Find the locus of point P(x, y) that moves so that its perpendicular distance from the line 8x+6y-1=0 is always 5 units.

Answer 1

6 y = -8 x + 1 is original line
slope = -8/6 = - 4/3

we want a parallel line, so same slope
y = -4x/3 + b
so what is b ?

line perpendicular to these has slope = 3/4
find a point on original line
for example (0,1/6)
now perpendicular line through that point is
1/6 = (3/4)(0) + c
c = 1/12
so
y = (3/4) x + 1/12)
where does that hit parallel line?

12 y = 9 x + 1
3 y = -4 x + 3 b

12 y = 9 x + 1
12 y = -16 x + 12 b
--------------------
0 = 25 x -11 b

x = 11 b/25
y = 31 b/75

so distance from (0,1/6) to (11b/25, 31b/75) = 5

25 = (11b/25)^2 + (31 b/75 - 1/6)^2

first check my arithmetic, then solve for b

Answer 2

The line 5 units away from the given line will cross the y-axis somewhere.

Pick a point (0,y)
Its distance from the line is

|6y-1|/10 = 5
6y-1=50 ==> y = 17/2
6y-1=-50 ==> y = -49/6

Now you know that the desired line contains either (0,17/2) or (0,-49/6)
Its slope is -4/3, so now you can use the point-slope form of the line.

Naturally, there are two lines parallel to the given line at a distance of 5 units.

I hope our answers agree ...