4.2: The Equation of a Line in Slope – Intercept Form
The Equation of a Line from a Graph
Introduction
In this lesson you will learn how to use the slope and the \begin{align*}y-\end{align*}intercept of a line to write the equation of the line. The equation will be written in the form \begin{align*}y=mx+b\end{align*} where ‘\begin{align*}m\end{align*}’ is the slope, ‘\begin{align*}b\end{align*}’ is the \begin{align*}y-\end{align*}intercept and \begin{align*}(x, y)\end{align*} are the coordinates of a point on the straight line. The form \begin{align*}y=mx+b\end{align*} is known as the slope-intercept form of the equation. You will learn how to determine the slope of a line from an equation that is written in standard form. The standard form of an equation 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. You will learn to write the equation of a line that has been graphed on a Cartesian grid as well as how to determine the slope and/or the \begin{align*}y-\end{align*}intercept from information that is given about a line. These calculations will then be used to create the equation of the line.
Objectives
The lesson objectives for the equation of a line in slope-intercept form are:
- Understanding how to determine the equation of the line from a graph
- Understanding how to determine the slope and/or the \begin{align*}y-\end{align*}intercept of a line from given information in order to write the equation of the line
- Understanding how to determine the slope of a line from an equation that is written in standard form.
- Understanding the equation of special lines
Introduction
In the above graph, 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 line. The slope is \begin{align*}m = \frac{\text{rise}}{\text{run}}\end{align*}. The slope of this line is found by running to the right one unit and rising upward three units. Therefore the slope of the line is \begin{align*}m = \frac{3}{1}\end{align*}. To write the equation of a line in slope-intercept form, the slope of the line ‘\begin{align*}m\end{align*}’ and the \begin{align*}y-\end{align*}intercept ‘\begin{align*}b\end{align*}’ must be known values. Thus, the equation of the above line is \begin{align*}\boxed{y = 3x + 2}\end{align*}
Watch This
Khan Academy Equation of a Line
Guidance
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 the variable ‘\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 has been solved for the variable ‘\begin{align*}y\end{align*}’ and is now in the form \begin{align*}y=mx+b\end{align*}. The slope of the line is the coefficient of ‘\begin{align*}x\end{align*}’ and in this equation \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*}
Example A
Write the equation for the line that passes through the points \begin{align*}A (3, 4)\end{align*} and \begin{align*}B (8, 2)\end{align*}.
The first step is to determine the slope of the line:
\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*}
The next step is to determine the \begin{align*}y-\end{align*}intercept of the line. This can be done by using the slope-intercept form for the equation of the line.
\begin{align*}y & = mx + b \\ ({\color{red}4}) & = {\color{red}-} \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.
You know only one point that the line passes through. Sketch the graph of the line that passes through the given point and is parallel to the \begin{align*}x-\end{align*}axis.
You learned in the previous lesson that the slope of a line that is parallel to the \begin{align*}x-\end{align*}axis is zero.
\begin{align*}y & = mx + b && m = 0 \text{ and } (x, y) = (6, -4)\\ ({\color{red}-4}) & = ({\color{red}0})({\color{red}6}) + b && \text{Solve for }`b'.\\ ({\color{red}-4}) & = 0 + b\\ -4 & = b && \text{The } y-\text{intercept is } (0, -4)\\ \\ y & = mx + b && m=0 \text{ and the } y-\text{intercept is } (0, -4)\\ y & = ({\color{red}0})x + ({\color{red}-4})\\ & \boxed{y = -4} && \text{This is the equation of the line.}\end{align*}
Any line that is parallel to the \begin{align*}x-\end{align*}axis will have \begin{align*}y =\end{align*} the \begin{align*}y-\end{align*}intercept, as its equation. This is the equation of one of the special lines. If you remember this, you can simply write the equation of the line and not have to show all of the work that is shown above.
(b) You know only one point that the line passes through. Sketch the graph of the line that passes through the given point and is parallel to the \begin{align*}y-\end{align*}axis.
You learned in the previous lesson that the slope of a line that is parallel to the \begin{align*}y-\end{align*}axis is undefined. From the graph, you can see that the line will never cross the \begin{align*}y-\end{align*}axis. Therefore, it is not possible to determine a \begin{align*}y-\end{align*}intercept for the line. A line that is parallel to the \begin{align*}y-\end{align*}axis will always have \begin{align*}x =\end{align*} the \begin{align*}x-\end{align*}coordinate of the point through which it passes, as its equation. This is also one of the special lines. 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*}.
\begin{align*}-3x + 6y + 18 & = 0 && \text{is written in standard form. To determine}\\ & && \text{the } y-\text{intercept, set } x = 0 \text{ and solve for }`y'\\ -3({\color{red}0}) + 6y + 18 & = 0\\ 6y + 18 - 18 & = 0 - 18\\ 6y & = -18\\ \frac{6y}{6} & = \frac{-18}{6}\\ y & = -3\end{align*}
The \begin{align*}y-\end{align*}intercept is (0, –3). The line also passes through the point (–2, 5).
\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\\ & \boxed{y = -4x-3} \text{ is the equation of the line.}\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
1. Write the equation for each of the lines graphed below. Write the equation in slope-intercept form and in standard form.
i)
ii)
2. Write the equation for the line that passes through the point (–4, 7) and is perpendicular of the \begin{align*}y-\end{align*}axis.
3. Write the equation of the line, in standard form, that passes through the points (2, 8) and has the same \begin{align*}x-\end{align*}intercept as the line \begin{align*}5x-4y+30=0\end{align*}
Answers
1. i) Two points on the graph are (–2, 6) and (8, 0). The first step is to determine the slope of the line either by counting or by using the slope formula. Use the formula to determine the slope of this line.
Designate the first and the second point
\begin{align*} \begin{pmatrix} x_1, & y_1\\ -2, & 6 \end{pmatrix} \qquad \begin{pmatrix} x_2, & y_2\\ 8, & 0 \end{pmatrix}\end{align*}
Use the formula to calculate the slope
\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}\end{align*}
Express the slope in simplest form.
\begin{align*}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 \begin{align*}y-\end{align*}intercept is \begin{align*}\left ( 0, \frac{24}{5} \right )\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.
\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\end{align*}
Set the equation equal to zero.
\begin{align*}5y + 3x {\color{red}-24} & = 24 {\color{red}-24}\\ 5y + 3x - 24 & = {\color{red}0}\end{align*}
Write the equation in the form \begin{align*}Ax +By +C = 0\end{align*}
\begin{align*}\boxed{3x+5y-24=0}\end{align*}
ii)
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*}. Remember to run to the right first. To run from –10 to +20 is 30 units. Next, rise from –8 up to +2 which is 10 units. Fill the values into the formula and express the answer in simplest form.
\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 \begin{align*}y-\end{align*}intercept of the line is \begin{align*}\left ( 0, -\frac{14}{3} \right )\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*}
To write the equation in standard form, the equation \begin{align*}y - y_1 = m(x-x_1)\end{align*} can be used. The value of the slope and the coordinates of one point are needed to determine the equation of the line.
\begin{align*}m & = \frac{1}{3} \ and \ \begin{pmatrix} x_1, & y_1\\ 20, & 2 \end{pmatrix}\\ y- {\color{red}2} & = {\color{red}\frac{1}{3}} (x - {\color{red}20}) && \text{Multiply both sides of the equation by } 3\\ {\color{red}3} (y-2) & = {\color{red}3} \left [ \frac{1}{3} (x-20) \right ]\\ {\color{red}3} (y-2) & = {\color{red}\cancel{3}} \left [ \frac{1}{\cancel{3}} (x-20) \right ]\\ 3y - 6 & = 1(x-20)\\ 3y - 6 & = x - 20 && \text{Set the equation equal to zero}\\ 3y - 6 + 20 & = x - 20 + 20\\ 3y + 14 & = x\\ -x + 3y + 14 & = x - x\\ -x + 3y + 14 & = 0\end{align*}
The coefficient of ‘\begin{align*}x\end{align*}’ cannot be a negative number. Multiply both sides of the equation by –1. This will change the signs of every term in the equation.
\begin{align*}\boxed{x-3y-14=0}\end{align*}
The equation of the line in standard form is \begin{align*}\boxed{x-3y-14=0}\end{align*}.
2. 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*}
3. \begin{align*}5x - 4y + 30 & = 0 && \text{Calculate the } x-\text{intercept of this line. Set } y = 0 \text{ and solve for }`x'.\\ 5x - 4({\color{red}0}) + 30 & = 0\\ 5x + 30 & = 0\\ 5x + 30 - 30 & = 0-30\\ 5x & = - 30\\ \frac{5x}{5} & = \frac{-30}{5}\\ x & = -6\end{align*}
The \begin{align*}x-\end{align*}intercept of the line \begin{align*}5x-4y+30=0\end{align*} is (–6, 0).
Use the \begin{align*}y-\end{align*}intercept (–6, 0) and the point (2, 8) to determine the slope of the line.
Designate the first and the second point
\begin{align*}\begin{pmatrix} x_1, & y_1\\ -6, & 0 \end{pmatrix} \qquad \begin{pmatrix} x_2, & y_2\\ 2, & 8 \end{pmatrix}\end{align*}
Use the formula to calculate the slope
\begin{align*}m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{8-0}{6 -- 2}\\ m & = \frac{8-0}{6+2}\\ m & = \frac{8}{8}\\ m & = 1\end{align*}
Write the equation in standard form using the point (–6, 0) and the slope of 1.
\begin{align*}& \qquad \ \ y - y_1 =m(x-x_1)\\ & \qquad \quad y - {\color{red}0} = {\color{red}1} (x - {\color{red}-6})\\ & \qquad \quad y - {\color{red}0} = {\color{red}1} (x + {\color{red}6})\\ & \qquad \qquad \ \ y = x + 6 && \text{Set the equation equal to zero.}\\ & \qquad \quad y - 6 = x + 6 - 6\\ & \qquad \quad y -6 = x\\ & -x + y -6 = x - x\\ & -x + y -6 = 0 && \text{Change all signs}\\ & \quad \boxed{x - y + 6 = 0}\end{align*}
The equation of the line, in standard form is \begin{align*}\boxed{x-y+6=0}\end{align*}
Summary
In this lesson you have learned how to write the equation of a line in slope-intercept form. This form is \begin{align*}y = mx + b\end{align*}. You also learned that the equation of a line that is parallel to the \begin{align*}x-\end{align*}axis is \begin{align*}y =\end{align*} the \begin{align*}y-\end{align*}coordinate of the point through which the line passes. In addition you have learned that the equation of a line that is parallel to the \begin{align*}y-\end{align*}axis is \begin{align*}x=\end{align*} the \begin{align*}x-\end{align*}coordinate of the point through which the line passes. Another way to write the equation of a line is to write it in standard form. Standard form is \begin{align*}Ax + By + C = 0\end{align*}.
Problem Set
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.
Answers
For each of the following graphs...
- Two points on the line are (0, –5) and (5, 5).
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 0, & -5 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 5, & 5 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{5--5}{5-0}\\ m & = \frac{5+5}{5-0}\\ m & = \frac{10}{5}\\ m & = 2\end{align*}
The slope of the line is \begin{align*}m = 2\end{align*} and the \begin{align*}y-\end{align*}intercept is (0, –5).
The equation of the line in slope intercept form is: \begin{align*}\boxed{y = 2x - 5}\end{align*}
- Two points on the line are (0, –8) and (8, –2).
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 0, & -8 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 8, & -2 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{-2--8}{8-0}\\ m & = \frac{-2+8}{8-0}\\ m & = \frac{6}{8}\\ m & = \frac{3}{4}\end{align*}
The slope of the line is \begin{align*}m = \frac{3}{4}\end{align*} and the \begin{align*}y-\end{align*}intercept is (0, –8).
The equation of the line in slope intercept form is: \begin{align*}\boxed{y = \frac{3}{4}x - 8}\end{align*}
- Two points on the line are (0, –1) and (12, –4).
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 0, & -1 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 12, & -4 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{-4--1}{12-0}\\ m & = \frac{-4+1}{12-0}\\ m & = \frac{-3}{12}\\ m & = \frac{-1}{4}\end{align*}
The slope of the line is \begin{align*}m = \frac{-1}{4}\end{align*} and the \begin{align*}y-\end{align*}intercept is (0, –1).
The equation of the line in slope intercept form is: \begin{align*}\boxed{y = - \frac{1}{4}x - 1}\end{align*}
Determine the equation of the line...
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ -3, & 1 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ -3, & -7 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{-7-1}{-3--3}\\ m & = \frac{-7-1}{-3+3}\\ m & = \frac{-8}{0}\\ m & = \text{undefined}\end{align*}
A line that has a slope that is undefined is a line that is parallel to the \begin{align*}y-\end{align*}axis. The equation of this line is \begin{align*}\boxed{x = -3}\end{align*}.
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ -8, & 4 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 2, & -6 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{-6-4}{2--8}\\ m & = \frac{-10}{10}\\ m & = -1\\ \\ y & = mx + b\\ 4 & = -1(-8) + b\\ 4 & = 8 + b\\ 4-8 &= 8-8+b\\ -4 & = b\end{align*}
\begin{align*}m = -1\end{align*} and the \begin{align*}y-\end{align*}intercept is (0, –4). The equation of this line is: \begin{align*}\boxed{y = -1x-4}\end{align*}
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 0, & 5 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 4, & -3 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{-3-5}{4-0}\\ m & = \frac{-8}{4}\\ m & = -2\end{align*}
\begin{align*}m = -2\end{align*} and the \begin{align*}y-\end{align*}intercept is (0, 5). The equation of this line is: \begin{align*}\boxed{y = -2x + 5}\end{align*}
For each of the following real world problems...
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 320, & 124 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 600, & 164 \end{pmatrix}\\ m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{164-124}{600-320}\\ m & = \frac{40}{280}\\ m & = \frac{1}{7}\\ \\ y - y_1 & = m(x-x_1)\\ y - 124 & = \frac{1}{7}(x-320)\\ 7(y-124) & = 7 \left [ \frac{1}{7} (x-320) \right ]\\ 7(y-124) & = \cancel{7} \left [ \frac{1}{\cancel{7}} (x-320) \right ]\\ 7(y-124) & = 1 (x-320)\\ 7y-868 & = x-320\\ 7y - 868 + 320 & = x - 320 + 320\\ 7y - 548 & = x\\ 7y - x - 548 & = x - x\\ 7y - x - 548 & = 0\\ -x + 7y - 548 & = 0\\ x - 7y + 548 & = 0\end{align*}
When you write an equation for a word problem, it is more meaningful to use variables that represent the dependent and independent quantities of the word problem.
\begin{align*}\boxed{d - 7c + 548 = 0}\end{align*}
\begin{align*}& \begin{pmatrix} x_1, & y_1\\ 1, & 1.65 \end{pmatrix} \begin{pmatrix} x_2, & y_2\\ 2, & 2.95 \end{pmatrix}\\ & m = \frac{y_2 - y_1}{x_2 - x_1}\\ & m = \frac{2.95-1.65}{2-1}\\ & m = \frac{1.30}{1}\\ & \\ & \qquad \qquad \quad y - y_1 = m(x-x_1)\\ & \qquad \qquad \ y - 1.65 = 1.30(x-1)\\ & \qquad \qquad \ y - 1.65 = 1.30x-1.30\\ & \quad \ y - 1.65 + 1.30 = 1.30x-1.30 + 1.30\\ & \qquad \qquad \quad y - 0.35 = 1.30x\\ & \quad \ y - 0.35 -1.30x = 1.30x - 1.30x\\ & \quad \ y - 0.35 -1.30x = 0\\ & -1.30x + y - 0.35 = 0\\ & \quad \ \ 1.30x - y + 0.35 = 0\\ & \quad \boxed{1.30v - p + 0.35 = 0}\end{align*}
Summary
In this lesson you have learned that the slope of a line and the \begin{align*}y-\end{align*}intercept of the line are the two facts that you must know in order to write the equation of a line. These two values can then be written into the slope-intercept form for a linear function. This form is \begin{align*}y = mx + b\end{align*} where ‘\begin{align*}m\end{align*}’ is the slope of the line and ‘\begin{align*}b\end{align*}’ is the \begin{align*}y-\end{align*}coordinate of the \begin{align*}y-\end{align*}intercept.
You have also learned that the equation of a line can be written in another form that is known as standard form. 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. This form can be determined by using the formula \begin{align*}y - y_1 = m(x-x_1)\end{align*} where \begin{align*}m\end{align*} is the slope of the line and \begin{align*}(x_1, y_1)\end{align*} are the coordinates of any point on the line. You also learned that when writing an equation in standard form, the value of ‘\begin{align*}A\end{align*}’ cannot be a negative value.
You also learned that the equation of a line that is parallel to the \begin{align*}x-\end{align*}axis is \begin{align*}y=\end{align*} the \begin{align*}y-\end{align*}coordinate of the point through which the line passes. In addition you have learned that the equation of a line that is parallel to the \begin{align*}y-\end{align*}axis is \begin{align*}x=\end{align*} the \begin{align*}x-\end{align*}coordinate of the point through which the line passes.
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