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Section BF-2 Forms of Linear Equations

Worksheet BF-2 Quick Notes

For this outcome we will be looking at relations in two variables. We will look at how to visualize points, lines, and other curves, and how to identify intercepts. We will look at linear equations in two variables and discuss the different forms that can be used to describe them. We will also look at slope (rate of change) and how that can be used to work with linear equations. Specifically, for this outcome, students should be able to:
Slope
Slope (or the steepness) of a line can be thought of as rise over run. Give, two points on the line, \((x_1, y_1)\) and \((x_2, y_2)\) then slope can be written as: \(m=\frac{y_2-y_1}{x_2-x_1}\)
Note: No slope \(\ne\) Zero slope \(\ne\) Undefined slope. These all represent different things.
Every line has a slope, so the term no slope does not make sense when referring to a line. Zero slope refers to horizontal lines where \(y_2=y_1\) so \(m=0\text{.}\) Undefined slope refers to vertical lines where \(x_2=x_1\) so \(m\) is undefined.
Positive slope means that the line increases as \(x\) increases or that the line rises to the right. Negative slope means that the line decreases as \(x\) increases or that the line falls to the right
If \(m_1\) is the slope of line 1, and \(m_2\) is the slope of line 2. Then:
Forms of Linear Equations
There are three main forms for the equation of a line.
Standard form is \(Ax+BY=C\) where \(A, B\text{,}\) and \(C\) are integers and \(A\gt 0\text{.}\)
The point-slope form is useful when you know the slope \(m\) and a point, \((x_1,y_1)\text{,}\) on the line. The point-slope form is given as \(y-y_1=m(x-x1)\)
The slope-intercept form is most commonly used to graph a line. For this form, we need the slope \(m\) and the y-intercept \((0,b)\text{.}\) The slope-intercept form is given by \(y=mx+b\text{.}\)
There are also two special cases for lines; horizontal and vertical lines, which correspond to our specific slopes. For a horizontal line \(m=0\) so any form of the equation can be reduced to \(y=b\) where \((0,b)\) is the y-intercept. For a vertical line, \(m\) is undefined. So any form of the equation, can be reduced to \(x=a\) where \((a,0)\) is the x-intercept.

Section BF-2 Videos

Slopes and Forms
Point-Slope Form
Slope-Intercept Form
Standard Form
Vertical and Horizontal Lines
Parallel Lines
Perpendicular Lines

Section BF-2 Rubric

Worksheet Worksheet

Before attempting the BF-2 outcome, you should review the following rubric. This shows the criteria you will be graded on and the expectations to earn each mastery level.
Table 11. BF2 Rubric
Criteria M P R N
Simplifying Algebraic Expressions All answers are fully simplified. Exact answers are given, not approximations. Radical and exponential expressions are used appropriately and are simplified in the correct manner No algebraic errors OR some expressions are not simplified. There are some algebraic errors and some expressions were not simplified. There are some algebraic errors and some expressions were not simplified.
Basics of Graphing Graph is labeled with units for each axis. The graph includes all of the important features (intercepts, asymptotes, deleted points, etc). Β The graph is appropriately drawn to accommodate all features. Graph is labeled with units for each axis and includes most of the important features or the graph has all of the important features but is not labeled. Graph is not labeled and has some of the important features. No graph was given or the given graph is not labeled and does not have the correct features of the function.
Problem Structure and Mathematical Communication Problems have a clear beginning, middle, and end. Work progresses clearly from one step to the next. The work provided uses appropriate methods and notation and provides a clear solution. Problems have clear beginnings and ends. Work progresses from one step to the next. The work provided uses algebraic methods, correct notation, and provides a solution. Parts of the mathematical structure are missing, or steps in the progress of the solution are missing. There are errors in the mathematical notation. The work provided uses algebraic methods and provides a solution. There is not a clear structure to the solution, notation is used incorrectly, and it is unclear what methods are being used.
Linear Equations in Two Variables Correctly used the given information to rewrite the line in the requested form.Β  The original and new slopes are both correct. Used the given information to rewrite the line.Β  Either the original slope or the new slope has errors.Β  The solution is mostly written in the requested form, but there are some errors. Used the given information to rewrite the line.Β  There are errors in the rewritten form, and the equation is not given in the requested form.Β  Both the original and the new slope have errors. No progress beyond rewriting the problem. Not enough evidence to show knowledge of how to find or use slope to rewrite a linear equation.

Worksheet BF-2 Sample Outcomes

This outcome covers slope and forms of linear equations. You will be given 2 problems in this outcome. Given information about a line, should be able to write the equation in one of the three forms: slope-intercept, point-slope, or standard form. You should be able to graph the line and find the intercepts. Given a line and a point, you should be able to find the equation of the line parallel or perpendicular to it, and write the equation in the appropriate form. You should be able to write out complete steps to solve these types of problems by hand, without the use of a calculator.
For the quiz on this outcome you would be given 20 minutes to complete 2 problems. Here are samples of the types of problems you will encounter:

1.

Find the equation of the line, in slope-intercept form that passes through the point \((5, -3)\) with slope \(=\frac{2}{3}\text{.}\) Give a sketch of the line with labeled intercepts.
Hint.
Slope intercept form is \(y=mx+b\)
Answer.
\(y=\frac{2}{3}x-\frac{19}{3}\)
Solution.
Since the slope is \(m=\frac{2}{3}\) and the given point is \((5, -3)\) we can plug these into the point-slope equation \(y-y_1=m(x-x_1)\) to get \(y+3=\frac{2}{3}(x-5)\text{.}\) Simplifying this equation gives us \(y=\frac{2}{3}x-\frac{19}{3}\text{.}\)

2.

Find the equation of the line, in standard form that passes through the points \((-2, 5)\) and \((8, 1)\text{.}\) Give a sketch of the line with labeled intercepts.
Hint.
Find the slope using the two points, then use point-slope form. Use the point-slope form to write the equation, then rearrange to standard form.
Answer.
\(2x+5y=21\)
Solution.
Using the slope formula, we find the slope to be \(m=\frac{1-5}{8-(-2)}=\frac{-4}{10}=-\frac{2}{5}\text{.}\) Using the point-slope form with the point \((-2, 5)\text{,}\) we get \(y-5=-\frac{2}{5}(x-(-2))\text{.}\) Simplifying and rearranging to standard form gives us \(2x+5y=21\text{.}\)

3.

Consider the line \(4x-3y=8\text{.}\) Find the equation of the line (in standard form) that is parallel to the given line that passes through the point \((-3,8)\text{.}\)
Hint.
Find the slope of the given line, then the parallel line has the same slope. Use the point-slope form with the given point.
Answer.
\(4x-3y=36\)
Solution.
The given line \(4x-3y=8\) can be rewritten in slope-intercept form as \(y=\frac{4}{3}x-\frac{8}{3}\text{.}\) The slope of this line is \(m_1=\frac{4}{3}\text{.}\) The slope of a line parallel to this one is \(m_2=\frac{4}{3}\text{.}\) Using the point-slope form with the point \((-3, 8)\text{,}\) we get \(y-8=\frac{4}{3}(x-(-3))\text{.}\) Simplifying and rearranging to standard form gives us \(4x-3y=36\text{.}\)

4.

Consider the line \(8x-3y=15\text{.}\) Find the equation of the line (in slope-intercept form) that is perpendicular to the given line that passes through the point \((2, 1)\text{.}\)
Hint.
Find the slope of the given line, then find the negative reciprocal to get the slope of the perpendicular line. Use the point-slope form with the given point.
Answer.
\(y=-\frac{3}{8}x+\frac{7}{4}\)
Solution.
The given line \(8x-3y=15\) can be rewritten in slope-intercept form as \(y=\frac{8}{3}x-5\text{.}\) The slope of this line is \(m_1=\frac{8}{3}\text{.}\) The slope of a line perpendicular to this one is \(m_2=-\frac{3}{8}\text{.}\) Using the point-slope form with the point \((2, 1)\text{,}\) we get \(y-1=-\frac{3}{8}(x-2)\text{.}\) Simplifying and rearranging to slope-intercept form gives us \(y=-\frac{3}{8}x+\frac{7}{4}\text{.}\)