Gradient (or slope) of a Line, and Inclination

Application of slope of a line - road gradient

Application: Road sign, indicating a steep gradient.
A

15 %

road gradient is equivalent to

m = 0.15

The gradient (also known as slope) of a line is defined as

$g r a d i e n t = h o r i z o n t a l r u n v e r t i c a l r i s e $

In the following triangle, the gradient of the line is given by:

​ b ​ ​ a ​ ​

In general, for the line joining the points (x₁, y₁) and (x₂,y₂), we have:
slope of a line diagram

We can now write the fomula for the slope of a line.

Gradient of a Line Formula

We see from the diagram above, that the gradient (usually written m) is given by:

$m = x 2 - x 1 y 2 - y 1 $

Interactive graph - slope of a line

You can explore the concept of slope of a line in the following JSXGraph (it's not a fixed image).
Drag either point A or point B to investigate how the gradient formula works. The numbers will update as you interact with the graph.
Notice what happens to the sign (plus or minus) of the slope when point B is above or below A.

0,0

– o + ← ↓ ↑ →

slope m
= rise / run
= 7 / 10
= 0.7

(-9, -4)

(1, 3)

(1, -4)

y₂ − y₁
= 3 − -4
= 7

x₂ − x₁
= 1 − -9
= 10

-5

-10

-5

You can move the graph up-down, left-right if you hold down the "Shift" key and then drag the graph.
Sometimes the explanation boxes overlap. It can't be helped!
If you get lost, you can always refresh the page.

Example

Find the slope of the line joining the points (-4, -1) and (2, -5).

Answer

Positive and Negative Slopes

In general, a positive slope indicates the value of the dependent variable increases as we go left to right:
graph of line with positive slope

[The dependent variable (usually x) in the above graph is the y-value.]

A negative slope means that the value of the dependent variable (usually y) is decreasing as we go left to right:
graph of line with negative slope

Inclination

We have a line with slope m and the angle that the line makes with thex-axis is α.
From trigonometry, we recall that the tan of angle α is given by:

$tan α = a d j a c e n t o p p o s i t e $

Now, since slope is also defined as opposite/adjacent, we have:
graph of inclination

This gives us the result:

tan α = m

Then we can find angle α using

α = arctan m
(That is, α = tan^-1m)

This angle α is called the inclination of the line.

Perpendicular Distance from a Point to a Line

Example using perpendicular distance formula

(BTW - we don't really need to say 'perpendicular' because the distance from a point to a line always means the shortest distance.)
This is a great problem because it uses all these things that we have learned so far:

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

$d = \sqrt A 2 + B 2 ∣ A m + B n + C ∣ $

There are some examples using this formula following the proof.

Proof of the Perpendicular Distance Formula

Let's start with the line Ax + By + C = 0 and label it DE. It has slope

- ​ B ​ ​ A ​ ​

We have a point P with coordinates (m, n). We wish to find the perpendicular distance from the point P to the line (that is, distance

P Q

We now do a trick to make things easier for ourselves (the algebra is really horrible otherwise). We construct a line parallel to DE through (m, n). This line will also have slope

- ​ B ​ ​ A ​ ​

, since it is parallel to DE. We will call this line FG.
perpendicular and parallel constructions

Now we construct another line parallel to PQ passing through the origin.
This line will have slope

​ A ​ ​ B ​ ​

, because it is perpendicular to DE.
Let's call it line RS. We extend it to the origin

(0, 0)

.
We will find the distance RS, which I hope you agree is equal to the distance PQ that we wanted at the start.
perpendicular through origin

Since FG passes through (m, n) and has slope

- ​ B ​ ​ A ​ ​

, its equation is

y - n = - ​ B ​ ​ A ​ ​ (x - m)

y = ​ B ​ ​ - A x + A m + B n ​ ​

.
Line RS has equation

y = ​ A ​ ​ B ​ ​ x .

Line FG intersects with line RS when

$ A B x = B - A x + A m + B n $

Solving this gives us

$x = A 2 + B 2 A (A m + B n) $

So after substituting this back into

y = ​ A ​ ​ B ​ ​ x

, we find that point R is

$( A 2 + B 2 A (A m + B n) , A 2 + B 2 B (A m + B n) )$

Point S is the intersection of the lines

y = ​ A ​ ​ B ​ ​ x

and Ax + By + C = 0,which can be written

y = - ​ B ​ ​ A x + C ​ ​

.This occurs when (that is, we are solving them simultaneously)

$- B A x + C = A B x$

Solving for x gives

$x = A 2 + B 2 - A C $

Finding y by substituting back into

$y = A B x$

gives

$y = A B ( A 2 + B 2 - A C ) = A 2 + B 2 - B C $

So S is the point

$( A 2 + B 2 - A C , A 2 + B 2 - B C )$

The distance RS, using the distance formula,

d = \sqrt ​ (x ​ 2 ​ ​ - x ​ 1 ​ ​) ​ 2 ​ ​ + (y ​ 2 ​ ​ - y ​ 1 ​ ​) ​ 2 ​ ​ ​ ​ ​

d = \sqrt ​ (​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ - A C ​ ​ - ​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ A (A m + B n) ​ ​) ​ 2 ​ ​ + (​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ - B C ​ ​ - ​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ B (A m + B n) ​ ​) ​ 2 ​ ​ ​ ​ ​

= \sqrt ​ ​ (A ​ 2 ​ ​ + B ​ 2 ​ ​) ​ 2 ​ ​ ​ ​ {- A (A m + B n + C)} ​ 2 ​ ​ + {- B (A m + B n + C)} ​ 2 ​ ​ ​ ​ ​ ​ ​

= \sqrt ​ ​ (A ​ 2 ​ ​ + B ​ 2 ​ ​) ​ 2 ​ ​ ​ ​ (A ​ 2 ​ ​ + B ​ 2 ​ ​) (A m + B n + C) ​ 2 ​ ​ ​ ​ ​ ​ ​

= \sqrt ​ ​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ (A m + B n + C) ​ 2 ​ ​ ​ ​ ​ ​ ​

= ​ \sqrt ​ A ​ 2 ​ ​ + B ​ 2 ​ ​ ​ ​ ​ ​ ​ ∣ A m + B n + C ∣ ​ ​

The absolute value sign is necessary since distance must be a positive value, and certain combinations of A, m , B, n and C can produce a negative number in the numerator.
So the distance from the point (m, n) to the line Ax + By + C = 0 is given by: