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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.**

Explore

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.

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

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

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

Pop Quiz

The applet shows linear equations showing a relationship between the variables $x$ and $y.$ Determine whether each 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 this applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.

Discussion

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 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

### Solution

The distance $d$ traveled by the train from Izabella's hometown is given by the following equation.

$d−22.5=90(t−0.25) $

In this equation, $t$ is the time in hours since the train departed from Izabella's hometown. Graph the equation on a coordinate plane. Begin by identifying the point used to write the equation. Then, use the method for graphing a linear equation in point-slope form.

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.

Discussion

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

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["y-7892=140(x-2012)"]}}

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["y-3420=125(x-2017)","y-4170=125(x-2023)"]}}

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.

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=x_{2}−x_{1}y_{2}−y_{1} $

SubstitutePoints

Substitute $(2017,3420)$ & $(2023,4170)$

$m=2023−20174170−3420 $

SubTerm

Subtract term

$m=6750 $

CalcQuot

Calculate quotient

$m=125$

$y−3420=125(x−2017) $

Example

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["y-3140=580(x-5)","y-8360=580(x-14)"]}}

b What was Izabella's initial distance from the city center?

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"meters","answer":{"text":["240"]}}

c How long does it take for Izabella to be $10000$ meters from the city center? Round to the nearest minute.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Around","formTextAfter":"minutes","answer":{"text":["17"]}}

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.
$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstitutePoints

Substitute $(5,3140)$ & $(14,8360)$

$m=14−58360−3140 $

$m=580$

$y−y_{1}y−3140 =m(x−x_{1})⇓=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 $x=0.$

c To find how long it took Izabella to reach the distance of $10000$ meters, the equation needs to be solved for $x$ when $y=10000.$

$y−3140=580(x−5)$

Substitute

$y=10000$

$10000−3140=580(x−5)$

▼

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

$x=17$

Example

Izabella is taking a break from all that horseback riding and she visits a local museum called
### Answer

### Solution

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.

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. a **Example solution:** $B+116=22(t−2)$

b $A=56t−1719$

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

$LHS+116=RHS+116$

$B+116=22(t−2)$

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 $A=56t−1719.$
Example

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.
### Answer

### Solution

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.
The resulting equation in standard form is $m−6t=-75.$ Note that this is only one of many possible equivalent equations in standard form.

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. a **Example solution:** $m+0.2=1.8(t+4)$

b **Example solution:** $m−6t=-75$

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

$LHS+0.2=RHS+0.2$

$m+0.2=1.8(t+4)$

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.
$m+9=6(t−11)$

Distr

Distribute $6$

$m+9=6t−66$

SubEqn

$LHS−9=RHS−9$

$m=6t−75$

SubEqn

$LHS−6t=RHS−6t$

$m−6t=-75$

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−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.
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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