If real numbers a and b satisfy a^2 + b^2 = 64, what is the distance between the lines ax + by = 1 and ax + by = 417 ?
To find the distance between two parallel lines, we can consider any point on one line and find the perpendicular distance to the other line. Let's first find the slope of the lines simultaneously:
The equations of the lines are:
Line 1: ax + by = 1
Line 2: ax + by = 417
Rewriting these equations in slope-intercept form (y = mx + c), we get:
Line 1: y = -(a/b)x + 1/b
Line 2: y = -(a/b)x + 417/b
The slope of both lines is -a/b since the coefficient of x is the same in both equations.
Now, let's consider a point (x0, y0) on Line 1. The distance between this point and Line 2 can be found using the formula for the perpendicular distance between a point and a line.
The formula for the perpendicular distance (d) between a point (x0, y0) and a line with equation ax + by + c = 0 is given by:
d = |ax0 + by0 + c| / √(a^2 + b^2)
For Line 1, a = -a/b, b = 1, and c = 1/b. Substituting these values into the formula, we get:
d = |-a(x0) + y0/b + 1/b| / √((-a/b)^2 + 1^2)
Since we have a^2 + b^2 = 64, we can simplify the formula further:
d = |-a(x0) + y0/b + 1/b| / √(64/b^2 + 1)
Note that we are assuming b ≠ 0, as we cannot divide by zero.
Hence, the distance between the lines ax + by = 1 and ax + by = 417 is given by the formula:
d = |-a(x0) + y0/b + 1/b| / √(64/b^2 + 1)
To find the distance, we need to locate a specific point on Line 1. Since we are dealing with real numbers, we can choose any real value for x0 and find the corresponding y0 using the equation of Line 1. Let's say we choose x0 = 0, then y0 = 1/b.
Substituting these values in the formula, we get:
d = |-a(0) + (1/b)/b + 1/b| / √(64/b^2 + 1)
= |1/b + 1/b| / √(64/b^2 + 1)
= 2/b / √(64/b^2 + 1)
= 2√(b^2) / b√(64 + b^2)
= 2|b| / √(64 + b^2)
Therefore, the distance between the lines ax + by = 1 and ax + by = 417 is given by:
d = 2|b| / √(64 + b^2)
Note that the distance will depend on the value of b.