Two lines r = Ax + By + C and s = Dx + Ey + F will be perpendicular, if their slopes m_{r} ym_{s} they are inverse and changed sign, that is:

m_{r} = - 1 / m_{s}

The above is known as the **theorem on slopes of perpendicular lines**.

**Theorem on slopes of perpendicular lines**

Two lines with slope m_{r} ym_{s} they are perpendicular if and only if:

m_{r} M_{s }= - 1 om_{r} = - 1 / m_{s}

**Examples**:

- Find a perpendicular line ar = x - 2y - 1 = 0 y that passes through point A (2,6).

The slope of r = x - 2y - 1 = 0 is:

2y = x - 1

y = 1 / 2x - 1/2

y = 1/2 (x - 1)

y = m_{r}∙ (x - 1)

m_{r }= 1/2

As m_{r} M_{s }= - 1, we have to:

m_{s }= – 2

Therefore, the equation of the perpendicular line ary that passes through that passes A (2,6), will be:

y - 6 = - 2 (x - 2)

2x + y - 10 = 0

- Find the equation of the perpendicular line ar = 6x + 4y - 1 = 0 and that passes through point A (-1, -3).

The slope of r = 6x + 4y - 1 = 0 is:

4y = - 6x + 1

y = - 3 / 2x + ¼

y = -3/2 (x - 1/6)

m_{r }= -3/2

As m_{r} M_{s }= - 1, we have to:

m_{s }= 2/3

Therefore, the equation of the perpendicular line ary that passes through that passes A (-1, -3), will be:

y + 3 = 2/3 (x + 1)

2x - 3y - 7 = 0

**Demonstration**

Let be two lines y = m_{r}x and y = m_{s}x that pass through the origin. When we draw the vertical line x = 1, we see that it cuts to:

- r at point A (x, m
_{r}x) = (1, m_{r}) - s in B (x, m
_{s}x) = (1, m_{s}).

These will be perpendicular if and only if the angle 0AB is right (90 °).

Applying the Pythagorean theorem, we have:

| AB |^{2 }= | 0A |^{2 }+ | 0B |^{2}

Where:

| 0A | = √ [1 + (m_{r})^{2}]

| 0B | = √ [1 + (- m_{s})^{2}] = √ [1+ (m_{s})^{2}]

| AB | = m_{r }+ (- m_{s}) = m_{r }- m_{s}

Thus:

(m_{r }- m_{s})^{2 }= {√ [1 + (m_{r})^{2}]}^{2}+ {√ [1+ (m_{s})^{2}]}^{2 }

(m_{r})^{2} - 2 m_{r}m_{s }+ (m_{s})^{2 }= 1 + (m_{r})^{2 }+ 1 (m_{s})^{2 }

- 2 m_{r}m_{s }= 2

m_{r}m_{s }= – 1

What we wanted to demonstrate