4. Point-Slope Form of a Line
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Chapter 7
4.
Point-Slope Form of a Line
The point-slope form is a useful tool for working with linear equations and their graphs. This lesson explains how to write linear equations in point-slope form and use them to represent relationships between variables. It also explores the process of graphing linear equations using this form, enabling students to visualize and interpret linear relationships effectively. These skills are essential for solving problems in algebra and understanding how linear models apply to real-world scenarios, such as predicting trends and analyzing data.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Point-Slope Form of a Line
Slide of 13
There are different forms in which equations of lines can be written. This lesson will focus on one of those ways called the Point-Slope Form. How this form is created, its characteristics, and ways to convert between this and other forms of equations will be explored.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Exploring the Information Given by the Slope and a Point on the Line
The following applet shows the graph of a line and its slope. By clicking anywhere on the line, a point together with its coordinates will appear.
- Is it possible to write an equation for the line using the slope and a point on the line?
- What happens if this information is substituted into the Slope Formula?
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Discussion
Introducing Point-Slope Form of a Line
There is a way of writing an equation of a line when only its slope and one point that lies on the line are known. This form of an equation gets its name from two pieces of information about the line — a point and a slope.
Concept
Point-Slope Form
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).
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Pop Quiz
Identifying the Point-Slope Form of a Linear Equation
The applet shows linear equations showing a relationship between the variables x and y. Determine whether each equation is written in point-slope form.
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Identifying the Essential Parts of Equations in Point-Slope Form
To get familiar with the point-slope form, it is essential to identify the parts of its composition. In this applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.
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Discussion
Graphing a Linear Equation 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.
1
Plot the Point Given by the Equation
expand_more The point-slope form gives a point through which the line passes. This point needs to be identified first in order to graph it on the coordinate plane. Consider the given equation and compare it with the general equation in point-slope form.
y- y_1&=m(x- x_1) y- 1&=2(x- 2)
The point used to write the given equation is ( 2, 1). This point will be drawn on the coordinate plane.
2
Use the Slope to Find a Second Point
expand_more Next, a second point on the line can be found by using the slope.
y-y_1&= m(x-x_1) y-1&= 2(x-2)
In this case, the given equation has a slope of 2, which can be written as 21. Therefore, a second point can be plotted by going 1 unit to the right and 2 units up.
3
Draw a Line Through the Two Points
expand_more Finally, the line described by the equation in the point-slope form will be found by drawing a line through the two plotted points.
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Example
Graphing Equations in Point-Slope Form
Izabella is leaving her hometown behind for the city of Mathville. Her passion awaits her in Mathville. She is traveling by train at the speed of 90 kilometers per hour.
Answer
Hint
Begin by identifying the point used to write the equation. Then, use the method for graphing a linear equation in point-slope form.
Solution
To graph the equation, first identify the point used to write the equation. Then, apply the method for graphing an equation in point-slope form. A linear equation in point-slope form is written in the following way. y-y_1=m(x-x_1)
In this equation, (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. Using this information, find the specific point and the slope of the given equation.
d- 22.5= 90(t- 0.25)
The slope is m= 90 and the given point is ( 0.25, 22.5). Since both time and distance must be non-negative, the graph of the equation must be in the first quadrant. The x-axis will represent the time and the y-axis the distance. Plot the point ( 0.25, 22.5) on the coordinate plane.
Next, find a second point on the line by using the slope. Because the given equation has a slope of 90, plot a second point by moving 1 unit to the right and 90 units up.
Finally, draw a line through the points to find the line described by the given equation.
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Discussion
Writing a Linear Equation in Point-Slope Form
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.
1
Calculate the Slope
expand_more First, the slope of the line can be calculated by substituting the given points into the Slope Formula. Note that if the slope is given, this step is not needed. If only the line is provided, select two points on the line whose coordinates are easy to identify.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( -1,5) & ( 1,1)
m=1- 5/1-( -1)
▼
Evaluate right-hand side
m=-2
In this case, the slope of the line that passes through the given points is - 2.
2
Select One Point on the Line
expand_more Next, one point on the line needs to be selected. Ideally, one of the points used in the previous step is chosen, but it can be any other point on the line whose coordinates are known. Out the given points, (1,1) will be used.
3
Substitute Values
expand_more Finally, once the slope and a point on the line are known, the equation can be written by substituting these values into the general equation in point-slope form. Here, m= -2 and (x_1,y_1)=( 1, 1).
y-y_1&=m(x-x_1) &⇓ y- 1&= -2(x- 1)
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Example
Population of Mathville
As Izabella pulled into Mathville, a sign greeted her stating that population of the town is 7892. That number was recorded in the year 2012.
a The population of the town increases by around 140 citizens every year. Write the equation in point-slope form where x is the year and y is the population. Explain the answer.
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b The population of the nearest town also grew linearly since the year 2012. After five years, the population was 3420, and in the year 2023, the population became 4170. Use the two points to write an equation for the population in point-slope form.
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Hint
a Recall the general equation of the point-slope form of a line.
b Use the known data to form two points and substitute them into the Slope Formula. Then, use the slope and either of the points to write the equation in point-slope form.
Solution
a Start by recalling the point-slope form of an equation.
y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of the point. The population of the town in 2012 was 7892 people. This data gives the point ( 2012, 7892). It is also given that the population of the town increases by 140 citizens, which means that the slope is 140. y- 7892= 140(x- 2012) This way, the equation in point-slope form for the given equation was written.
b It is known that 5 years after 2012 the population of the nearest town to Mathville was 3420. This can be represented by the point (2017,3420). The other known point is (2023,4170). These two points can be substituted into the Slope Formula to determine the slope.
m=y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 2017,3420) & ( 2023,4170)
m=4170- 3420/2023- 2017
SubTerm
Subtract term
m=750/6
CalcQuot
Calculate quotient
m= 125
The slope is 125. Now, either of the known points can be substituted along with the slope into the formula for the point-slope form of an equation. For example, ( 2017, 3420). y- 3420= 125(x- 2017)
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Example
Using Point-Slope Form to Find Missing Information
In Mathville, Izabella can kindle her spirit by getting to ride the horse of her dreams — this is her passion. The given graph describes the distance Izaballa is from the city center, in meters, as she rides the horse over a certain time, in minutes.
Use the given graph to find the following information.
a Find the equation of the line in point-slope form.
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b What was Izabella's initial distance from the city center?
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c How long does it take for Izabella to be 10 000 meters from the city center? Round to the nearest minute.
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Solution
a The equation of a line that passes through the point (x_1,y_1) and has a slope of m can be written in point-slope form.
y-y_1=m(x-x_1) The given graph identifies two points on the line — either point can be used to write the equation. The slope of the line, however, needs to be calculated first. The slope can be calculated by substituting both of the given points into the Slope Formula.
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, ( 5, 3140) will be used. y-y_1&=m(x-x_1) &⇓ y- 3140&= 580(x- 5)
b Izabella's initial distance from the city center is at the point where x=0. It means that the initial distance can be found by solving the equation found in Part A for y when x= 0.
Izabella's initial distance from the city center is 240 meters.
c To find how long it took Izabella to reach the distance of 10 000 meters, the equation needs to be solved for x when y= 10 000.
y-3140=580(x-5)
Substitute
y= 10 000
10 000-3140=580(x-5)
▼
Solve for x
SubTerm
Subtract term
6860=580(x-5)
RearrangeEqn
Rearrange equation
580(x-5)=6860
DivEqn
.LHS /580.=.RHS /580.
x-5=11.827586...
AddEqn
LHS+5=RHS+5
x=16.827586...
RoundInt
Round to nearest integer
x=17
It took Izabella around 17 minutes to reach a distance of 10 000 meters or 10 km away from the city center.
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Example
Translating Between Slope-Intercept and Point-Slope Forms
Izabella is taking a break from all that horseback riding and she visits a local museum called The Incredible World Inside of Horses.
She found a screen allowing her to scroll side to side to see illustrations of horse organs. She would then read some amazing facts about them.
a The way a horse heart pumps out blood can be represented by the following equation.
B=22t-160
Transform this equation from slope-intercept form to point-slope form. b The amount of air that a horse breaths out, on average, can be modeled by the following equation.
A-129=56(t-33)
Transform from point-slope form to slope-intercept form.Answer
a Example solution: B+116=22(t-2)
b A=56t-1719
Solution
a The equation that represents how a horse heart pumps out blood is given in slope-intercept form.
B=22t-160 To transform this equation into point-slope form, recall that point-slope form looks like this. y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of a point on the line. To transform the given equation, split the constant term into two constant terms so that one of these terms has as common factor the slope of 22. Then, factor out the slope and move the remaining constant term to the left-hand side of the equation.
B=22t-160
Rewrite
Rewrite 160 as 44+116
B=22t-(44+116)
Distr
Distribute - 1
B=22t-44-116
SplitIntoFactors
Split into factors
B=22t-22* 2-116
FactorOut
Factor out 22
B=22(t-2)-116
AddEqn
LHS+116=RHS+116
B+116=22(t-2)
Therefore, an example equation in point-slope form is B+116=22(t-2). Keep in mind that this is only one of many possible equivalent equations in point-slope form.
b The following equation in point-slope form is given.
A-129=56(t-33) To rewrite this equation into slope-intercept form, isolate A on one side of the equation. To isolate A, start by removing the parentheses on the right-hand side by distributing the factor of 56 to each of the terms inside. Then, add 129 to both sides of the equation.
The resulting equation in slope-intercept form is A=56t-1719.
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Example
Translating Between General and Point-Slope Forms
Later in the museum there was a section explaining some common gestures made by horseback riders. For example, how many muscles are activated when riders wave or high-five fellow riders.
a When waving, the number of muscles m activated is given by the following equation.
5m-9t=35
Transform this equation from standard form to point-slope form where m depends on t. b When giving a high-five, the number of muscles activated is given by the following equation.
m+9=6(t-11)
Transform from point-slope form to standard form.Answer
a Example solution: m+0.2=1.8(t+4)
b Example solution: m-6t=- 75
Solution
a The equation that describes the relationship between the number of muscles m activated when waving is given in standard form.
5m-9t=35 In this equation, m depends on t. To transform this equation into point-slope form, remember that point-slope form looks like this. y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of a point on the line. Recall that m is the dependent variable and t is the independent variable. To rewrite the equation into point-slope form, start by dividing both sides of the equation by 5.
5m-9t=35
AddEqn
LHS+9t=RHS+9t
5m=9t+35
DivEqn
.LHS /5.=.RHS /5.
m=1.8t+7
Rewrite
Rewrite 7 as 7.2-0.2
m=1.8t+7.2-0.2
FactorOut
Factor out 1.8
m=1.8(t+4)-0.2
AddEqn
LHS+0.2=RHS+0.2
m+0.2=1.8(t+4)
This shows that an example equation in point-slope form is m+0.2=1.8(t+4). Keep in mind that this is only one of many possible equivalent equations in point-slope form.
b The following equation in point-slope form is given.
m+9=6(t-11) To rewrite it into standard form, start by removing the parentheses on the right-hand side by multiplying 6 by each of the terms inside. Next, move all variable terms to one side of the equation and all constant terms to the other side.
The resulting equation in standard form is m-6t=- 75. Note that this is only one of many possible equivalent equations in standard form.
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Closure
Infinite Equations in the Point-Slope Form
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.
- If the equation is in slope-intercept form, evaluate the equation for some x-value to find a point on the line. Then, use this point to find its equivalent equation in point-slope form.
- If the equation is in point-slope form, isolate the y-variable and simplify to find its equivalent equation in slope-intercept form.
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Point-Slope Form of a Line
Exercises
Use the graph to write the equation of the line in point-slope form.
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We are given the following graph. Let's start by identifying the coordinates of the points.
To write the equation of the line in point-slope form, let's recall the general equation of this form. y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of a point on the line. We can find the slope of the line by substituting the two known points into the Slope Formula.
The slope of the line is 2. Finally, we can use the slope of 2 and either of the points, like ( 5, 3), to write the equation in point-slope form. y-y_1&=m(x-x_1) &⇓ y- 3&= 2(x- 5)
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Solution
Use the graph to write the equation of the line in point-slope form.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["y-1=-0.75(x-4)","y-4=-0.75x"]}}
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We are given the following graph. Let's start by identifying the coordinates of the points.
To write the equation of the line in point-slope form, let's recall the general equation of this form. y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of a point on the line. We can find the slope of the line by substituting the two known points into the Slope Formula.
We found that the slope is - 0.75. Finally, we can use the slope of - 0.75 and either of the points, like ( 4, 1), to write the equation in point-slope form. y-y_1&=m(x-x_1) &⇓ y- 1&= - 0.75(x- 4)
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Solution
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We are given the equation in point-slope form. y+17=23(x-9) Let's recall the general equation of the point-slope form. y-y_1=m(x-x_1) We need to rewrite the given equation to match the general equation. y+17&=23(x-9) &⇕ y-(- 17)&=23(x-9) By comparing this equation with the general equation of point-slope form, we can identify the coordinates of the points (x_1,y_1) on the line. y- y_1&=m(x- x_1) ⇓ y-( - 17)&=23(x- 9) Therefore, the point on the line is (9,- 17).
We are given the following graph.
We are asked to find the slope of the line. To do this, let's determine the coordinates of the points on the line using the coordinate plane.
Now, we can substitute the coordinates of the points into the Slope Formula.
Therefore, the slope of the line is 0.5.
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Solution
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We are given the equation in point-slope form. y-5.2=2.5(x-6.3) Let's recall the general equation of the point-slope form. y-y_1=m(x-x_1) By comparing this equation with the general equation of point-slope form, we can identify the coordinates of the points (x_1,y_1) on the line. y- y_1&=m(x- x_1) ⇓ y-( 5.2)&=23(x- 6.3) Therefore, the point on the line is (6.3,5.2).
We are given the following graph.
We are asked to find the slope of the line. To do this, let's determine the coordinates of the points on the line using the coordinate plane.
Now, we can substitute the coordinates of the points into the Slope Formula.
Therefore, the slope of the line is - 1.8.
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Solution
Which graph corresponds to the following equation? y-4.2=3.5(x-1.7)
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We are asked to graph the following equation. y-4.2=3.5(x-1.7) To do so, we need to identify the point on the line and then apply the method for graphing an equation in point-slope form. A linear equation in point-slope form is written in the following way. y-y_1=m(x-x_1) In this equation, (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. Using this information, we can find the specific point and the slope of the given equation. y- 4.2= 3.5(x- 1.7) As we can see, the slope is m= 3.5 and the given point is ( 1.7, 4.2). Let's start making our graph by plotting the point ( 1.7, 4.2) on the coordinate plane.
Next, we can plot a second point on the line by using the slope. Because the given equation has a slope of 3.5, we can plot a second point by going 1 unit to the right and 3.5 units up.
Finally, we can draw a line through the points to graph the given equation.
This corresponds to option B.
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Solution
Point-Slope Form of a Line
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