兩條直線斜率分別是l1:y=k1x+b1,l2:y=k2x+b2,並且k1k2=-1,k1<0,k2>0

與x軸正半軸的夾角θ1有tanθ1=k1<0,θ1∈(π/2,π)

l2與x軸正半軸的夾角θ2有tanθ2=k2>0,θ2∈(0,π/2)

θ1-θ2∈(0,π)

tanθ1tanθ2=-1

tan(θ1-θ2)=(tanθ1-tanθ2)/(1+tanθ1tanθ2)=∞

θ1-θ2=π/2

P(x,y)A(a,b),B(c,d)分別在l1,l2上且都不是P點

[PA]=(a-x,b-y)

[PB]=(c-x,d-y)

[PA].[PB]=(a-x)(c-x)+(b-y)(d-y)

由於k1=(b-y)/(a-x);k2=(d-y)/(c-x)

k1k2=(b-y)(d-y)/(a-x)(c-x)=-1

(b-y)(d-y)=(a-x)(c-x)

因而[PA].[PB]=0