Gradient of a Line
In Mathematics, the gradient of a line is indicated with the letter m. It is simply a measure of how steep the line is.
A real life example is when you are considering two mountains to climb. If one mountain is steeper than the other, it means that the steeper mountain is more difficult to climb than the less steep one.
The aim of this lesson is to learn how to calculate the gradient of a line and use the knowledge gained to find the equation of a line in our next lesson.
Since gradient is a measure of how steep a line is, the best place to start our lesson is the cartesian plane.
On the cartesian plane, in order to draw a line, you will need either one of these:
When two (2) lines are located on the cartesian plane, the gradient of the line is given as below:
\[Gradient, m = \frac{y_2 - y_1}{x_2 - x_1}\]
Where:
\(y\)2 \(-\) \(y\)1 \(\Rightarrow\) the change or difference in the \(y\) components of the two points or coordinates given, and
\(x\)2 \(-\) \(x\)1 \(\Rightarrow\) the change or difference in the \(x\) components of the two points or coordinates given.
This implies that:
\[Gradient, m = \frac{Change \ in \ y}{Change \ in \ x}\]
Let's use the formular we have just learnt to solve some questions.
Try and solve the questions yourself.
Example 1
Find the gradient of the line which passes through the points \(A(1, 1)\) and \(B(7, 2)\)
Solution
Points given \(\Rightarrow \ A(1, 1) \ and \ B(7, 2)\)
Gradient, \(m = \frac{Change \ in \ y}{Change \ in \ x}\)
\(\hspace{2.4cm} m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(\hspace{2.4cm} m = \frac{2 - 1}{7 - 1}\)
\(\hspace{2.4cm} m = \frac{1}{6}\)
\(\therefore\) the gradient of the line is \(\frac{1}{6}\)
Example 2
Find the gradient of the line which passes through the points \(P(-2, 4)\) and \(Q(3, 5)\).
Solution
Points given \(\Rightarrow \ P(-2, 4) \ and \ B(3, 5)\)
Gradient, \(m = \frac{Change \ in \ y}{Change \ in \ x}\)
\(\hspace{2.4cm} m = \frac{y2 - y1}{x2 - x1}\)
\(\hspace{2.4cm} m = \frac{5 - 4}{3 - -2}\)
\(\hspace{2.4cm} m = \frac{5 - 4}{3 + 2}\)
\(\hspace{2.4cm} m = \frac{1}{5}\)
\(\therefore\) the gradient of the line is \(\frac{1}{5}\)
Example 2
Find the gradient of the line which passes through the points \(C(3, -2)\) and \(D(-3, 4)\).
Solution
Points given \(\Rightarrow \ C(3, -2) \ and \ D(-3, 4)\)
Gradient, \(m = \frac{Change \ in \ y}{Change \ in \ x}\)
\(\hspace{2.4cm} m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(\hspace{2.4cm} m = \frac{-2 - 4}{3 - -3}\)
\(\hspace{2.4cm} m = \frac{-2 - 4}{3 + 3}\)
\(\hspace{2.4cm} m = \frac{-6}{6}\)
\(\hspace{2.4cm} m = -1\)
\(\therefore\) the gradient of the line is -1
THE GRADIENT WHEN THE EQUATION OF THE LINE IS GIVEN
Sometimes you will be given the equation of the line for you to find the gradient.
The equation of a line is always written in the form:
\[y = mx + c\]
When the equation is written in this form, the gradient \(m\) is the coefficient of \(x\) and the \(c\) in the equation is called the y-intercept, that is, the point where the line cuts the \(y-axis\).
If the equation of the line is not given in this form, you will need to make y the subject of the equation in order to find your gradient.
Let's look at some examples:
Example 4
Find the gradient of the equation \(y= 5x-13\).
Solution
Equation given \(\Rightarrow \ y= 5x-13\)
Equation of a line \(\Rightarrow \ y= mx + c\)
Comparing the two equations,
\(\Rightarrow\) Gradient, \(m = 5\)
\(\therefore\) the gradient is 5.
Example 5
Find the gradient of the equation \(2x - 8y + 3 = 0\).
Solution
Equation given \(\Rightarrow \ 2x - 8y + 3 = 0\)
Equation of a line \(\Rightarrow \ y= mx + c\)
Making \(y\) the subject,
\(\Rightarrow 2x - 8y + 3 = 0\)
\(\Rightarrow \hspace{0.9cm} -8y+3 = -2x\)
\(\Rightarrow \hspace{1.95cm} -8y = -2x-3\)
\(\Rightarrow \hspace{1.98cm} \frac{-8y}{-8}= \frac{-2x}{-8}-\frac{3}{-8}\)
\(\Rightarrow \hspace{2.65cm} y= \frac{1}{4}x+\frac{3}{8}\)
The coefficient of \(x\) is \(\frac{1}{4}\)
\(\therefore\) the gradient of the line is \(\frac{1}{4}\)
Example 6
Find the gradient of the equation \(y=-3x+12\).
Solution
Equation given \(\Rightarrow \ y=-3x+12\)
Equation of a line \(\Rightarrow \ y= mx + c\)
The coefficient of \(x\) is \(-3\)
\(\therefore\) the gradient of the line is \(-3\)
THE GRADIENT FROM A GRAPH
Kindly check back later for the lesson on this topic