Sign In
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.
| | 13 Theory slides |
| | 10 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 anywhere on the line, a point together with its coordinates will appear.
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.
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).
The applet shows linear equations showing a relationship between the variables x and y. Determine whether each 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 this applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.
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 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.
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.
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.
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.
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.
Substitute ( 2017,3420) & ( 2023,4170)
Subtract term
Calculate quotient
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)
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.
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)
Izabella's initial distance from the city center is 240 meters.
y= 10 000
Subtract term
Rearrange equation
.LHS /580.=.RHS /580.
LHS+5=RHS+5
Round to nearest integer
It took Izabella around 17 minutes to reach a distance of 10 000 meters or 10 km away from the city center.
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.
B=22t-160
Transform this equation from slope-intercept form to point-slope form.A-129=56(t-33)
Transform from point-slope form to 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.
Rewrite 160 as 44+116
Distribute - 1
Split into factors
Factor out 22
LHS+116=RHS+116
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.
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.
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.
5m-9t=35
Transform this equation from standard form to point-slope form where m depends on t.m+9=6(t-11)
Transform from point-slope form to 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.
LHS+9t=RHS+9t
.LHS /5.=.RHS /5.
Rewrite 7 as 7.2-0.2
Factor out 1.8
LHS+0.2=RHS+0.2
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.
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.
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 cost of producing toy owls equals $7 plus an initial fee that is fixed. The cost of producing 16 toy owls is $138.
Write the equation that describes the situation in point-slope form and then translate it to standard form.
Let's start by recalling the general equation of point-slope form. y-y_1=m(x-x_1) Here, m is the slope and (x_1,y_1) are the coordinates of the point on the line. We know that the cost of producing one toy owl is $ 7. This means that the slope is 7. Also, we are told that the cost of producing 16 toy owls is $138. This gives the point (16,138). Slope:& 7 Point:& ( 16, 138) Let's use this information to write the equation describing the situation in point-slope form. y-y_1&=m(x-x_1) &⇓ y- 138&= 7(x- 16) Now, we need to translate this equation to slope-intercept form. First, we can recall the standard form. Ax+By=C Let's rewrite the equation so that it matches the standard form.
We found the equation in standard form describing the given situation to be the following. y-7x=26