The Relation between the Slopes of Two Mutually Perpendicular Straight Lines

Question

In coordinate geometry, we know that the product of the slopes of two mutually perpendicular straight lines is generally equal to -1. But if we consider the slopes of these two lines y = k (parallel to the X-axis) and x = k' (parallel to the Y-axis), this relation does no longer hold. What is wrong? I have not found any satisfactory explanation. Can you help?

Answer 1

The lines $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ have slopes $m_1=-\dfrac{a_1}{b_1}$ and $m_2=-\dfrac{a_2}{b_2}$ respectively. As you have noticed, we have a problem when either $b_1=0$ or $b_2=0$.

Note that, when $b_1 \ne 0$ and $b_2 \ne 0$, then

\begin{align} m_1 \cdot m_2 = -1 &\implies -\dfrac{a_1}{b_1} \cdot -\dfrac{a_2}{b_2} = -1 \\ &\implies \dfrac{a_1 \cdot a_2}{b_1 \cdot b_2} = -1 \\ &\implies a_1 \cdot a_2 = -b_1 \cdot b_2 \\ &\implies a_1 \cdot a_2 + b_1 \cdot b_2 = 0 \end{align}

The nice thing about the equation $$a_1 \cdot a_2 + b_1 \cdot b_2 = 0$$

is that is also works when $a_1$ or $b_1$ or $a_2$ or $b_2$ is equal to $0$.

There is a geometrical interpretation of this.

Consider the line $ax+by=c$ which passes through ther point $(x_0, y_0)$. It can be shown that all points $(x,y)$ on that line sasify the vector equation $(x,y) = (x_0,y_0)+t(a,-b)$ where $t \in \mathbb R$ and the vector $\overrightarrow v = (a,-b)$ indicates the $``$direction$"$ of the line $ax+by=c$. It is known that two vectors are perpendicular if their dot product is $0$. Hence the two lines $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ are perpendicular if and only if $0 = (a_1,-b_1)\circ (a_2,-b_2) = a_1 \cdot a_2 + b_1 \cdot b_2$

There is a practical application of this property. The line perpendicular to the line $ax+by=c$ must be of the form $bx-ay=d$ for some number $d$ since $a\cdot b + b \cdot (-a) = 0$.

For example, the equation of the line passing through the point $(1,2)$ and perpendicular to the line $3x+4y=5$ is $4x-3y=4(1)-3(2)=-2$.