Perpendicular equations

Perpendicular equationsTwo 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

demonstration perpendicular lines

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:

demonstration perpendicular lines| 0A | = √ [1 + (mr)2]

demonstration perpendicular lines| 0B | = √ [1 + (- ms)2] = √ [1+ (ms)2]

demonstration perpendicular lines| 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