE PERPENDICULAR TO L IS OF THE FORM X+Y=K(..RULE IS EXCHANGE COEFFICIENTS OF X AND Y AND MAKE ONE NEGATIVE AND EQUATE THE RESULT TO A CONSTANT TO BE FOUND OUT) THIS IS TO GO THROUGH A ..HENCE.. 3+(-1)=K=2..........HENCE EQN. OF REQD.LINE IS X+Y=2 b The foot of the perpendicular from A to the line LET B(P,Q) BE THE FOOT OF PERPENDICULAR FROM A ON TO THE LINE. SLOPE OF AB =(Q+1)/(P-3)...SINCE THIS IS PERPENDICULAR TO OUR LINE SLOPE =1.HENCE (Q+1)/(P-3)=1...............OR.....Q+1=P-3 P-Q=4.....................I BUT B IS ON LINE ...SO.... P+Q=2.........................II ADDING EQN.I AND EQN.II 2P=6 P=3 Q=2-P=2-3=-1 HENCE FOOT OF PERPENDICULAR IS (3,-1) c The distance of A from the line IT IS EQUAL TO AB .HENCE AB=SQRT{(3-3)^2+(1+1)^2}=SQRT(2^2)=2 Hope someone can help, thank you