Gradients of Perpendicular Lines

If a line intersects another one such that the two are perpendicular,and the gradient of the other is known, it is a piece of cake to work out the unknown gradient.

The gradients of perpendicular lines have a product of -1.

Our line from the previous post has gradient 12-6/8-3 = 6/5 Its perpendicular bisector will therefore have gradient of (using the common letter m to stand for unknown gradient):

m₁xm2=-1

6/5xm₂ = -1

m₂= -1 ÷ 6/5

At this point use the fraction division rule:

m₂= -1 x 5/6

= -5/6

The gradient of the perpendicular bisector is -5/6.

The really cool bit is this. If you have two co-ordinates of the first line, say in the post below (3,6) and (8,12); then you can now easily find the equation of the perpendicular bisector.

If it bisects the first line it must pass through the mid-point: x1+x2/2, y1+y2/2

or 3+8/2, 6+12/2

which is: (5 1/2, 9)

Plug what you know of this line into y - y₁ = m(x - x₁).

Or: y - 9 = -5/6(x - 11/2)

I've turned 5 1/2 into 11/2 for ease of calculation.

y -9 = -1/3x + 55/12

y = -1/3x + 163/12

This is a bit messy so multiply through by 12:

12y = -4x + 163



Lovely.

NOTE: I updated this post because it was wrong: it is still a bit more tricky than most of the questions you will get at C1. Most of these type questions resolve into a line with a constant of no more than about 20. But theoretically one like this could come up.