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.
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
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 ...
To 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, we can use the concept of perpendicular distance from a point to a line.
The equation of a line in slope-intercept form is represented by y = mx + c, where m is the slope and c is the y-intercept. In this case, we can rearrange the given line equation to this form:
8x + 6y - 1 = 0
6y = -8x + 1
y = (-4/3)x + 1/6
The line equation is y = (-4/3)x + 1/6, so the slope (m) of this line is -4/3.
The perpendicular distance (d) from a point (x₁, y₁) to a line y = mx + c is given by the formula:
d = |y₁ - mx₁ - c| / sqrt(1 + m²)
In this case, we want the distance to be 5 units. So, let's substitute these values into the formula and solve for the locus.
5 = |y - (-4/3)x - 1/6| / sqrt(1 + (-4/3)²)
Simplifying the equation, we have:
5 = |y + (4/3)x - 1/6| / sqrt(1 + 16/9)
5 = |9y + 12x - 1| / sqrt(25/9)
5 = |9y + 12x - 1| / (5/3)
5 * (5/3) = |9y + 12x - 1|
25/3 = |9y + 12x - 1|
Now we have two cases to consider, based on the absolute value:
Case 1: 9y + 12x - 1 = 25/3
Solving this equation, we get:
9y + 12x = 82/3
Case 2: 9y + 12x - 1 = -25/3
Solving this equation, we get:
9y + 12x = -47/3
Both of these equations represent the locus of point P(x, y) that moves along the lines, maintaining a perpendicular distance of 5 units from the original line 8x + 6y - 1 = 0.