{{ 'ml-label-loading-course' | message }}

{{ tocSubheader }}

{{ 'ml-toc-proceed-mlc' | message }}

{{ 'ml-toc-proceed-tbs' | message }}

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.intro.summary }}

Show less Show more Lesson Settings & Tools

| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |

| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |

| {{ 'ml-lesson-time-estimation' | message }} |

Among the ways of writing the equation of a line, *point-slope form* highlights the slope and a point on the line that is not the $y-$intercept. This lesson will explore how to write the *point-slope form* of a line and some of its applications. It will also show how to transform an equation in *point-slope form* to an equation in slope-intercept form and vice versa. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Explore

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.

Think about the following questions.

- 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?

Discussion

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.

Concept

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.

The point-slope form can be derived by using the Slope Formula. To do so, $(x,y)$ — which represents any point on the line — is substituted for $(x_{2},y_{2})$ into the formula.

$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstituteII

$x_{2}=x$, $y_{2}=y$

$m=x−x_{1}y−y_{1} $

MultEqn

$LHS⋅(x−x_{1})=RHS⋅(x−x_{1})$

$m(x−x_{1})=y−y_{1}$

RearrangeEqn

Rearrange equation

$y−y_{1}=m(x−x_{1})$

Pop Quiz

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.

Pop Quiz

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.

Discussion

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.

Method

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. *expand_more*
*expand_more*
*expand_more*

$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

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}y−1 =m(x−x_{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

Next, a second point on the line can be found by using the slope.

$y−y_{1}y−1 =m(x−x_{1})=2(x−2) $

In this case, the given equation has a slope of $2,$ which can be written as $12 .$ 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

Finally, the line described by the equation in the point-slope form will be found by drawing a line through the two plotted points.

Example

Izabella learned that the following equation gives the relationship between degrees Fahrenheit and degrees Celsius. ### Answer

### Hint

### Solution

$Fahrenheit to Celsius C=95 (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.
**Graph: **

Begin by identifying the point used to write the equation. Then, use the method for graphing a linear equation in point-slope form.

To make the graph, the point used to write the equation needs to be identified first. Then, the method for graphing an equation in point-slope form can be applied. A linear equation in point-slope form has the following form.

$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, the specific point and the slope of the given equation can be found.
$C=C−0= 95 (F−32)⇕95 (F−32) $

As can be seen, the slope is $m=95 $ and the given point is $(32,0).$ All the information needed for applying the method for graphing an equation in point-slope form has been found. The $x-$axis will represent the Fahrenheit degrees and the $y-$axis the Celsius degrees. Then, the point $(32,0)$ will be plotted on the coordinate plane.
Next, a second point on the line can be found by using the slope. Because the given equation has a slope of $95 ,$ 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.

Example

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.
### Answer

### Hint

### Solution

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.
This means that LaShay started her analysis after the bus traveled $180$ kilometers.

Answer the following questions to help LaShay finish her analysis.

a Find an equation in point-slope form that models the distance $d$ traveled by the bus after $t$ hours.

b Graph the equation that describes the movement of the bus.

c How far has the bus driven $10$ hours after LaShay started her analysis?

d How far had the bus driven before LaShay started analyzing its movement?

a $d−500=80(t−4)$

b **Graph: **

c $980$ Kilometers

d $180$ Kilometers

a The rate of change represents the line's slope, and the ordered pair given by the time and distance traveled is a point on the line.

b Begin by plotting the point given by the equation. Then, using the slope, find a second point.

c Evaluate the equation found in Part A when $t=10.$

d To find the distance where LaShay started her analysis, evaluate the equation found in Part A when $t=0.$

a Because the bus moves at a constant rate, its movement can be described with a linear equation in point-slope form. An equation in point-slope form through the point $(x_{1},y_{1})$ and slope $m$ has the following form.

$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}d−500 =m(t−t_{1})⇓=80(t−4) $

b To graph the equation that models the movement of the bus, the specific point given by the equation needs to be plotted first. This point is $(4,500).$ Moreover, the graph will be only on the first quadrant since distance and time are non-negative quantities.

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 $180 .$ 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.

c To find the distance traveled after $10$ hours, the equation from Part A can be evaluated when $t=10$ and solved for $d.$

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.

d The distance at which LaShay started her analysis can be calculated by evaluating the previous equation when $t=0.$

$d−500=80(t−4)$

Substitute

$t=0$

$d−500=80(0−4)$

▼

Solve for $d$

SubTerm

Subtract term

$d−500=90(-4)$

MultPosNeg

$a(-b)=-a⋅b$

$d−500=-360$

AddEqn

$LHS+500=RHS+500$

$d=-360+500$

AddTerms

Add terms

$d=180$

Discussion

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.

Method

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. *expand_more*
*expand_more*
*expand_more*
Note that, unlike the slope-intercept form, an *infinite* number of equations written in point-slope form can represent the same line. In other words, when written in point-slope form, any line does not have only one unique equation.

$(-1,5)and(1,1) $

Three steps are needed to find the equation in point-slope form. 1

Calculate the Slope

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.
In this case, the slope of the line that passes through the given points is $-2.$

$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstitutePoints

Substitute $(-1,5)$ & $(1,1)$

$m=1−(-1)1−5 $

▼

Evaluate right-hand side

$m=-2$

2

Select One Point on the Line

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

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}y−1 =m(x−x_{1})⇓=-2(x−1) $

Example

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.

a Find the equation of the line in point-slope form.

b What is the height at which Kevin starts descending?

c How long does it take Kevin to land?

a **Example Equation: **$y−1500=-600(x−4)$

b $3900$ feet

c $6.5$ minutes

a The equation of a line through the point $(x_{1},y_{1})$ and slope $m$ has the following form.

$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.
$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstitutePoints

Substitute $(2,2700)$ & $(4,1500)$

$m=4−21500−2700 $

▼

Evaluate right-hand side

SubTerms

Subtract terms

$m=2-1200 $

MoveNegNumToFrac

Put minus sign in front of fraction

$m=-21200 $

CalcQuot

Calculate quotient

$m=-600$

$y−y_{1}y−2700 =m(x−x_{1})⇓=-600(x−2) $

b When $x=0$, Kevin is at the height where he started descending. It means that the initial height can be found by solving the equation found in Part A for $y$ when $x=0.$

$y−2700=-600(x−2)$

Substitute

$x=0$

$y−2700=-600(0−2)$

▼

Solve for $y$

SubTerm

Subtract term

$y−2700=-600(-2)$

MultNegNegOnePar

$-a(-b)=a⋅b$

$y−2700=1200$

AddEqn

$LHS+2700=RHS+2700$

$y=3900$

c When Kevin landed, he was at the height of $y=0.$ Therefore, to find how long it took Kevin to land, the equation needs to be solved for $x$ when $y=0.$

$y−2700=-600(x−2)$

Substitute

$y=0$

$0−2700=-600(x−2)$

▼

Solve for $x$

SubTerm

Subtract term

$-2700=-600(x−2)$

DivEqn

$LHS/(-600)=RHS/(-600)$

$-600-2700 =x−2$

DivNegNeg

$-b-a =ba $

$6002700 =x−2$

AddEqn

$LHS+4=RHS+4$

$6002700 +2=x$

RearrangeEqn

Rearrange equation

$x=6002700 +2$

NumberToFrac

$a=600600⋅a $

$x=6002700 +6001200 $

AddFrac

Add fractions

$x=6003900 $

ReduceFrac

$ba =b/300a/300 $

$x=213 $

WriteDec

Write as a decimal

$x=6.5$

Example

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 |
---|---|

$y+2=21 (x+1)$ | |

$y=51 x+31 ,$ $(1,158 )$ | |

$y−30=5(x+1)$ | |

$y=-2x+3,$ $(1,-1)$ |

Equation | Equivalent Equation |
---|---|

$y+2=21 (x+1)$ | $y=21 x−23 $ |

$y=51 x+31 ,$ $(1,158 )$ | $y−158 =51 (x−1)$ |

$y−30=5(x+1)$ | $y=5x+35$ |

$y=-2x+3,$ $(1,-1)$ | $y+1=-2(x−1)$ |

For equations in slope-intercept form, its equivalent equation will be found by isolating the $y$ variable. On the other hand, the given point jointly with the slope will be used to find the equivalent equation of the equations in slope-intercept form.

$y+2=21 (x+1)$

Distr

Distribute $21 $

$y+2=21 x+21 ⋅1$

MultByOne

$a⋅1=a$

$y+2=21 x+21 $

SubEqn

$LHS−2=RHS−2$

$y=21 x+21 −2$

NumberToFrac

$a=22⋅a $

$y=21 x+21 −24 $

SubTerm

Subtract term

$y=21 x−23 $

$y−y_{1}y−158 =m(x−x_{1})⇓=51 (x−1) $

$y−(-1)y+1 =-2(x−1)⇕=-2(x−1) $

With the information found, the table can be now completed.

Equation | Equivalent Equation |
---|---|

$y+2=21 (x+1)$ | $y=21 x−23 $ |

$y=51 x+31 ,$ $(1,158 )$ | $y−158 =51 (x−1)$ |

$y−30=5(x+1)$ | $y=5x+35$ |

$y=-2x+3,$ $(1,-1)$ | $y+1=-2(x−1)$ |

Pop Quiz

The following applet provides some information about a linear equation. Identify the equation in the point-slope form that corresponds to the given information.

Closure

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.

$Point-Slope Form y$