# Perpendicular equations Two lines r = Ax + By + C and s = Dx + Ey + F will be perpendicular, if their slopes mr yms they are inverse and changed sign, that is:

mr = - 1 / ms

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

## Theorem on slopes of perpendicular lines

Two lines with slope mr yms they are perpendicular if and only if:

mr Ms = - 1 omr = - 1 / ms

Examples:

1. 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 = mr∙ (x - 1)

mr = 1/2

As mr Ms = - 1, we have to:

ms = – 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

1. 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)

mr = -3/2

As mr Ms = - 1, we have to:

ms = 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 = mrx and y = msx that pass through the origin. When we draw the vertical line x = 1, we see that it cuts to:

• r at point A (x, mrx) = (1, mr)
• s in B (x, msx) = (1, ms).

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 + (mr)2] | 0B | = √ [1 + (- ms)2] = √ [1+ (ms)2] | AB | = mr + (- ms) = mr - ms

Thus:

(mr - ms)2 = {√ [1 + (mr)2]}2+ {√ [1+ (ms)2]}2

(mr)2 - 2 mrms + (ms)2 = 1 + (mr)2 + 1 (ms)2

- 2 mrms =  2

mrms =  – 1

What we wanted to demonstrate