Concerning $m_1m_2=-1$

Question

Given two straight lines $$y=m_1x+c$$ and $$y=m_2x+d$$

When two lines are perpendicular to each other the product of their gradients, $m_1m_2=-1.$

How can I show that $$m_1m_2=-1?$$

Answer 1

Let $l_1$ denote the first line $y = m_1x+c$, and let $l_2$ denote the second line $y = m_2x+d$. Whether these two lines are perpendicular or not depends only on their slopes, so in the following, we may set $c=d=0$.

Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be two points on the line $l_1$. Then $$m_1 = \frac{y_2-y_1}{x_2-x_1}.$$ Assume that the two lines are perpendicular. Then $P' = (-y_1,x_1)$ and $Q'=(-y_2,x_2)$ are points on the line $l_2$ (the points have been rotated $90^{\circ}$ counter-clockwise about the origin).

The slope of line $l_2$ is then seen to be $$m_2 = \frac{x_2-x_1}{-y_2-(-y_1)}=\frac{x_2-x_1}{y_1-y_2} = -1/m_1.$$ Thus $$m_1m_2 = -1.$$

Answer 2

By direction vectors

• $y=m_1x+c\implies v=(1,m_1)$
• $y=m_2x+c\implies w=(1,m_2)$

then by dot product

$$v\cdot w = 1+m_1m_2=0\implies m_1m_2=-1$$

Or as an alternative

• $m_1= \tan \theta_1$
• $m_2= \tan \theta_2=\tan (\theta_1+\pi/2)= -\cot \theta_1=-\frac 1 {\tan \theta_1}$

then

$$m_1m_2=\tan \theta_1\left(-\frac 1 {\tan \theta_1}\right)=-1$$

Answer 3

The angle $\theta$ between 2 lines is given by $$\tan(\theta)=|\frac{m_1-m_2}{1+m_1m_2}|$$ Suppose $m_1m_2\ne -1$, then $\tan(\theta)$ is defined and hence $\theta$ is not $90^{\circ}$ which is a contradiction. Thus we must have $m_1m_2=-1$.

Answer 4

Think about rotating line $y=m_1x+c$ 90° clockwise about the origin. This will result in $x$ becoming $-y$ and $y$ becoming $x$ so the equation of the rotated line will be $-x=m_1y+c$ or $y=-\frac{1}{m_1}x-\frac{c}{m_1}$.

Answer 5

Similar to Gimusi's answer, hopefully a bit different :

$y=m_1x +c_1$, or

$m_1x -y+c_1=0$.

Normal to $f(x,y) = m_1x-y +c_1=0:$

$\vec n = (m_1, -1,0)$

Line with direction vector $\vec n$:

$\vec r (t) =t \vec n + \vec r_0$.

Let $\vec r_0=(0,0,0)$.

$x(t)= t m_1, y(t)= -t1 $.

$y = -x/m_1.$