Apply distance formula and substitute y in terms of x from the equation of the given line in the distance formula. Quadratic equation will be formed in x. Solve to get 2 values of x and then find two values of y.

Points on the line 3x - 4y - 1 = 0 at a distance of 5 units from the point (3,2)

To find the points on the line 3x - 4y - 1 = 0 that are at a distance of 5 units from the point (3,2), we can use the distance formula and solve for the unknowns.

Step 1: Write the equation of the line in slope-intercept form
To make the calculations easier, let's rewrite the equation of the line 3x - 4y - 1 = 0 in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

3x - 4y - 1 = 0
-4y = -3x + 1
y = (3/4)x - 1/4

So, the equation of the line in slope-intercept form is y = (3/4)x - 1/4.

Step 2: Find the equation of the perpendicular line passing through (3,2)
To find the equation of the perpendicular line passing through the point (3,2), we first need to determine the slope of the given line. The slope of the given line is the negative reciprocal of (3/4), which is -4/3.

Using the point-slope form of a line, we can write the equation of the perpendicular line passing through (3,2):

y - 2 = (-4/3)(x - 3)
3y - 6 = -4x + 12
3y = -4x + 18
y = (-4/3)x + 6

So, the equation of the perpendicular line passing through (3,2) is y = (-4/3)x + 6.

Step 3: Find the intersection points of the two lines
To find the intersection points of the two lines, we can solve the system of equations:

y = (3/4)x - 1/4
y = (-4/3)x + 6

Solving these equations simultaneously, we have:

(3/4)x - 1/4 = (-4/3)x + 6
Multiply through by 12 to eliminate fractions:
9x - 3 = -16x + 72
25x = 75
x = 3

Substituting the value of x back into either equation, we have:

y = (3/4)(3) - 1/4
y = 9/4 - 1/4
y = 8/4
y = 2

So, the intersection point of the two lines is (3,2).

Step 4: Find the points on the line that are 5 units away from (3,2)
Now that we have the intersection point, we can find the points on the line 3x - 4y - 1 = 0 that are 5 units away from (3,2). We can use the distance formula:

d = √[(x2 - x1)^2 + (y2 - y1)^2]

Given that the distance is 5 units, we can substitute the values of (x