### Theory:

Let $$l_1$$ and $$l_2$$ be two non-vertical lines.

The slope of line $$l_1$$ is $$m_1$$, and line $$l_2$$ is $$m_2$$.

Let the inclination of $$l_1$$ be $$\theta_1$$ and $$l_2$$ be $$\theta_2$$.

Then, $$m_1 = \tan \theta_1$$ and $$m_2 = \tan \theta_2$$.

Assume $$l_1$$ and $$l_2$$ are perpendicular lines.

In $$\Delta ABC$$:

$$\angle A = \theta_1$$ and $$\angle C = 90^\circ$$

Then, $$\angle B = 180^\circ - \angle A - \angle C$$ [by Angle sum property].

$$\angle B = 180^\circ - \theta_1 - 90^\circ$$

$$\angle B =90^\circ - \theta_1$$

Measuring slope of $$l_2$$ through angle $$\theta_2$$ and $$90^\circ - \theta_1$$ are opposite of each other.

$$\Rightarrow \tan \theta_2 = - \tan (90^\circ - \theta_1)$$

$$\Rightarrow \tan \theta_2 = - \frac{\sin(90^\circ - \theta_1)}{\cos(90^\circ - \theta_1)}$$

[Using the trigonometric identities $$\sin (90^\circ - A) = \cos A$$ and $$\cos (90^\circ - A) = \sin A$$]

$$\Rightarrow \tan \theta_2 = - \frac{\cos \theta_1}{\sin \theta_1}$$

$$\Rightarrow \tan \theta_2 = - \cot \theta_1$$

$$\Rightarrow \tan \theta_2 = - \frac{1}{\tan \theta_1}$$

$$\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1$$

$$\Rightarrow m_1 \cdot m_2 = -1$$

Conversely:

Let $$l_1$$ and $$l_2$$ be two non-vertical lines with slopes $$m_1$$ and $$m_2$$, respectively, such that $$m_1 \cdot m_2 = -1$$.

$$\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1$$

$$\Rightarrow \tan \theta_1 = - \frac{1}{\tan \theta_2}$$

$$\Rightarrow \tan \theta_1 = - \cot \theta_2$$

$$\Rightarrow \tan \theta_1 = - \tan (90^\circ - \theta_2)$$ [Using $$\cot (90^\circ - A) = \tan A$$]

$$\Rightarrow \tan \theta_1 = \tan (-(90^\circ - \theta_2))$$ [Using $$\tan (-A) = - \tan A$$]

$$\Rightarrow \tan \theta_1 = \tan (\theta_2 - 90^\circ)$$

$$\Rightarrow \theta_1 = \theta_2 - 90^\circ$$ [since $$0^\circ \le \theta_1, \theta_2 \le 180^\circ$$]

$$\Rightarrow \theta_2 = \theta_1 + 90^\circ$$ - - - - - (I)

An exterior angle of a triangle is equal to the sum of two opposite interior angles.

$$\Rightarrow \theta_2 = \angle A + \angle C$$

$$\Rightarrow \theta_2 = \theta_1 + \angle C$$ - - - - - (II)

On comparing (I) and (II), we get:

$$\angle C = 90^\circ$$

Thus, the lines $$l_1$$ and $$l_2$$ are perpendicular.
Two lines are perpendicular if and only if the slopes are negative reciprocal of each other. That is, $$m_1 \cdot m_2 = -1$$.
Example:
The slope of the line $$p$$ is $\frac{-20}{5}$, and the slope of the line $$q$$ is $\frac{8}{32}$. Are the lines $$p$$ and $$q$$ are perpendicular?

Solution:

Let $$m_1$$ be the slope of $$p$$, and $$m_2$$ be the slope of $$q$$.

If $$m_1 \cdot m_2 = -1$$, then the lines are perpendicular.

$$m_1 =$$ $\frac{-20}{5}$ $$=$$ $$-4$$

$$m_2 =$$ $\frac{8}{32}$ $$=$$ $\frac{1}{4}$

$$m_1 \cdot m_2 = - 4 \ \times$$ $\frac{1}{4}$

$$m_1 \cdot m_2 = - 1$$

Therefore, $$p$$ and $$q$$ are perpendicular lines.