|
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 you
Answer 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
|
|
|
| |
