The relation between a, b, a' and b' such that the two lines ax + by = c and a'x + b'y = c' are perpendicular is

Question

The relation between a, b, a' and b' such that the two lines ax + by = c and a'x + b'y = c' are perpendicular is (A) aa' - bb' = 0 (B) aa' + bb' = 0 (C) ab + a'b' = 0 (D) ab - a'b' = 0

Tardigrade Admin · Accepted Answer

Slope of the line ax + by = c is (-a/b), and the slope of the line a'x + b'y = c' is (-a'/b') The lines are perpendicular if ((-a/b))((-a'/b')) = - 1 or aa' + bb' = 0

Q. The relation between a, b, a' and b' such that the two lines ax + by = c and a'x + b'y = c' are perpendicular is

aa' - bb' = 0

aa' + bb' = 0

ab + a'b' = 0

ab - a'b' = 0

Slope of the line $a x + b y = c$ is $\frac{- a}{b}$ , and the slope of the line $a^{'} x + b^{'} y = c^{'}$ is $\frac{- a ^{'}}{b ^{'}}$
The lines are perpendicular if $(\frac{- a}{b}) (\frac{- a ^{'}}{b ^{'}}) = - 1$ or $a a^{'} + b b^{'} = 0$

Q. The relation between a, b, a' and b' such that the two lines ax + by = c and a'x + b'y = c' are perpendicular is

aa' - bb' = 0

aa' + bb' = 0

ab + a'b' = 0

ab - a'b' = 0

Slope of the line ax+by=c is b−a​, and the slope of the line a′x+b′y=c′ is b′−a′​ The lines are perpendicular if (b−a​)(b′−a′​)=−1 or aa′+bb′=0

Slope of the line $a x + b y = c$ is $\frac{- a}{b}$ , and the slope of the line $a^{'} x + b^{'} y = c^{'}$ is $\frac{- a ^{'}}{b ^{'}}$
The lines are perpendicular if $(\frac{- a}{b}) (\frac{- a ^{'}}{b ^{'}}) = - 1$ or $a a^{'} + b b^{'} = 0$