Learn How to Graph Point-Slope Form and Master Writing Equations in Point Slope Form

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.

Coordinate Plane
Linear Equations
Solving Linear Equations
Slope-Intercept Form
Slope Formula
Rate of Change

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 on the line, a point can be seen jointly with its coordinates.

an applet showing the graph of a line and its slope. Any point on the line can be chosen to see 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

Finding the Equation of a Line in Point-Slope Form

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

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 $(x_{1}, y_{1}) .$

It is worth mentioning that the point-slope form can only be written for non-vertical lines.

Why

Derivation of the Formula

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 = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstituteII

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

m = \frac{y - y _{1}}{x - x _{1}}

MultEqn

$LHS \cdot (x - x_{1}) = RHS \cdot (x - x_{1})$

m (x - x_{1}) = y - y_{1}

RearrangeEqn

Rearrange equation

y - y_{1} = m (x - x_{1})

Pop Quiz

Identifying the Point-Slope Form of a Linear 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.

Linear equation written in different forms

Pop Quiz

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 the following applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.

An apple that generates linear equations in point-slope form and asked for its components

Discussion

Visualizing the Graph of a Linear Equation in Point-Slope Form

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

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.

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.

The point given by the point-slope form plotted on a coordinate plane

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

\frac{2}{1} .

Therefore, a second point can be plotted by going

1

unit to the right and

2

units up.

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.

A line with the equation y-1=2(x-2) passing through the points (2,1) and (3,3).

Example

Graphing Equations in Point-Slope Form to Visualize Relationships Between Variables

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

\underline{Fahrenheit to Celsius} C = \frac{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.

Tadeo got 23 C degrees and Magdalena got 25 C degrees.

Help Izabella to create the graph and have a better understanding of the given equation.

Answer

Graph:

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 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 = \frac{5}{9} (F - 32) ⇕ \frac{5}{9} (F - 32)

As can be seen, the slope is

m = \frac{5}{9}

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 $\frac{5}{9},$ 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

Modeling Daily Situations Into Equations in Point-Slope Form

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 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?

Answer

d - 500 = 80 (t - 4)

b Graph:

980

Kilometers

180

Kilometers

Hint

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 .

Solution

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 $\frac{8 0}{1} .$ 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 .

d - 500 = 80 (t - 4)

Substitute

$t = 10$

d - 500 = 80 (10 - 4)

▼

Solve for

d

SubTerm

Subtract term

d - 500 = 80 (6)

Multiply

d - 500 = 480

AddEqn

$LHS + 500 = RHS + 500$

d = 980

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.

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

$a (- b) = - a \cdot b$

d - 500 = - 360

AddEqn

$LHS + 500 = RHS + 500$

d = - 360 + 500

AddTerms

Add terms

d = 180

This means that LaShay started her analysis after the bus traveled

180

kilometers.

Discussion

How Much Information Is Needed to Write an Equation in Point-Slope Form?

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

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.

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.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

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

m = \frac{1 - 5}{1 - ( - 1 )}

▼

Evaluate right-hand side

SubNeg

$a - (- b) = a + b$

m = \frac{1 - 5}{1 + 1}

AddSubTerms

Add and subtract terms

m = \frac{- 4}{2}

CalcQuot

Calculate quotient

m = - 2

In this case, the slope of the line that passes through the given points is

- 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.

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)

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.

Example

Using Point-Slope Form to Find Missing Information

Kevin went parachuting. The following graph describes his descent in terms of the time until he landed.

A line describing the descending of a parachutist

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?

Answer

a Example Equation:

y - 1500 = - 600 (x - 4)

3900

feet

6.5

minutes

Hint

a Use the given points to find the slope of the line.

b Solve the equation found in Part A for

y

when

x = 0 .

c Solve the equation found in Part A for

x

when

y = 0 .

Solution

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 = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

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

m = \frac{1 5 0 0 - 2 7 0 0}{4 - 2}

▼

Evaluate right-hand side

SubTerms

Subtract terms

m = \frac{- 1 2 0 0}{2}

MoveNegNumToFrac

Put minus sign in front of fraction

m = - \frac{1 2 0 0}{2}

CalcQuot

Calculate quotient

m = - 600

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} 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

$- a (- b) = a \cdot b$

y - 2700 = 1200

AddEqn

$LHS + 2700 = RHS + 2700$

y = 3900

The height where Kevin started descending is

3900

feet.

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

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

\frac{- 2 7 0 0}{- 6 0 0} = x - 2

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

\frac{2 7 0 0}{6 0 0} = x - 2

AddEqn

$LHS + 4 = RHS + 4$

\frac{2 7 0 0}{6 0 0} + 2 = x

RearrangeEqn

Rearrange equation

x = \frac{2 7 0 0}{6 0 0} + 2

NumberToFrac

$a = \frac{6 0 0 \cdot a}{6 0 0}$

x = \frac{2 7 0 0}{6 0 0} + \frac{1 2 0 0}{6 0 0}

AddFrac

Add fractions

x = \frac{3 9 0 0}{6 0 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 3 0 0}{b / 3 0 0}$

x = \frac{1 3}{2}

WriteDec

Write as a decimal

x = 6.5

It took Kevin

6.5

minutes to land.

Example

Matching Equivalent Equations in Point-Slope Form and Slope-Intercept Form

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
$y + 2 = \frac{1}{2} (x + 1)$
$y = \frac{1}{5} x + \frac{1}{3},$ $(1, \frac{8}{1 5})$
$y - 30 = 5 (x + 1)$
$y = - 2 x + 3,$ $(1, - 1)$

Help Kriz complete the table.

Answer

Equation	Equivalent Equation
$y + 2 = \frac{1}{2} (x + 1)$	$y = \frac{1}{2} x - \frac{3}{2}$
$y = \frac{1}{5} x + \frac{1}{3},$ $(1, \frac{8}{1 5})$	$y - \frac{8}{1 5} = \frac{1}{5} (x - 1)$
$y - 30 = 5 (x + 1)$	$y = 5 x + 35$
$y = - 2 x + 3,$ $(1, - 1)$	$y + 1 = - 2 (x - 1)$

Hint

If the equation is in point-slope form, isolate the $y$ variable to find its equivalent equation. Conversely, if the equation is in slope-intercept form, use the given point and the slope to write its equivalent equation.

Solution

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.

First Equation

The first equation is in point-slope form. Then, the

y

variable will be isolated to find its equivalent equation in slope-intercept form.

y + 2 = \frac{1}{2} (x + 1)

Distr

Distribute $\frac{1}{2}$

y + 2 = \frac{1}{2} x + \frac{1}{2} \cdot 1

MultByOne

$a \cdot 1 = a$

y + 2 = \frac{1}{2} x + \frac{1}{2}

SubEqn

$LHS - 2 = RHS - 2$

y = \frac{1}{2} x + \frac{1}{2} - 2

NumberToFrac

$a = \frac{2 \cdot a}{2}$

y = \frac{1}{2} x + \frac{1}{2} - \frac{4}{2}

SubTerm

Subtract term

y = \frac{1}{2} x - \frac{3}{2}

Second Equation

The second equation is in slope-intercept form and has a slope of

\frac{1}{5} .

Therefore, using the point

(1, \frac{8}{1 5}),

an equation in point-slope form can be written.

y - y_{1} y - \frac{8}{1 5} = m (x - x_{1}) ⇓ = \frac{1}{5} (x - 1)

Third Equation

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 .

y - 30 = 5 (x + 1)

Distr

Distribute $5$

y - 30 = 5 x + 5

SubEqn

$LHS - 30 = RHS - 30$

y = 5 x + 35

Fourth Equation

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) y + 1 = - 2 (x - 1) ⇕ = - 2 (x - 1)

Completing the Table

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

Equation	Equivalent Equation
$y + 2 = \frac{1}{2} (x + 1)$	$y = \frac{1}{2} x - \frac{3}{2}$
$y = \frac{1}{5} x + \frac{1}{3},$ $(1, \frac{8}{1 5})$	$y - \frac{8}{1 5} = \frac{1}{5} (x - 1)$
$y - 30 = 5 (x + 1)$	$y = 5 x + 35$
$y = - 2 x + 3,$ $(1, - 1)$	$y + 1 = - 2 (x - 1)$

Pop Quiz

Using Different Information to Find Equations in the Point-Slope Form

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

An applet that asks to identify the equation in point-slope form that matches the given information

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.

\underline{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.

an applet showing the graph of the line and its equation depending on the point selected

Moreover, transforming between point-slope and slope-intercept forms can be addressed by the following procedure.

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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Catch-Up and Review

Why

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

First Equation

Second Equation

Third Equation

Fourth Equation

Completing the Table