Practice with Lines
A Linear Relation can be expressed using the equation of a line:
- \(y\) is the dependent variable
- \(x\) is the independent variable
- \(m\) is the slope or rate of change
- \(b\) is the \(y\)-intercept (the value of \(y\) when \(x = 0\))
A linear relation can be represented graphically, algebraically, or using tables. Algebraically, a linear relation can be expressed in either standard or slope-intercept form.
You can identify a linear equation since it always has a degree of \(1\), can be represented algebraically as a straight line, and contains constant first differences).
In the diagram below, the line crosses the \(y\)-axis at \(y = 4\). Every time \(x\) increases by \(1\), \(y\) increases by \(2\).
The equation of the given line is:
When graphing a line with a given slope and \(y\)-intercept, you should ALWAYS begin at the \(y\)-intercept and move in the direction of the slope.
Using the description above, we can identify the variables as \(m = 5\) and \(b = -3\) respectively.
We can now plug these values into the equation of a line:
\(y = mx + b\)
\(y = 5x -3\)
Therefore, we can determine that the equation of the line is \(\boldsymbol{y = 5x -3}\).
First, we can rearrange the equation so that the \(y\)-value is on the left side and everything else is on the right side:
\(12x - 3y = 9\)
\(-3y = -12x + 9\)
Next, we can divide the entire equation by the coefficient of \(y\) (in this case, \(-3\)):
\(\cfrac{-3y}{\textcolor{red}{-3}} = \cfrac{-12x}{\textcolor{red}{-3}} + \cfrac{9}{\textcolor{red}{3}}\)
\(y = 4x + -3\)
Therefore, we can determine that \(\boldsymbol{m = 4}\) and \(\boldsymbol{b = -3}\).
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Slope
You are able to calculate the Slope by calculating the Rise over the Run. The rise represents vertical distance between the two points. The run represents the horizontal distance between the two points. The slope can be determined algebraically as such:
This formula calculates the change in \(y\) (rise) divided by the change in \(x\) (run) to measure the steepness (slope).
In order to determine the slope, select 2 separate points. Then plug the \(x\) and \(y\) values into the equation to calculate the slope. There are various different types of slopes depending on the resulting \(x\) and \(y\) values:
If \(m > 0\), then the line will have a positive slope. This means that the two variables are positively related; that is, when \(x\) increases, so does \(y\), and when \(x\) decreases, \(y\) also decreases.
If \(m < 0\), then the line will have a negative slope. This means that the two variables are negatively related; that is, when \(x\) increases, \(y\) decreases, and when \(x\) decreases, \(y\) increases:
If \(m = \cfrac{0}{x}\), this will result in a slope of zero, resulting in a horizontal line. This means there is a constant relationship between \(x\) and \(y\); the \(y\) value does not change when the value of \(x\) changes. This can be expressed algebraically as \(y = b\) where \(b\) represents the \(y\)-intercept:
If \(m = \cfrac{y}{0}\), this will result, in an undefined slope, resulting in a vertical line. This means that the \(x\) value does not change when the \(y\) value changes. This can be expressed algebraically as \(x = a\) where \(a\) represents the \(x\)-intercept:
Example
Determine the slope of a line with the points \((4, -1)\) and \((6, 2)\).
\(m = \cfrac{y₂-y₁}{x₂-x₁}\)
\(m = \cfrac{2-(-1)}{6-4}\)
\(m = \cfrac{3}{2}\)
\(m = 1.5\)
Thus, the slope of the line is \(\boldsymbol{1.5}\).
| x Values | 6 | 4 | 2 | |||
|---|---|---|---|---|---|---|
| y Values | -5 | -2 | 1 |
First, we will use 2 of the coordinates already found in the table, \((6,-5)\) and \((4,-2)\). We will insert the coordinate values into the slope equation to get the slope value:
\(m = \cfrac{y₂-y₁}{x₂-x₁}\)
\(m = \cfrac{-2-(-5)}{4-6}\)
\(m = -\cfrac{3}{2}\)
\(m = -1.5\)
We will use this \(m\) value \((-1.5)\) along with one set of coordinate values to determine the \(y\)-intercept (or \(b\)).
In this case, we will use \((6,-5)\):
\(y = mx + b\)
\(-5 = (-1.5)(6) + b\)
\(-5 = -9 + b\)
\(b = -5 + 9\)
\(b = 4\)
We will be able to determine the remainder of the table values using the following linear equation:
We can first determine the \(y\)-value using the \(x\)-value: \(0\), \(-2\) and \(-4\) respectively:
\(y₁ = (-1.5)(0) + 4\)
\(y₁ = 4\)
We can next determine the \(y\)-value using the \(x\)-value \(-2\):
\(y₂ = (-1.5)(-2) + 4\)
\(y₂ = 3 + 4 = 7\)
\(y₂ = 7\)
We can then determine the \(y\)-value using the \(x\)-value \(-4\):
\(y₃ = (-1.5)(-4) + 4\)
\(y₃ = 6 + 4 = 10\)
\(y₃ = 10\)
Finally, we can fill in the empty cells of the table:
| x Values | 6 | 4 | 2 | 0 | -2 | -4 |
|---|---|---|---|---|---|---|
| y Values | -5 | -2 | 1 | 4 | 7 | 10 |
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Intercepts
An Intercept is a point where the line crosses an axis.
- An x-intercept is the point where the line crosses the x-axis
- A y-intercept is the point where the line crosses the y-axis
NOTE: the \(x\)-value of the \(y\)-intercept and the \(y\)-value of the \(x\)-intercept are \(0\). In order to solve for the \(x\)-intercept or the \(y\)-intercept, we must set the opposite value to \(0\).
Process for Solving x-Intercepts
Outlined below are the steps required to determine the \(x\)-intercepts:
- Arrange the equation into the form \(y = mx + b\)
- If the coefficient of \(y\) (or \(a\) ≠ \(1\), divide the entire equation by \(a\)
- Set \(y = 0\)
- Move the variables and constants onto separate sides
- Divide both sides by the coefficient of \(x\) (or \(b\)) to solve the \(x\)-intercept
Process for Solving y-Intercepts
Outlined below are the steps required to determine the \(y\)-intercepts:
- Arrange the equation into the form \(y = mx + b\)
- Set \(x = 0\)
- Divide both sides by the coefficient of y (or a) to solve the y-intercept
Example
Find the intercepts for the equation \(y = 2x + 3\)
To find the \(x\)-intercept, set \(y = 0\). Then, solve for \(x\):
\(0 = 2x + 3\)
\(\cfrac{-2x}{\textcolor{red}{-2}} = \cfrac{3}{\textcolor{red}{-2}}\)
\(x = -\cfrac{3}{2}\)
To find the \(y\)-intercept, set \(x = 0\). Then, solve for \(y\):
\(y = 2(0) + 3\)
\(y = 3\)
Therefore, we can determine that the \(x\) and \(y\)-intercepts are \(\boldsymbol{\left(-\cfrac{3}{2}, 0\right)}\) and \(\boldsymbol{(0,3)}\) respectively.
To find the \(x\)-intercept, set the \(y = 0\). Then, solve for \(x\):
\(–10(0) = -5x + 120\)
\(\cfrac{5x}{\textcolor{red}{5}} = \cfrac{120}{\textcolor{red}{5}}\)
\(x = 20\)
To find the \(y\)-intercept, set the \(x = 0\). Then, solve for \(y\):
\(–10y = -5(0) + 120\)
\(\cfrac{-10y}{\textcolor{red}{10}} = \cfrac{120}{\textcolor{red}{-10}}\)
\(y = -12\)
Therefore, we can determine that the \(x\) and \(y\)-intercepts are \(\boldsymbol{(20, 0)}\) and \(\boldsymbol{(0,-12)}\) respectively.