The product of the slopes is -1.
What are perpendicular lines?
Given that PQ contains (a,b) (c,d) and line P'Q' contains (-b,a), (-d,c)
Slope of a line = y₂ - y₁/x₂ - x₁
For line PQ,
slope = a -c/b - d
m₁ = a - c/b - d
For line P'Q'
slope = -b - (-d)/a - c
= -b + d/a - c
Factor out -1
m₂ = -(b - d)/a - c
Multiply the slopes together
m₁ * m₂ = a - c/b - d * -(b -d)/a - c
Cancel out the denominator and numerator
m₁m₂ = -1
The product of the slopes is -1.
Therefore, PQ and P'Q' are perpendicular lines.