What is the equation of a line which is parallel to the line 3x - 4y = 7 and is at a distance of 2 units from it?
Correct: A
The equation of a line parallel to 3x - 4y = 7 will have the same slope, hence it can be written as 3x - 4y = k. The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by |c1 - c2| / sqrt(a^2 + b^2). Here, the given line is 3x - 4y - 7 = 0, so a = 3, b = -4, and c1 = -7. The line we're looking for is 3x - 4y - k = 0, so c2 = -k. The distance formula gives us |(-7) - (-k)| / sqrt(3^2 + (-4)^2) = 2. This simplifies to |k - 7| / 5 = 2, which gives us k - 7 = ±10. For k - 7 = 10, k = 17 and for k - 7 = -10, k = -3. So, the equations are 3x - 4y = 17 and 3x - 4y = -3, which can be rearranged as 3x - 4y = 7 + 10 and 3x - 4y = 7 - 10, thus 3x - 4y = 7 + 8/5 is not an option but 3x - 4y = 7 + 10 or 3x - 4y = 7 - 10 are, none of which exactly match the given choices but demonstrate the method.