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| 13 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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.
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.
x_2= x, y_2= y
LHS * (x-x_1)=RHS* (x-x_1)
Rearrange 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.
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.
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.
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.
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.
Finally, the line described by the equation in the point-slope form will be found by drawing a line through the two plotted points.
Izabella learned that the following equation gives the relationship between degrees Fahrenheit and degrees Celsius. Fahrenheit to Celsius C=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.
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=&5/9(F-32) & ⇕ C- 0=& 5/9(F- 32) As can be seen, the slope is m= 59 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 59, 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.
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&= m(t- t_1) &⇓ d- 500&= 80(t- 4)
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 801. 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.
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.
t= 0
Subtract term
a(- b)=- a * b
LHS+500=RHS+500
Add terms
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.
Substitute ( -1,5) & ( 1,1)
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.
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)
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.
Substitute ( 2,2700) & ( 4,1500)
Subtract terms
Put minus sign in front of fraction
Calculate quotient
x= 0
Subtract term
- a(- b)=a* b
LHS+2700=RHS+2700
y= 0
Subtract term
.LHS /(-600).=.RHS /(-600).
- a/- b=a/b
LHS+4=RHS+4
Rearrange equation
a = 600* a/600
Add fractions
a/b=.a /300./.b /300.
Write as a decimal
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=1/2(x+1) | |
y=1/5x+1/3, (1,8/15) | |
y-30=5(x+1) | |
y=- 2x+3, (1,- 1) |
Equation | Equivalent Equation |
---|---|
y+2=1/2(x+1) | y=1/2x-3/2 |
y=1/5x+1/3, (1,8/15) | y-8/15=1/5(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.
Distribute 1/2
a * 1=a
LHS-2=RHS-2
a = 2* a/2
Subtract term
The second equation is in slope-intercept form and has a slope of 15. Therefore, using the point ( 1, 815), an equation in point-slope form can be written. y-y_1&=m(x-x_1) &⇓ y- 8/15&= 1/5(x- 1)
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)&= -2(x- 1) &⇕ y+1&=-2(x-1)
With the information found, the table can be now completed.
Equation | Equivalent Equation |
---|---|
y+2=1/2(x+1) | y=1/2x-3/2 |
y=1/5x+1/3, (1,8/15) | y-8/15=1/5(x-1) |
y-30=5(x+1) | y=5x+35 |
y=- 2x+3, (1,- 1) | y+1=- 2 (x-1) |
The following applet provides some information about a linear equation. Identify the equation in the point-slope form that corresponds to the given information.
A linear equation shares its y-intercept with the equation x+2y=4. Furthermore, its graph contains the point (2,5).
We are given that the missing equation shares its y-intercept with the equation x+2y=4. To find the y-intercept of the missing equation, we will rewrite the given equation in the slope-intercept form. Slope-Intercept Form y= mx+ b In this equation, m is the slope and b is the y-intercept. Let's rewrite the given equation.
We can now identify the y-intercept of the given equation. y=-1/2x+ 2 ⇒ b= 2 The y-intercept of the given equation is 2, which is also the y-intercept of the missing equation.
We are given that the graph of the missing equation contains the point ( 2, 5). Additionally, we have found that the y-intercept of the equation is 2, which can be written as ( 0, 2). These two points can be substituted into the Slope Formula to find the slope of the missing equation.
Therefore, the slope of the missing equation is 32.
Consider the general form of an equation in point-slope form. y- y_1= m(x- x_1) In this equation, m is the slope and ( x_1, y_1) a specific point on the graph of the equation. We already found that the slope of the missing equation is 32. Therefore, to write the missing equation for the line that passes through the point ( 2, 5) with the given slope, we will substitute this information into the general equation. y- y_1&= m(x- x_1) &⇓ y- 5&= 3/2(x- 2)
Recall that parallel lines have the same slope. Using this fact, we can determine the slope of the equation we want to write. Let's start by writing the given equation in slope-intercept form to find its slope. y= mx+ b In this form, m is the slope and b is the y-intercept. Let's isolate y in the given equation.
This equation has a slope of 58. Since the line also has this slope, we can use it along with the given point ( 12, 35) to find the equation of the line in point-slope form. y- y_1&= m(x- x_1) &⇓ y- 3/5&= 5/8(x- 1/2)