Perpendicular straight lines

Two straight lines $r$ and $s$ are perpendicular if and only if the angle between them is of ${90}^{\circ }$. This is equivalent to the fact that the cosine of the angle is equal to $0$ ($\cos \widehat{(r,s)}=0$) and so that the scalar product of its director vectors is equal to $0$.

If we have the straight lines $Ax+By+C=0$ and $A^{\prime }x+B^{\prime }y+C^{\prime }=0$,the director vectors of the above mentioned straight lines are $\overrightarrow{u}=(-B,A)$ and $\overrightarrow{v}=(-B^{\prime },A^{\prime })$.

Therefore, if in coordinates we impose that the scalar product of two vectors is $0$ we have: $$\overrightarrow{u}\cdot \overrightarrow{v}=0\iff u_1\cdot v_1+u_2\cdot v_2=0\iff -B\cdot (-B^{\prime })+A\cdot A^{\prime }=0\iff$$ $$\iff B\cdot B^{\prime }+A\cdot A^{\prime }=0\iff A\cdot A^{\prime }=-B·B^{\prime }\iff {\displaystyle \frac{A}{B}=-\frac{B^{\prime }}{A^{\prime }}}$$

Therefore we already have a way of verifying if two vectors, and therefore two straight lines, are perpendicular to its components.

If we remember as well that ${\displaystyle m_1=\frac{-A}{B}}$ and ${\displaystyle m_2=\frac{-A^{\prime }}{B^{\prime }}}$ are the slopes of $r$ and $s$, then the perpendicularity condition is equivalent to: $m_1={\displaystyle -\frac{1}{m_2}}$

Let's remember finally that if we have a vector $\overrightarrow{v}=(v_1,v_2)$, a $\overrightarrow{w}$ perpendicular to $\overrightarrow{v}$ is $\overrightarrow{w}=(-v_2,v_1)$.