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 $$A'x + B'y +C '= 0$$,the director vectors of the above mentioned straight lines are $$\overrightarrow{u}= (-B, A)$$ and $$\overrightarrow{v}= (-B', A ')$$.

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

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 $$\displaystyle m_2=\frac{-A'}{B'}$$ 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 $$\overrightarrow{w}=(-v_2,v_1)$$.