If a,b, and c are real numbers...?

Question

If a,b, and c are real numbers, what is the only value of a for which the graphs of y = ax+b and x=ay+c are perpendicular to each other?

Answer 1

two lines are perpendicular when the product of their slopes is -1

if you can write a line in the form:

y=mx+b

the slope is m, the coefficient of x

so, for the line y=ax+b the slope is a

for the line x=ay+c, we rearrange to write:

y=1/a x -c/a  and the slope is 1/a

the product of a x 1/a =1 and is never -1, so there is no value of a for which these lines are perp

Answer 2

0

y = 0x + b = b = constant ──► Horizontal line

x = 0y + c = c = constant ──► Vertical line

Since there is a uniqueness condition (in the problem statement), if you find 1 solution, you've found the ONLY solution.

EDIT: Kuiperbelt2003 got it right, for the most part.  His or her mistake was in this step:

x = ay + c ──► y = (x-c) / a

Because the coefficients are real numbers, any one CAN be 0.  However, dividing by zero is an undefined operation, so the equation  y=(x-c)/a applies only to the nonzero real numbers.