Recall the gradient (slope) of a line is a measure of how steep the line is, is often represented by the letter $m$m and has the following properties:
We can see that between any two points on a graph we can form a right-angled triangle with a horizontal distance or 'run' and vertical distance or 'rise' and we can use two points to calculate the gradient using the following formula:
$m$m | $=$= | $\frac{rise}{run}$riserun |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
Another way to describe the steepness of the line is by using the angle the line is inclined at, that is the angle the line makes with the positive direction of the $x$x-axis. For the line shown below, $\theta$θ is the angle of inclination.
If we wanted to find the gradient, we could calculate:
$m$m | $=$= | $\frac{rise}{run}$riserun |
$=$= | $\frac{AB}{CB}$ABCB |
Look at angle $\theta$θ, what is $\tan\theta$tanθ?
$\tan\theta$tanθ | $=$= | $\frac{Opposite}{Adjacent}$OppositeAdjacent |
$=$= | $\frac{AB}{CB}$ABCB |
This is the value of the gradient we worked out just before! Hence:
$\tan\theta$tanθ | $=$= | $m$m |
Find the angle the line $y=3x+2$y=3x+2 makes with the positive direction of the $x-axis$x−axis.
Does the value of the $y$y-intercept effect the angle of inclination? No, all lines with the same gradient are parallel and make the same angle with the $x$x-axis. Illustrated below.
So from the equation we just need the value of the gradient, $m=3$m=3, and use this in the equation $\tan\theta=m$tanθ=m, as follows:
$\tan\theta$tanθ | $=$= | $m$m |
$\tan\theta$tanθ | $=$= | $3$3 |
$\theta$θ | $=$= | $\tan^{-1}\left(3\right)$tan−1(3) |
$\theta$θ | $=$= | $71.565^{\circ}$71.565∘ (to 3 decimal places) |
So the angle of inclination is approximately $71.565^{\circ}$71.565∘.
Find the gradient of the line that has an angle of inclination of $120^\circ$120°.
Can you picture this line? Should the gradient be positive or negative? What happens to $\tan\theta$tanθ outside the first quadrant? A line that makes an angle of $120^\circ$120° with the positive direction of the $x$x-axis would be the same as making an angle of $-60^\circ$−60° or $300^\circ$300° or $480^\circ$480°. Try using your calculator to find the gradient for each of these angles like shown below:
$m$m | $=$= | $\tan\theta$tanθ |
$=$= | $\tan\left(120^\circ\right)$tan(120°) | |
$=$= | $-\sqrt{3}$−√3 |
So the gradient of the line is $-\sqrt{3}$−√3.
Did you get the same gradient for each of the angles? Our equation does hold for all angles and we will look deeper into this in Chapter 5, when we explore the unit circle, but in fact all the trigonometric ratios can be extended for angles larger than $90^\circ$90° and for negative angles.
Since many angles give the same gradient, what angle does the calculator return when you use the formula to find the angle of inclination of a line with gradient $-\sqrt{3}$−√3? Does the calculator always give angles in a certain range? Experiment and find out.
Find the gradient of a line that is inclined at an angle of $50$50° to the positive $x$x-axis . Write your answer correct to one decimal place.
The given line has an angle of inclination of $\theta$θ with the positive $x$x-axis.
Determine the gradient of the line.
Hence solve for $\theta$θ. Give your answer in degrees to two decimal places.
Two points $A$A$\left(1,-2\right)$(1,−2) and $B$B$\left(9,30\right)$(9,30) lie on a line that makes an angle of $\theta$θ degrees with the positive $x$x-axis.
Determine the gradient $m$m of the line.
Find $\theta$θ in degrees to two decimal places.