Finding Gradients

The gradient of a line that passes through two known points, say $\left(x_1,y_1\right)$(x1​,y1​) and $\left(x_2,y_2\right)$(x2​,y2​) on the cartesian plane can be found easily.  Gradient is a measure of steepness. It is the ratio of a line's rising (or falling) to its running. 

If, over a distance of $8$8 metres, a driveway rises $2$2 metres, then its gradient is said to be the ratio $\frac{2}{8}=\frac{1}{4}=0.25$28​=14​=0.25. It is also defined as the tangent of the angle of rise as shown in this simple diagram.

Consider the following example of a line passing through two points $\left(-3,7\right)$(−3,7) and $\left(5,9\right)$(5,9) as shown here:

Looking at the two $y$y values of the two points, the rise is clearly $2$2. We could either use a formula for rise which might look like $Rise=y_2-y_1=9-7=2$Rise=y2​−y1​=9−7=2 or simply notice that there is a gap of $2$2 between the two values. 

Looking at the two $x$x values we could again either use the formula $Run=x_2-x_1=5-\left(-3\right)=8$Run=x2​−x1​=5−(−3)=8 or simply notice that the gap between $-3$−3 and $5$5 is $8$8. 

We also realise that the line is rising and this means that the gradient is positive. 

The gradient, often denoted by the letter $m$m is simply the ratio given by:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{2}{8}=0.25$m=y2​−y1​x2​−x1​​=28​=0.25

From the fact that $\tan\theta=0.25$tanθ=0.25 we can use a scientific calculator to show that $\theta=\tan^{-1}\left(0.25\right)=14^\circ2'$θ=tan−1(0.25)=14°2′, which gives some sense to the steepness of the rise.

Note that if the line is falling, the line's gradient will be negative. In such cases the acute angle the line makes with the  $x$x axis will be shown on the calculator as a negative angle. Adding $180^\circ$180°  to this will reveal the obtuse angle of inclination the line makes with the axis.

For example if the gradient was given by $m=-0.25$m=−0.25, then $\theta=180^\circ+\tan^{-1}\left(-0.25\right)$θ=180°+tan−1(−0.25), which simplifies to $\theta=165^\circ58'$θ=165°58′. 

Suppose we consider the line given by $5x-2y=20$5x−2y=20. By putting $x=0$x=0 we see that $y=-10$y=−10 (note the negative sign here). Also, by putting $y=0$y=0, we find that $x=4$x=4. This means that the $x$x and $y$y intercepts are $4$4 and $-10$−10 respectively.  The situation is shown here:

Note that the rise and run can be determined from the $x$x and $y$y intercepts. The positive gradient of the line shown is given as $m=\frac{10}{4}=2.5$m=104​=2.5. 

 

Worked Examples

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Question 4