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| 13 Theory slides |
| 10 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.
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.
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.
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.
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)
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
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= 10 000
Subtract term
Rearrange equation
.LHS /580.=.RHS /580.
LHS+5=RHS+5
Round to nearest integer
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.Rewrite 160 as 44+116
Distribute - 1
Split into factors
Factor out 22
LHS+116=RHS+116
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.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
Use the graph to write the equation of the line in point-slope form.
We are given the following graph. Let's start by identifying the coordinates of the points.
To write the equation of the line in point-slope form, let's recall the general equation of this form. 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. We can find the slope of the line by substituting the two known points into the Slope Formula.
The slope of the line is 2. Finally, we can use the slope of 2 and either of the points, like ( 5, 3), to write the equation in point-slope form. y-y_1&=m(x-x_1) &⇓ y- 3&= 2(x- 5)
Use the graph to write the equation of the line in point-slope form.
We are given the following graph. Let's start by identifying the coordinates of the points.
To write the equation of the line in point-slope form, let's recall the general equation of this form. 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. We can find the slope of the line by substituting the two known points into the Slope Formula.
We found that the slope is - 0.75. Finally, we can use the slope of - 0.75 and either of the points, like ( 4, 1), to write the equation in point-slope form. y-y_1&=m(x-x_1) &⇓ y- 1&= - 0.75(x- 4)
What is the slope of the line given the following graph?
We are given the equation in point-slope form. y+17=23(x-9) Let's recall the general equation of the point-slope form. y-y_1=m(x-x_1) We need to rewrite the given equation to match the general equation. y+17&=23(x-9) &⇕ y-(- 17)&=23(x-9) By comparing this equation with the general equation of point-slope form, we can identify the coordinates of the points (x_1,y_1) on the line. y- y_1&=m(x- x_1) ⇓ y-( - 17)&=23(x- 9) Therefore, the point on the line is (9,- 17).
We are given the following graph.
We are asked to find the slope of the line. To do this, let's determine the coordinates of the points on the line using the coordinate plane.
Now, we can substitute the coordinates of the points into the Slope Formula.
Therefore, the slope of the line is 0.5.
What is the slope of the line given the following graph?
We are given the equation in point-slope form. y-5.2=2.5(x-6.3) Let's recall the general equation of the point-slope form. y-y_1=m(x-x_1) By comparing this equation with the general equation of point-slope form, we can identify the coordinates of the points (x_1,y_1) on the line. y- y_1&=m(x- x_1) ⇓ y-( 5.2)&=23(x- 6.3) Therefore, the point on the line is (6.3,5.2).
We are given the following graph.
We are asked to find the slope of the line. To do this, let's determine the coordinates of the points on the line using the coordinate plane.
Now, we can substitute the coordinates of the points into the Slope Formula.
Therefore, the slope of the line is - 1.8.
Which graph corresponds to the following equation? y-4.2=3.5(x-1.7)
We are asked to graph the following equation. y-4.2=3.5(x-1.7) To do so, we need to identify the point on the line and 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, we can find the specific point and the slope of the given equation. y- 4.2= 3.5(x- 1.7) As we can see, the slope is m= 3.5 and the given point is ( 1.7, 4.2). Let's start making our graph by plotting the point ( 1.7, 4.2) on the coordinate plane.
Next, we can plot a second point on the line by using the slope. Because the given equation has a slope of 3.5, we can plot a second point by going 1 unit to the right and 3.5 units up.
Finally, we can draw a line through the points to graph the given equation.
This corresponds to option B.