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 $\overrightarrow{w}=(-v_2,v_1)$.

Find the equation of the perpendicular straight line to $r: y = 2x - 5$ that crosses point $A = (1, 2)$

The given straight line has slope $m = 2$. Therefore we want a straight line with slope $\displaystyle m' =-\frac{1}{2}$.

This way, using the equation slope-point we will have that the straight line that we are looking for is: $$y - 2 = \displaystyle -\frac{1}{2}(x - 1)$$