Answer to Question #140633 in Geometry for Joe

Question

The product of the slopes of and the line perpendicular to through C is -1 in all cases. So, I can conclude that any two perpendicular lines have slopes that are negative reciprocals of each other.

Answer 1

As per the given question,

The gradient of line A $=\tan \theta$

The gradient of line B$=\tan \phi$

$\theta = 90+\phi$

$\tan\theta=\tan(90+\phi)$

$=\frac{\sin(90+\phi)}{\cos(90+\phi)}$

$=\frac{\cos(\phi)}{-\sin\phi}$

$=-\cot(\phi)$

Hence, $\tan \theta. \tan \phi =-1$

Hence we can say that the product of the slope of the perpendicular lines is -1.