4.3: Equations of Lines from Two Points
Write the equation of the line that has the same slope as \begin{align*}3x+2y-8=0\end{align*}
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Khan Academy Equation of a Line
Guidance
Equations of Lines in Slope-Intercept Form
In order to find the equation of a line, you must know its slope (\begin{align*}m\end{align*}
\begin{align*}y = mx + b\end{align*}
Always remember that as long as you know the slope and the y-intercept, you can write the equation of the line. Sometimes you will have to do some work to determine either the slope or the y-intercept, as will be shown in the examples.
In the above graph, the line crosses the \begin{align*}y\end{align*}
Equations of Horizontal and Vertical Lines
A line that is parallel to the \begin{align*}x\end{align*}
A line that is parallel to the \begin{align*}y\end{align*}
Equations of Lines in Standard Form
The equation of a line can also be written in another form that is known as standard form. Standard form is \begin{align*}Ax + By + C = 0\end{align*}
\begin{align*}3x + 2y - 8 & = 0\\
3x {\color{red}-3x} + 2y - 8 & = 0 {\color{red}-3x}\\
2y - 8 & = -3x\\
2y - 8 {\color{red}+8} & = -3x {\color{red}+8}\\
2y & = -3x + 8\\
\frac{2y}{{\color{red}2}} & = \frac{-3x}{{\color{red}2}} + \frac{8}{{\color{red}2}}\\
& \boxed{y = \frac{-3}{2}x + 4}\end{align*}
The equation has been solved for \begin{align*}y\end{align*}
Occasionally you will be given the equation of a line given in slope-intercept form and want to rewrite this equation in standard form. To do this, multiply all terms by the denominators of any fractions to get rid of the fractions and then move all variables and constants to one side of the equation in order to set it equal to 0.
Example A
Write the equation for the line that passes through the points \begin{align*}A (3, 4)\end{align*}
Solution:
Remember that to find the equation you need to figure out the slope (\begin{align*}m\end{align*}
\begin{align*}\begin{pmatrix}
x_1, & y_1 \\
3, & 4
\end{pmatrix} \qquad \begin{pmatrix}
x_2, & y_2 \\
8, & 2
\end{pmatrix}\end{align*}
\begin{align*}m & = \frac{y_2 - y_1}{x_2 - x_1}\\
m & = \frac{2 - 4}{8 - 3}\\
m & = -\frac{2}{5}\end{align*}
Next, determine the \begin{align*}y\end{align*}
\begin{align*}y & = mx + b \\ ({\color{red}4}) & = \left ( {\color{red}-\frac{2}{5}} \right )({\color{red}3}) + b && \text{Use one of the given points for } (x, y) \text{ and } \left( -\frac{2}{5} \right ) \text{ for } `m'.\\ 4 & = -\frac{6}{5} + b && \text{Solve for } `b'.\\ 4 + \frac{6}{5} & = - \frac{6}{5} + \frac{6}{5} + b\\ 4 + \frac{6}{5} & = b && \text{To add these numbers, a common denominator is necessary.}\\ \left ( \frac{4}{1} \right ) \left ( \frac{5}{5} \right ) + \frac{6}{5} & = b\\ \frac{20}{5} + \frac{6}{5} & = b\\ \frac{26}{5} & = b && \text{The } y \text{-intercept is } \left ( 0, \frac{26}{5} \right )\end{align*}
The equation for the line is \begin{align*}\boxed{y = - \frac{2}{5} x + \frac{26}{5}}\end{align*}
Example B
(a) Write the equation of the line passing through the point (6, -4) and parallel to the \begin{align*}x\end{align*}-axis.
(b) Write the equation of the line passing through the point (3, -2) and parallel to the \begin{align*}y\end{align*}-axis.
Solution:
(a) A line that is parallel to the \begin{align*}x\end{align*}-axis will always have the equation \begin{align*}y=a\end{align*}, where \begin{align*}a\end{align*} is the \begin{align*}y-\end{align*}coordinate of the point through which the line passes. Therefore, the equation of this line is \begin{align*}\boxed{y = -4}\end{align*}.
(b) A line that is parallel to the \begin{align*}y\end{align*}-axis will always have the equation \begin{align*}x=c\end{align*}, where \begin{align*}c\end{align*} is the \begin{align*}x-\end{align*}coordinate of the point through which the line passes. Therefore, the equation of this line is \begin{align*}\boxed{x = 3}\end{align*}.
Example C
Write the equation of the line that passes through the point (-2, 5) and has the same \begin{align*}y\end{align*}-intercept as: \begin{align*}-3x +6y +18 = 0\end{align*}.
Solution:
In order to find the equation of any line, we need to figure out the slope and the intercept. First, find the y-intercept.
\begin{align*}-3x + 6y + 18 & = 0 && \text{is written in standard form. To determine}\\ & && \text{the } y \text{-intercept, solve for }`y'\\ -3x + 3x + 6y + 18 - 18 & = 3x-18\\ 6y & = 3x-18\\ \frac{6y}{6} & = \frac{3}{6}x-\frac{18}{6}\\ y & = \frac{1}{2}x-3\\\end{align*}
You can see now that \begin{align*}b=-3\end{align*}. The line also passes through the point (-2, 5). You can use this point along with the y-intercept to help find the slope.
\begin{align*}y & = mx + b && \text{Fill in } -3 \text{ for }`b' \text{ and } (x, y) = (-2, 5)\\ ({\color{red}5}) & = m({\color{red}-2}) + ({\color{red}-3})\\ 5 & = -2{\color{red}m} - 3 && \text{Solve for } `m'.\\ 5 + 3 & = -2m - 3 + 3\\ 8 & = - 2m\\ \frac{8}{-2} & = \frac{-2m}{-2}\\ -4 & = m\end{align*}
\begin{align*} \boxed{y = -4x-3} \end{align*} is the equation of the line.
Concept Problem Revisited
Write the equation of the line that has the same slope as \begin{align*}3x+2y-8=0\end{align*} and passes through the point (-6, 7).
To determine the slope of \begin{align*}3x+2y-8=0\end{align*}, solve the equation for ‘\begin{align*}y\end{align*}’.
\begin{align*}3x + 2y - 8 & = 0\\ 3x {\color{red}-3x} + 2y - 8 & = 0 {\color{red}-3x}\\ 2y - 8 & = -3x\\ 2y - 8 {\color{red}+8} & = -3x {\color{red}+8}\\ 2y & = -3x + 8\\ \frac{2y}{{\color{red}2}} & = \frac{-3x}{{\color{red}2}} + \frac{8}{{\color{red}2}}\\ & \boxed{y = \frac{-3}{2}x + 4}\end{align*}
The equation is now in the form \begin{align*}y=mx+b\end{align*}. The slope of the line is \begin{align*}m = -\frac{3}{2}\end{align*}. The line can now be sketched on the Cartesian grid.
The point (-6, 7) was plotted first and then the slope was applied – run two units to the right and then move downwards three units. The line crosses the \begin{align*}y\end{align*}-axis at the point (0, -2). This is the \begin{align*}y\end{align*}-intercept of the graph. The equation for the line has a slope of \begin{align*}-\frac{3}{2}\end{align*} and a \begin{align*}y\end{align*}-intercept of -2.
The equation of the line is \begin{align*}\boxed{y = -\frac{3}{2}x-2}\end{align*}
Vocabulary
- Slope – Intercept Form
- The slope-intercept form is one method for writing the equation of a line. The slope-intercept form is \begin{align*}y = mx + b\end{align*} where \begin{align*}m\end{align*} refers to the slope and \begin{align*}b\end{align*} identifies the \begin{align*}y\end{align*}-intercept.
- Standard Form
- The standard form is another method for writing the equation of a line. The standard form is \begin{align*}Ax + By + C = 0\end{align*} where \begin{align*}A\end{align*} is the coefficient of \begin{align*}x\end{align*}, \begin{align*}B\end{align*} is the coefficient of \begin{align*}y\end{align*} and \begin{align*}C\end{align*} is a constant.
Guided Practice
Write the equation for each of the lines graphed below in slope-intercept form and in standard form.
1.
2.
3. Write the equation for the line that passes through the point (-4, 7) and is perpendicular of the \begin{align*}y\end{align*}-axis.
Answers:
1. Two points on the graph are (-2, 6) and (8, 0). First use the formula to determine the slope of this line.
- \begin{align*}m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{0-6}{8 -- 2}\\ m & = \frac{0-6}{8+2}\\ m & = -\frac{6}{10}\\ m&=-\frac{3}{5}\end{align*}
- Use the slope and one of the points to determine the \begin{align*}y\end{align*}-intercept.
\begin{align*}y & = mx + b\\ ({\color{red}0}) & = \left ( {\color{red}-\frac{3}{5}} \right ) ({\color{red}8}) + b && \text{Use one of the given points for } (x, y) \text{ and } \left ( -\frac{3}{5} \right ) \text{ for } `m'.\\ 0 & = -\frac{24}{5} + b && \text{Solve for }`b'.\\ 0 + \frac{24}{5} & = -\frac{24}{5} + \frac{24}{5} + b\\ \frac{24}{5} & = b\end{align*}
- The equation for the line in slope-intercept form is \begin{align*}\boxed{y =-\frac{3}{5}x + \frac{24}{5}}\end{align*}. To express the equation in standard form, multiply each term by 5 and set the equation equal to zero.
- \begin{align*}y & = -\frac{3}{5}x + \frac{24}{5}\\ {\color{red}5}(y) & = {\color{red}5} \left ( -\frac{3}{5}x \right ) + {\color{red}5} \left ( \frac{24}{5} \right ) \\ {\color{red}5}(y) & = {\color{red}\cancel{5}} \left ( -\frac{3}{\cancel{5}}x \right ) + {\color{red}\cancel{5}} \left ( \frac{24}{\cancel{5}} \right ) \\ 5y & = -3x + 24 && \text{Apply the zero principle to move } -3x \text{ to the left side of the equation.}\\ 5y {\color{red}+3x} & = - 3x {\color{red}+3x} + 24\\ 5y {\color{red}+3x} & = 24\\ 5y + 3x {\color{red}-24} & = 24 {\color{red}-24}\\ 5y + 3x - 24 & = {\color{red}0}\end{align*}
- \begin{align*}\boxed{3x+5y-24=0}\end{align*}
2. Two exact points on the graph are (-10, -8) and (20, 2). The slope of the line can be calculated by counting to determine the value of \begin{align*}m = \frac{\text{rise}}{\text{run}}\end{align*}.
- \begin{align*}m & = \frac{\text{rise}}{\text{run}}\\ m & = \frac{10}{30}\\ m & = \frac{1}{3}\end{align*}
- Now, use the slope and one of the points to calculate the \begin{align*}y\end{align*}-intercept of the line.
- \begin{align*}y&=mx+b && m=\frac{1}{3} \ and \begin{pmatrix} x, & y\\ 20, & 2 \end{pmatrix}\\ {\color{red}2} & = {\color{red}\frac{1}{3}(20)} + b\\ 2 & = \frac{20}{3} + b && \text{Solve for } `b'.\\ 2 - \frac{20}{3} & = \frac{20}{3} - \frac{20}{3} + b\\ 2 - \frac{20}{3} & = b && \text{A common denominator is needed.}\\ {\color{red}\left(\frac{3}{3}\right)} \frac{2}{1} - \frac{20}{3} & = b\\ \frac{6}{3} - \frac{20}{3} & = b\\ -\frac{14}{3} & = b\end{align*}
- The equation of the line in slope-intercept form is \begin{align*}\boxed{y = \frac{1}{3}x - \frac{14}{3}}\end{align*}. Multiply the equation by 3 and set the equation equal to zero to write the equation in standard form. The equation of the line in standard form is \begin{align*}\boxed{x-3y-14=0}\end{align*}.
3. Begin by sketching the graph of the line.
A line that is perpendicular to the \begin{align*}y\end{align*}-axis is parallel to the \begin{align*}x\end{align*}-axis. The slope of such a line is zero. The equation of this line is \begin{align*}\boxed{y = 7}\end{align*}
Practice
For each of the following graphs, write the equation of the line in slope-intercept form.
Determine the equation of the line that passes through the following pairs of points:
- (-3, 1) and (-3, -7)
- (-5, -5) and (10, -5)
- (-8, 4) and (2, -6)
- (14, 8) and (4, 4)
- (0, 5) and (4, -3)
For each of the following real world problems, write the linear equation in standard form that would best model the problem.
- The cost of operating a car for one month depends upon the number of miles you drive. According to a recent survey completed by drivers of mid size cars, it costs $124/month if you drive 320 miles/month and $164/month if you drive 600 miles/month.
- Designate two data values for this problem. State the dependent and independent variables.
- Write an equation to model the situation.
- A Glace Bay developer has produced a new handheld computer called the Blueberry. He sold 10 computers in one location for $1950 and 15 in another for $2850. The number of computers and the cost forms a linear relationship.
- Designate two data values for this problem. State the dependent and independent variables.
- Write an equation to model the situation.
- Shop Rite sells a one-quart carton of milk for $1.65 and a two-quart carton for $2.95. Assume there is a linear relationship between the volume of milk and the price.
- Designate two data values for this problem. State the dependent and independent variables.
- Write an equation to model the situation.
- Some college students, who plan on becoming math teachers, decide to set up a tutoring service for high school math students. One student was charged $25 for 3 hours of tutoring. Another student was charged $55 for 7 hours of tutoring. The relationship between the cost and time is linear.
- Designate two data values for this problem. State the dependent and independent variables.
- Write an equation to model the situation.
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Here you'll learn how to find the equation of a line in slope intercept form or standard form.
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