Sign In
This lesson focuses on two main mathematical concepts: graphing equations using point-slope form and writing equations in this form. It offers a step-by-step guide, complete with real-world examples, to help you understand these topics. Whether you're a student looking to improve your algebra skills or a teacher seeking a lesson for your class, this material serves as a valuable educational tool. It breaks down complex ideas into manageable pieces, making it easier for you to grasp the subject matter. The use cases include academic settings, test preparation, and even real-world applications like engineering or economics where understanding linear relationships is crucial.
| | 13 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The following applet shows the graph of a line and its slope. By clicking on the line, a point can be seen jointly with its coordinates.
A linear equation can be written in many ways. Depending on the information at hand, one can be more suitable than others. One case is when the slope and a point on the line are known. In this situation, the point-slope form may be the best option.
A linear equation with slope m through the point (x_1,y_1) is written in the point-slope form if it has the following form.
y-y_1 = m(x-x_1)
In this point-slope equation, (x_1,y_1) represents a specific point on the line, and (x,y) represents any point also on the line. Graphically, this means that the line passes through the point (x_1,y_1).
It is worth mentioning that the point-slope form can only be written for non-vertical lines.
x_2= x, y_2= y
LHS * (x-x_1)=RHS* (x-x_1)
Rearrange equation
The following applet shows linear equations representing the relationship between the variables x and y. Determine whether the given equation is written in point-slope form.
To get familiar with the point-slope form, it is essential to identify the parts of its composition. In the following applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.
Sometimes knowing the behavior of a linear equation by just looking at the equation can be complicated. In these cases, the graph can be helpful. Therefore, it is essential to have a straightforward way to graph equations in point-slope form.
When working with a linear equation in point-slope form, the graph’s line is described by the equation. The known point and slope of the line to is used to find another point. A line is then drawn through the points to create its graph. Consider the following equation. y-1=2(x-2) There are three steps to graph an equation written in point-slope form.
Izabella learned that the following equation gives the relationship between degrees Fahrenheit and degrees Celsius. Fahrenheit to Celsius C=5/9(F-32) Izabella knows that the equation is in point-slope form. However, she wonders if it is possible to have a graphical representation of this equation.
Help Izabella to create the graph and have a better understanding of the given equation.
Next, a second point on the line can be found by using the slope. Because the given equation has a slope of 59, a second point can be plotted by going 9 units to the right and 5 units up.
Finally, the line described by the given equation will be found by drawing a line through the points.
LaShay is traveling by bus to her hometown. After a certain distance, she feels bored and wants to have some fun analyzing the bus's movement. The bus driver tells her that the bus moves at a constant rate of 80 kilometers per hour. Four hours after LaShay started her analysis, the driver also tells her that the bus had traveled 500 kilometers.
y-y_1=m(x-x_1) Since the distance d traveled depends on the time t, d is the dependent variable and t the independent variable. Therefore, the above equation can be rewritten in terms of d and t. d- d_1= m(t- t_1) Here, ( t_1, d_1) represents the distance traveled at a specific time. In this situation, it is given that after 4 hours that LaShay started her analysis, the bus has traveled 500 kilometers. Writing this information as an ordered pair will give the specific point needed to write the equation in point-slope form. (t_1,d_1)=( 4, 500) Moreover, the rate of change represents the slope of the line. Therefore, since the rate of change is given as a constant 80 kilometers per hour, the equation will have a slope of 80. Substituting these values into the equation will give the equation of the movement of the bus. d- d_1&= m(t- t_1) &⇓ d- 500&= 80(t- 4)
Next, using the slope, a second point on the line can be found. The equation has a slope of 80, which can be written as 801. This means that a second point can be found by going 1 unit to the right and 80 units up.
Finally, connecting the points with a line will give the graph of the equation.
Note that in this situation, it makes sense that the distance traveled at t=0 is greater than 0 because LaShay started counting after the bus had traveled a certain distance.
After 10 hours, the bus has traveled 980 kilometers. Another way of finding the distance traveled is by looking at the graph of the equation.
However, this approach will not guarantee an exact solution. From the graph, it can be said that the distance traveled is nearly 1000 kilometers, while it was found to be 980 kilometers by using the equation. The graph gives an approximate answer, but the equation gives a precise answer.
t= 0
Subtract term
a(- b)=- a * b
LHS+500=RHS+500
Add terms
This means that LaShay started her analysis after the bus traveled 180 kilometers.
The point-slope form is a helpful tool because the information needed to write it is minimal. Once the equation is found, how it behaves at any point can be obtained. The following method describes how to find the point-slope form of a linear equation.
The slope and a point on a certain line are needed to write an equation in point-slope form. The slope can be found using two points on the line. That means this process begins by assuming that two points are given. After the slope is calculated, either point can be used to determine the equation. This method is illustrated using the following points that lie on a line. (-1,5) and (1,1) Three steps are needed to find the equation in point-slope form.
Substitute ( -1,5) & ( 1,1)
In this case, the slope of the line that passes through the given points is - 2.
Kevin went parachuting. The following graph describes his descent in terms of the time until he landed.
Use the given graph to find the following information.
y-y_1=m(x-x_1) From the given graph, two points on the line are already known. Either of them can be used to write the equation, but the slope of the line needs to be calculated first. This can be done by substituting both of these points into the Slope Formula.
Substitute ( 2,2700) & ( 4,1500)
Subtract terms
Put minus sign in front of fraction
Calculate quotient
Now that the slope is known, it can be substituted into the general equation in point-slope form. Additionally, since either of the given points can be chosen, ( 2, 2700) will be used. y-y_1&=m(x-x_1) &⇓ y- 2700&= -600(x- 2)
x= 0
Subtract term
- a(- b)=a* b
LHS+2700=RHS+2700
The height where Kevin started descending is 3900 feet.
y= 0
Subtract term
.LHS /(-600).=.RHS /(-600).
- a/- b=a/b
LHS+4=RHS+4
Rearrange equation
a = 600* a/600
Add fractions
a/b=.a /300./.b /300.
Write as a decimal
It took Kevin 6.5 minutes to land.
Kriz is doing their math homework assignment. They are asked to transform each equation on the left to its corresponding equation in point-slope or slope-intercept forms using the provided information. Moreover, for the equations in slope-intercept form, the point that it is provided should be used as a guide to finding its equivalent equation in point-slope form.
| Equation | Equivalent Equation |
|---|---|
| y+2=1/2(x+1) | |
| y=1/5x+1/3, (1,8/15) | |
| y-30=5(x+1) | |
| y=- 2x+3, (1,- 1) |
Help Kriz complete the table.
| Equation | Equivalent Equation |
|---|---|
| y+2=1/2(x+1) | y=1/2x-3/2 |
| y=1/5x+1/3, (1,8/15) | y-8/15=1/5(x-1) |
| y-30=5(x+1) | y=5x+35 |
| y=- 2x+3, (1,- 1) | y+1=- 2 (x-1) |
The first equation is in point-slope form. Then, the y variable will be isolated to find its equivalent equation in slope-intercept form.
Distribute 1/2
a * 1=a
LHS-2=RHS-2
a = 2* a/2
Subtract term
The second equation is in slope-intercept form and has a slope of 15. Therefore, using the point ( 1, 815), an equation in point-slope form can be written. y-y_1&=m(x-x_1) &⇓ y- 8/15&= 1/5(x- 1)
The third equation is written in point-slope form. Again, its equivalent equation in slope-intercept form will be found by solving the equation for y.
Finally, the fourth equation is in slope-intercept form. Using the slope 2 and the point ( 1, -1), its equivalent equation in point-slope form can be written. y-( -1)&= -2(x- 1) &⇕ y+1&=-2(x-1)
With the information found, the table can be now completed.
| Equation | Equivalent Equation |
|---|---|
| y+2=1/2(x+1) | y=1/2x-3/2 |
| y=1/5x+1/3, (1,8/15) | y-8/15=1/5(x-1) |
| y-30=5(x+1) | y=5x+35 |
| y=- 2x+3, (1,- 1) | y+1=- 2 (x-1) |
The following applet provides some information about a linear equation. Identify the equation in the point-slope form that corresponds to the given information.
In this lesson, the point-slope form of linear equations was explored. As its name suggests, this form communicates the slope and a point on the line described by the equation. Point-Slope Form y-y_1=m(x-x_1) Since a line has infinite points, any of these points can be used to find its equation. Therefore, a line has infinite equivalent equations in point-slope form. This can be visualized in the following applet. Select any point on the line to see how the equation varies depending on the point selected.
The revenue of a company has been increasing by $35 000 each year. In 2018, the revenue of the company was $655 000. Which equation gives the company's revenue y in thousands of dollars x years after 2000? A. & y=35x+655 B. & y-655=35(x-18) C. & x=35y+18 D. & y-18=35(x-655)
We will use the given information to write an equation in point-slope form. Consider the general form of an equation written in point-slope form. y- y_1= m(x- x_1) In this formula, ( x_1, y_1) is a specific point on the line, (x,y) represents any point on the line, and m is the slope of the line. We are told that the company's revenue increases by a constant rate of $35 000 each year. This will be the slope m of the equation. m= 35 000 We are also given that the company's revenue was $655 000 in 2018. Because x is the number of years since 2000 and y is the total revenue, ( 18, 655000) is a point on the line of the equation representing the company's revenue. We can now substitute this point and the slope into the slope-intercept form to determine the equation for the revenues. y- y_1= m(x- x_1) ⇓ y- 655 000 = 35 000(x- 18) Since we are asked to give the revenue in thousands of dollars, we will reduce our slope and y-coordinate numbers by 1000. y-655=35(x-18) This corresponds to option B.
We are asked to write an equation in point-slope form for Vincenzo's balance on the gift card as a function of books bought. A linear equation in point-slope form through the point ( x_1, y_1) with slope m has the following form. y- y_1= m(x- x_1) We are told that if he buys 3 books, the balance on the card will be $22.90. Additionally, buying 6 books will bring the balance down to $18.10. This information can be written as ordered pairs. (3,22.90) (6,18.10) We can use these points to determine the slope of the equation representing Vincenzo's balance. Let's substitute these points into the Slope Formula.
We can now substitute the slope and one of the points into the general form to determine an equation for the balance. It does not matter which point we use. For our calculation, let's use the point ( 3, 22.90). y- y_1= m(x- x_1) ⇓ y- 22.90= -1.60(x- 3)
Consider the equation found in the previous part. y-22.90= -1.60(x-3) This equation has a slope of m= -1.60, which means that buying a book reduces the money on the card by $1.60. Therefore, the price of a book at the bookstore is $1.60.
The initial amount in the gift card is given by the y-intercept of the graph of the equation. To find it, we need to rewrite the equation in the slope-intercept form.
Therefore, the initial amount in the gift card was $27.70.
We will calculate the ratio of the initial balance of the gift cards to the cost of one book to determine how many books Vincenzo can buy. 27.70/1.60≈ 17.31 Because only whole books can be bought, Vincenzo can buy 17 books.
Tearrik is driving from his home to his grandma's house. When he is 200 miles away, he looks at his watch. Later, when he is 180 miles away, he checks the time and notes that 6 minutes have passed. Write an equation in point-slope form representing the distance as a function of time that Tearrik is from his grandma's house.
We will write an equation in point-slope form to represent the distance Tearrik is from his grandma's house. Consider the general form of an equation in point slope form of a line that passes through the point ( x_1, y_1) with slope m. y- y_1= m(x- x_1) At time x=0, when Tearrik looks at his watch, he was 200 miles away. Then, 6 minutes later, he was 180 miles away. We can write this information as ordered pairs. (0,200) (6,180) We can now use these points to find the slope of the equation representing Tearrik's distance from his grandma's home.
We can find an equation in point-slope form that represents Tearrik's distance from his grandma's house as a function of time by substituting the slope and either of the identified points into the general formula. We can use either point, but let's arbitrarily use ( 6, 180). y- y_1= m(x- x_1) ⇓ y- 180= -10/3(x- 6)
To write a linear equation in point-slope form, we need to find the slope m and a point that is a solution to the equation. We are given that the sports club offers a monthly membership for $30.99. This represents the slope of the equation. m= 30.99 Additionally, since the total price for 6 months of membership is $221.99, the point ( 6, 221.99) is a solution to the equation. Using this information, we can write the equation in point-slope form. y - y_1 = m(x- x_1) ⇓ y- 221.99= 30.99(x- 6)
The slope-intercept form of a linear equation is y=mx+b, where m is the slope and b is the y-intercept. By isolating the variable y in the equation we found previously, we can find an equivalent equation written in slope-intercept form.
The initial sign-up fee is given by the y-intercept of the equation. This is because the y-intercept shows the membership cost after 0 months, so it only reflects the initial sign-up fee. We can use the equation found in Part B to identify the y-intercept.
y=30.99x+ 36.05
The y-intercept is 36.05, meaning that the sports club charges a sign-up fee of $36.05.