Question

find the equation of the locus of a point that moves so that its distance from the line 3x-4y+9=0 is always 5

Answer 1

let that point be P (x, y)

Answer 2

distance from a point to a line = \(\frac{Ax+By+C}{\sqrt{A^2 + B^2}}\)

Answer 3

can you put the equation for distance ?
from point P(x, y) to line 3x-4y+9=0

Answer 4

oh so it would be (3x-4y+9)/square root( 3^2 +(-4)^2)

Answer 5

(3x -4y+9)/5 =5
3x -4y+9 =25
3x-4y -16 =0

Answer 6

is that right ganeshi8?

Answer 7

btw i thought there was an ABSOLUTE SIGN around Ax + By + C for the formula

Answer 8

yep! u r right.. we must get two lines

Answer 9

|(3x -4y+9)|/5 =5

Answer 10

3x-4y+9 = 25, and, 3x-4y + 9 = -25

Answer 11

ngawww i get it !

Answer 12

it tricked me.... i was thinking there should be two lines at the very end.
nice catch :)

Answer 13

thanks again - LOL and about writing GAPS yesterday, it wasn't meant to be rude or anything, just expressing my gratitude for your help ;)

Answer 14

lol fine i get it nw :))