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.

Gradient

Hmm. I can see, from Matt's comment below, that I need to be a bit more specific. well it's early doors and though I'm giving it 110%, it's not always easy to see where I should pitch or focus this blog. I'm sure Matt will be delighted to know that I'm just going to let it evolve...

But a word on gradient.

Take a graph.

Any graph.

Put a straight line in it.

Any angle.



Take the left hand side of the line. What are its co-ordinates? In the picture above, the line begins at point (1,1). From that point, call it A, the line goes along and up. As you move along the line its co-ordinates rise in value. It moves along the x-axis and the y-axis.

Gradient is simply a measure of how y changes as you change x values. In other words, steepness.

Look at point B. It is at (5,4). From A to B you have gone 4 x-units along and 3 y-units up. On this line, wherever you are on the graph, if you go 4 x units along you will ALWAYS go 3 y units up.

Gradient, which is a general statement of this quantity, is calculated through this difference between the co-ordinates at A and B.

We divide the difference in y across the two points, by the difference in x.

y2-y1/x2-x1, meaning: the y co-ordinate of point B minus the y co-ordinate of point A, then divided by the same thing done for the x co-ordinates.

In this case,

= 4-1/5-1

= 3/4

The gradient is "three quarters".

In the picture below the line is slanting downwards. This will give it a negative gradient. Same method though.



Horizontal lines have a gradient of 0.

Vertical lines are a tricky one. If you think of a line going straight up and down, and try to apply this logic to it, you end up with a weird division.

Say you've got a vertical line through points (5,0) and up to (5, 5). If you do the y2-y1/x2-x1 calculation to find the gradient, you end up doing 5/0. You can't divide by 0, so you can't get a result. There isn't a gradient for a vertical line.

Another way of looking at it is that a vertical line cannot be a result of y = f(x) because it would be one-many - ie it's not a function.