# Gradients of Straight Line Graphs

GCSELevel 4-5Level 6-7Cambridge iGCSE

## Gradients of Straight Line Graphs

You will need to be able to work out how steep straights lines on an $x$$y$ graph are.

What a gradient specifically tells us is the rate of change of $y$ with respect to $x$. This is usually denoted, for a straight line, as a constant $m$. The higher the value of $m$ the “steeper” the line is. The closer to $0$, $m$ is, the shallower the slope.

As the gradient is the rate of change of $y$ with respect to $x$, if the gradient is $m$, then for every $1$ unit that $x$ increases, $y$ will increase by $m$ units.

If $m$ is negative, that means that while $x$ is increasing, $y$ is decreasing ($m=-1$ pictured).

Gradients can also be fractions, so if the gradient of a line is $\dfrac{2}{3}$, then for every increase of $3$ to $x$, $y$ is increased by $2$.

Level 4-5GCSECambridge iGCSE

## Finding the Gradient – Graphically

In order to find the gradient we must find the difference in $y$, denoted $\Delta y$, and the difference in $x$, denoted $\Delta x$. So, for gradient $m$:

$m$ $= \dfrac{\text{Difference in } y}{\text{Difference in } x}=$$\dfrac{\Delta y}{\Delta x}$

There are two ways we need to be able to find $\Delta y$ and $\Delta x$.

Firstly, graphically. The graph on the left shows how the differences can be interpreted. We can draw a right angled triangle that has the straight line as it’s hypotenuse. The width of this triangle is $\Delta x$, and it’s height is $\Delta y$.

So if each square in the graph is $1$ unit by $1$ unit, then $\Delta y$ $= 4$ and $\Delta x$$= 2$. Which would give us a gradient of $\dfrac{4}{2} = 2$.

Level 4-5GCSECambridge iGCSE

## Finding the Gradient – Given 2 Points

The other way we must be able to find the gradient of a straight line is by having points with coordinates.

Between any two points in the $x$$y$ plane, we can draw a straight line that goes between them. So we can use the coordinates of those two points to find $\Delta y$ and $\Delta x$, and as such, the gradient. So given two points on a straight line $(x_1, y_1)$ and $(x_2, y_2)$, then

$\Delta y$ $=$ $y_2$ $-$ $y_1$
$\Delta x$ $=$ $x_2$ $-$ $x_1$

and so for gradient $m$,

$m$ $=$ $\dfrac{\Delta y}{\Delta x}$ $=$ $\dfrac{y_2-y_1}{x_2-x_1}$.

Level 6-7GCSECambridge iGCSE

## Example 1: Finding a Gradient Graphically

Find the gradient of the red line.

[2 marks]

We can draw on our triangle, as shown by the dotted lines. The line passes through the points $(1,3)$ and $(4,5)$ so we can draw our triangle there.

We can count the units and see the triangle is $3$ units wide and $2$ units tall. This gives us

$\Delta y = 2$
$\Delta x = 3$

So the gradient, $m$ is

$m = \dfrac{\Delta y}{\Delta x} = \dfrac{2}{3}$.

Level 4-5GCSECambridge iGCSE

## Example 2: Finding a Gradient from Coordinates

A straight line passes through the points $(1,1)$ and $(0, 13)$.

Find the gradient to this straight line.

[2 marks]

We shall first find

$\Delta y = 13 - 1 = 12$
$\Delta x= 0 - 1 = -1$.

So the gradient $m$ is

$m = \dfrac{\Delta y}{\Delta x}= \dfrac{12}{-1} = -12$.

Level 6-7GCSECambridge iGCSE

## Gradients of Straight Line Graphs Example Questions

Drawing a triangle from two points on the line, we can see

which gives us
$\Delta y = 2$
$\Delta x = -1$

So,

$m=\dfrac{2}{-1}=-2$

Drawing a triangle from two points on the line, we can see

which gives us
$\Delta y = 4$
$\Delta x = 0.4$

So,

$m=\dfrac{4}{0.4}=10$

We know

$m=\dfrac{\Delta y}{\Delta x}$

So we must find $\Delta y$ and $\Delta x$

$\Delta y = 71-3.5= 67.5$
$\Delta x = 10-1=9$

So

$m=\dfrac{67.5}{9}=\dfrac{15}{2} = 7.5$

We know

$m=\dfrac{\Delta y}{\Delta x}$

So we must find $\Delta y$ and $\Delta x$

$\Delta y = 1-2= -1$
$\Delta x = 32-16=16$

So

$m=-\dfrac{1}{16}$.

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