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As students taking pre-calculus, you are probably aware that when two lines are perpendicular, the product of their slopes is -1. Let's see why this is.

Let one line be $\displaystyle y_1=m_1x+b_1$ and the other line be $\displaystyle y_2=m_2x+b_2$.

Now,we know the angle of inclination of a line is found from:

$\displaystyle m=tan(\theta)\,\therefore\,\theta=\tan^{-1}(m)$

Let $\displaystyle \theta_1$ be the angle of inclination of $\displaystyle y_1$ and $\displaystyle \theta_2$ be the angle of inclination of $\displaystyle y_2$.

Now, suppose $\displaystyle 0\le\theta_2\le\frac{\pi}{2}$ and $\displaystyle -\frac{\pi}{2}\le\theta_1\le0$.

If the two lines are perpendicular, then we must have:

$\displaystyle \theta_2=\theta_1+\frac{\pi}{2}$

Now, taking the tangent of both sides, we find:

$\displaystyle \tan\left(\theta_2 \right)=\tan\left(\theta_1+\frac{\pi}{2} \right)$

Using the identity $\displaystyle \tan\left(x+\frac{\pi}{2} \right)=-\cot(x)$ we have:

$\displaystyle \tan\left(\theta_2 \right)=-\cot\left(\theta_1 \right)$

$\displaystyle \tan\left(\theta_2 \right)=-\frac{1}{\tan\left(\theta_1 \right)}$

And so, we must then have:

$\displaystyle m_2=-\frac{1}{m_1}$

$\displaystyle m_1m_2=-1$

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...ndicular-lines-product-their-slopes-4218.html

Let one line be $\displaystyle y_1=m_1x+b_1$ and the other line be $\displaystyle y_2=m_2x+b_2$.

Now,we know the angle of inclination of a line is found from:

$\displaystyle m=tan(\theta)\,\therefore\,\theta=\tan^{-1}(m)$

Let $\displaystyle \theta_1$ be the angle of inclination of $\displaystyle y_1$ and $\displaystyle \theta_2$ be the angle of inclination of $\displaystyle y_2$.

Now, suppose $\displaystyle 0\le\theta_2\le\frac{\pi}{2}$ and $\displaystyle -\frac{\pi}{2}\le\theta_1\le0$.

If the two lines are perpendicular, then we must have:

$\displaystyle \theta_2=\theta_1+\frac{\pi}{2}$

Now, taking the tangent of both sides, we find:

$\displaystyle \tan\left(\theta_2 \right)=\tan\left(\theta_1+\frac{\pi}{2} \right)$

Using the identity $\displaystyle \tan\left(x+\frac{\pi}{2} \right)=-\cot(x)$ we have:

$\displaystyle \tan\left(\theta_2 \right)=-\cot\left(\theta_1 \right)$

$\displaystyle \tan\left(\theta_2 \right)=-\frac{1}{\tan\left(\theta_1 \right)}$

And so, we must then have:

$\displaystyle m_2=-\frac{1}{m_1}$

$\displaystyle m_1m_2=-1$

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...ndicular-lines-product-their-slopes-4218.html

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