Exercises marked with requires Mathleaks premium to view it's solution in the app.
Download Mathleaks app on Google Play or iTunes AppStore.

##### Sections

###### Communicate Your Answer

Exercise name | Free? |
---|---|

Communicate Your Answer 4 | |

Communicate Your Answer 5 |

###### Monitoring Progress

Exercise name | Free? |
---|---|

Monitoring Progress 1 | |

Monitoring Progress 2 | |

Monitoring Progress 3 | |

Monitoring Progress 4 | |

Monitoring Progress 5 | |

Monitoring Progress 6 |

###### Exercises

Exercise name | Free? |
---|---|

Exercises 1 Equations in point-slope form follow a specific format. y−y1=m(x−x1) In this form, m is the slope of the line and (x1,y1) is a point on the line. Let's modify the given equation just a little bit, without simplifying it. y−5=-2(x+5)⇔y−5=-2(x−(-5)) Now, we can easily identify the slope and one of the points. Slope: Point: -2 (-5,5) | |

Exercises 2 The formula for slope can be written as: m=x2−x1y2−y1, where m is the slope and (x1,y1) and (x2,y2) are two arbitrary points on the line. We have been given a slope of 4 and a point on the line, (3,-2). If we substitute this information into the slope formula, we have: 4=x2−3y2−(-2). We can simplify this equation to be in the point-slope form, y−y1=m(x−x1). First, we can remove the subscripts because there is only one of each variable. 4=x2−3y2−(-2)Rewrite y2 as y4=x2−3y−(-2)Rewrite x2 as x4=x−3y−(-2) Next we can simplify and use inverse operations to rearrange the equation to be in the correct form. 4=x−3y−(-2)a−(-b)=a+b4=x−3y+2LHS⋅(x−3)=RHS⋅(x−3)4(x−3)=y+2Rearrange equationy+2=4(x−3) As we have shown, the slope formula and the point-slope form are very closely related. We can easily find one when given the other. | |

Exercises 3 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (2,1) and the slope 2, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=2,y1=1,m=2y−1=2(x−2) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 4 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (3,5) and the slope -1, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=3,y1=5,m=-1y−5=-1(x−3) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 5 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (7,-4) and the slope -6, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=7,y1=-4,m=-6y−(-4)=-6(x−7)a−(-b)=a+by+4=-6(x−7) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 6 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (-8,-2) and the slope 5. We have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=-8,y1=-2,m=5y−(-2)=5(x−(-8))a−(-b)=a+by+2=5(x+8) Please note that this is only one of infinitely many possible equations for this line. Any point lying on the line could be used to create an equivalent equation. | |

Exercises 7 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (9,0) and the slope -3, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=9,y1=0,m=-3y−0=-3(x−9)Subtract termy=-3(x−9) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 8 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (0,2) and the slope 4, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=0,y1=2,m=4y−2=4(x−0)Subtract termy−2=4x Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 9 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (-6,6) and the slope 23, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=-6,y1=6,m=23y−6=23(x−(-6))a−(-b)=a+by−6=23(x+6) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 10 An equation in point-slope form follows a specific format. y−y1=m(x−x1) In this form, m is the slope and the point (x1,y1) lies on the line. We are given the point (5,-12) and the slope -52, so we have everything we need to create a point-slope form equation. y−y1=m(x−x1)Substitute x1=5,y1=-12,m=-52y−(-12)=-52(x−5)a−(-b)=a+by+12=-52(x−5) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form. | |

Exercises 11 Using the given points, (1,-3) and (3,1), we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (1,-3) and (3,1). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (1,-3) & (3,1)m=3−11−(-3)a−(-b)=a+bm=3−11+3Add and subtract termsm=24Calculate quotientm=2 Thus, the slope of the line is 2.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (1,-3). y−(-3)=2(x−1) ⇒y+3=2(x−1)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y+3=2(x−1)Distribute 2y+3=2x−2LHS−3=RHS−3y=2x−5 Now that the y-variable is isolated, our equation is in slope-intercept form. y=2x−5 | |

Exercises 12 Using the given points, (-4,0) and (1,-5), we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (-4,0) and (1,-5). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-4,0) & (1,-5)m=1−(-4)-5−0a−(-b)=a+bm=1+4-5−0Add and subtract termsm=5-5Calculate quotientm=-1 Thus, the slope of the line is -1.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-4,0). y−0=-1(x−(-4)) ⇒y=-1(x+4)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y=-1(x+4)Distribute -1y=-x−4 Now that the y-variable is isolated, our equation is in slope-intercept form. y=-x−4 | |

Exercises 13 Using the given points, (-6,4) and (-2,2), we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (-6,4) and (-2,2). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-6,4) & (-2,2)m=-2−(-6)2−4a−(-b)=a+bm=-2+62−4Add and subtract termsm=4-2Calculate quotientm=-21 Thus, the slope of the line is -21.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-6,4). y−4=-21(x−(-6)) ⇒y−4=-21(x+6)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−4=-21(x+6)Distribute -21y−4=-21x−3LHS+4=RHS+4y=-21x+1 Now that the y-variable is isolated, our equation is in slope-intercept form. y=-21x+1 | |

Exercises 14 Using the given points, (4,1) and (8,4), we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (4,1) and (8,4). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (4,1) & (8,4)m=8−44−1Subtract termsm=43 Thus, the slope of the line is 43.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (4,1). y−1=43(x−4)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−1=43(x−4)Distribute 43y−1=43x−3LHS+1=RHS+1y=43x−2 Now that the y-variable is isolated, our equation is in slope-intercept form. y=43x−2 | |

Exercises 15 Using the given points (7,2) and (2,12) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (7,2) and (2,12). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (7,2) & (2,12)m=2−712−2Subtract termsm=-510Calculate quotientm=-2 Thus, the slope of the line is -2.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line in point-slope form. Let's choose the point (7,2). y−2=-2(x−7)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−2=-2(x−7)Distribute -2y−2=-2x+14LHS+2=RHS+2y=-2x+16 Now that the y-variable is isolated, our equation is in slope-intercept form. y=-2x+16 | |

Exercises 16 Using the given points (6,-2) and (12,1) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (6,-2) and (12,1). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (6,-2) & (12,1)m=12−61−(-2)a−(-b)=a+bm=12−61+2Add and subtract termsm=63ba=b/3a/3m=21 Thus, the slope of the line is 21.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (6,-2). y−(-2)=21(x−6) ⇒y+2=21(x−6)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y+2=21(x−6)Distribute 21y+2=21x−3LHS−2=RHS−2y=21x−5 Now that the y-variable is isolated, our equation is in slope-intercept form. y=21x−5 | |

Exercises 17 Using the given points (6,-1) and (3,-7) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (6,-1) and (3,-7). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (6,-1) & (3,-7)m=3−6-7−(-1)a−(-b)=a+bm=3−6-7+1Add and subtract termsm=-3-6Calculate quotientm=2 Thus, the slope of the line is 2.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (6,-1). y−(-1)=2(x−6) ⇒y+1=2(x−6)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y+1=2(x−6)Distribute 2y+1=2x−12LHS−1=RHS−1y=2x−13 Now that the y-variable is isolated, our equation is in slope-intercept form. y=2x−13 | |

Exercises 18 Using the given points (-2,5) and (-4,-5) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (-2,5) and (-4,-5). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-2,5) & (-4,-5)m=-4−(-2)-5−5a−(-b)=a+bm=-4+2-5−5Add and subtract termsm=-2-10-b-a=bam=5 Thus, the slope of the line is 5.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-2,5). y−5=5(x−(-2)) ⇒y−5=5(x+2)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−5=5(x+2)Distribute 5y−5=5x+10LHS+5=RHS+5y=5x+15 Now that the y-variable is isolated, our equation is in slope-intercept form. y=5x+15 | |

Exercises 19 Using the given points (1,-9) and (-3,-9) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (1,-9) and (-3,-9). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (1,-9) & (-3,-9)m=-3−1-9−(-9)a−(-b)=a+bm=-3−1-9+9Add and subtract termsm=-40Calculate quotientm=0 Thus, the slope of the line is 0.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (1,-9). y−(-9)=0(x−1) ⇒y+9=0Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y+9=0LHS−9=RHS−9y=-9 Now that the y-variable is isolated, our equation is in slope-intercept form. y=-9 | |

Exercises 20 Using the given points (-5,19) and (5,13) we will write an equation in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line passes through the points (-5,19) and (5,13). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-5,19) & (5,13)m=5−(-5)13−19a−(-b)=a+bm=5+513−19Add and subtract termsm=10-6Put minus sign in front of fractionm=-106ba=b/2a/2m=-53 Thus, the slope of the line is -53.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-5,19). y−19=-53(x−(-5)) ⇒y−19=-53(x+5)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−19=-53(x+5)Distribute -53y−19=-53x−3LHS+19=RHS+19y=-53x+16 Now that the y-variable is isolated, our equation is in slope-intercept form. y=-53x+16 | |

Exercises 21 Notice that the given two points, f(2)=-2 and f(1)=1, are in function notation. To start, let's write these points as coordinate pairs. Recall that the input is the x-coordinate and the output is the y-coordinate.f(x)=y(x,y) f(2)=-2(2,-2) f(1)=1(1,1) We can't determine the y-intercept of the equation from the given points. Therefore, to write the equation of the line in slope-intercept form, we will follow three steps.First, we will find the slope of the equation by using the Slope Formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (2,-2) and (1,1). Let's substitute these points into the slope formula and find the slope m. m=x2−x1y2−y1Substitute (2,-2) & (1,1)m=1−21−(-2) Simplify right-hand side a−(-b)=a+bm=1−21+2Add and subtract termsm=-13Calculate quotient m=-3 The slope of the line is -3.Point-Slope Form Let's recall the point-slope form of a line. y−y1=m(x−x1) Here, m is the slope and (x1,y1) is a point on the line. We already know that the slope is -3. We can choose any of the given points and write the equation of the line. Let's use (2,-2). y−(-2)=-3(x−2)⇔y+2=-3(x−2)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable in the equation. y+2=-3(x−2)Distribute -3y+2=-3x+6LHS−2=RHS−2y=-3x+4 | |

Exercises 22 Notice that the given two points, f(5)=7 and f(-2)=0, are in function notation. To start, let's write these points as coordinate pairs. Remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(5)=7⇔(5,7)f(-2)=0⇔(-2,0) Now we are able to write an equation for function f in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (5,7) and (-2,0). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (5,7) & (-2,0)m=-2−50−7Subtract termsm=-7-7aa=1m=1 Thus, the slope of the line is 1.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (5,7). y−7=1(x−5) ⇒y−7=x−5Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−7=x−5LHS+7=RHS+7y=x+2 Thus, the y-variable is isolated and we have our equation in slope-intercept form. f(x)=x+2 | |

Exercises 23 Notice that the given two points, f(-4)=2 and f(6)=-3, are in function notation. To start, let's write these points as coordinate pairs. Remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(-4)=2⇔(-4,2)f(6)=-3⇔(6,-3) Now we are able to write an equation for function f in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (-4,2) and (6,-3). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-4,2) & (6,-3)m=6−(-4)-3−2a−(-b)=a+bm=6+4-3−2Add and subtract termsm=10-5ba=b/5a/5m=-21 Thus, the slope of the line is -21.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-4,2). y−2=-21(x−(-4)) ⇒y−2=-21(x+4)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−2=-21(x+4)Distribute -21y−2=-21x−2LHS+2=RHS+2y=-21x Thus, the y-variable is isolated and we have our equation in slope-intercept form. f(x)=-21x | |

Exercises 24 Notice that the given two points, f(-10)=4 and f(-2)=4, are in function notation. To start, let's write these points as coordinate pairs. Remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(-10)=4⇔(-10,4)f(-2)=4⇔(-2,4) Now we are able to write an equation for function f in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (-10,4) and (-2,4). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-10,4) & (-2,4)m=-2−(-10)4−4a−(-b)=a+bm=-2+104−4Add and subtract termsm=80Calculate quotientm=0 Thus, the slope of the line is 0.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-10,4). y−4=0(x−(-10)) ⇒y−4=0Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−4=0LHS+4=RHS+4y=4 Thus, the y-variable is isolated and we have our equation in slope-intercept form. f(x)=4 | |

Exercises 25 Notice that the given two points, f(-3)=1 and f(13)=5, are in function notation. To start, let's write these points as coordinate pairs. Remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(-3)=1⇔(-3,1)f(13)=5⇔(13,5) Now we are able to write an equation for function f in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow three steps.First, we will find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (-3,1) and (13,5). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (-3,1) & (13,5)m=13−(-3)5−1a−(-b)=a+bm=13+35−1Add and subtract termsm=164ba=b/4a/4m=41 Thus, the slope of the line is 41.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-3,1). y−1=41(x−(-3)) ⇒y−1=41(x+3)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable of the equation. y−1=41(x+3)Distribute 41y−1=41x+43LHS+1=RHS+1y=41x+43+1Write as a fractiony=41x+43+44Add fractionsy=41x+47 Thus, the y-variable is isolated and we have our equation in slope-intercept form. f(x)=41x+47 | |

Exercises 26 Notice that the given two points, f(-9)=10 and f(-1)=-2, are in function notation. To start, let's write these points as coordinate pairs. Recall that the input is the x-coordinate and the output is the y-coordinate.f(x)=y(x,y) f(-9)=10(-9,10) f(-1)=-2(-1,-2) We can't determine the y-intercept of the equation from the given points. Therefore, to write the equation of the line in slope-intercept form, we will follow three steps.First, we will find the slope of the equation by using the Slope Formula. Next, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Finding the Slope We know that the line of function f passes through the points (-9,10) and (-1,-2). Let's substitute these points into the slope formula and find the slope m. m=x2−x1y2−y1Substitute (-9,10) & (-1,-2)m=-1−(-9)-2−10 Simplify right-hand side a−(-b)=a+bm=-1+9-2−10Add and subtract termsm=8-12ba=b/4a/4m=2-3Put minus sign in front of fraction m=-23 The slope of the line is -23.Point-Slope Form Let's recall the point-slope form of a line. y−y1=m(x−x1) Here, m is the slope and (x1,y1) is a point on the line. We already know that the slope is -23. We can choose any of the given points and write the equation of the line. Let's use (-9,10). y−10=-23(x−(-9))⇔y−10=-23(x+9)Slope-Intercept Form Finally, we can write the equation of the line in slope-intercept form by isolating the y-variable in the equation. y−10=-23(x+9) Solve for y Distribute -23y−10=-23x−23(9)ca⋅b=ca⋅by−10=-23x−227LHS+10=RHS+10y=-23x−227+10a=22⋅ay=-23x−227+220Add fractions y=-23x−27 | |

Exercises 27 One way to determine if points are collinear is to calculate and compare the slope of the line between points in the data set. We can do this by using the slope formula. Note that it is enough to check each point with just one other point in the data set.Rate of change between...x2−x1y2−y1m (2,-1)&(4,5)4−25−(-1)3 (4,5)&(6,15)6−415−55 (6,15)&(8,29)8−629−157 (8,29)&(10,47)10−847−299 We see that the slope between each pair of points is different. Since the rate of change is inconsistent, we know that the points are not collinear. Therefore, it is also not possible to write a linear equation that represents y as a function of x. | |

Exercises 28 One way to determine if points are collinear is to calculate and compare the slope of the line between points in the data set. We can do this by using the slope formula. Note that it is enough to check each point with just one other point in the data set.Rate of change between...x2−x1y2−y1m (-3,16)&(-1,10)-1−(-3)10−16-3 (-1,10)&(1,4)1−(-1)4−10-3 (1,4)&(3,-2)3−1-2−4-3 (3,-2)&(5,-8)5−3-8−(-2)-3We see that the slope between each pair of points is m=-3. Since the rate of change is constant, we know that the points are collinear. To write an equation for this line, we will use the point-slope form. y−y1=m(x−x1) In this form, m is the slope of the line and the point (x1,y1) lies on the line. Let's substitute the slope we calculated above and one of the given points, (-3,16), into this equation. y−16=-3(x−(-3)) Note that any point that lies on the same line as the given points would create a valid equation for this line. To write a unique linear equation that represents y as a function of x, we will rewrite this point-slope form equation in slope-intercept form. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept. y−16=-3(x−(-3))a−(-b)=a+by−16=-3(x+3)Distribute -3y−16=-3x−9LHS+16=RHS+16y=-3x+7 | |

Exercises 29 One way to determine if points are collinear is to calculate and compare the slope of the line between points in the data set. We can do this by using the slope formula. Note that it is enough to check each point with just one other point in the data set.Rate of change between...x2−x1y2−y1m (0,1.2) & (1,1.4)1−01.4−1.20.2 (1,1.4) & (2,1.6)2−11.6−1.40.2 (2,1.6) & (4,2)4−22−1.60.2We see that the slope between each pair of points is m=0.2. Since the rate of change is constant, we know that the points are collinear. To write an equation for this line, we will use the point-slope form. y−y1=m(x−x1) In this form, m is the slope of the line and the point (x1,y1) lies on the line. Let's substitute the slope we calculated above and one of the given points, (4,2), into this equation. y−2=0.2(x−4) Note that any point that lies on the same line as the given points would create a valid equation for this line. To write a unique linear equation that represents y as a function of x, we will rewrite this point-slope form equation in slope-intercept form. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept. y−2=0.2(x−4)Distribute 0.2y−2=0.2x−0.8LHS+2=RHS+2y=0.2x+1.2 | |

Exercises 30 One way to determine if points are collinear is to calculate and compare the slope of the line between points in the data set. We can do this by using the slope formula. Note that it is enough to check each point with just one other point in the data set.Rate of change between...x2−x1y2−y1m (1,18)&(2,15)2−115−18-3 (2,15)&(4,12)4−212−15-23 (4,12)&(8,9)8−49−12-43 We see that the slope between each pair of points is different. Since the rate of change is inconsistent, we know that the points are not collinear. Therefore, it is also not possible to write a linear equation that represents y as a function of x. | |

Exercises 31 As we can see, the original solver tries to write the equation in point-slope form. Notice that the given two points, g(5)=4 and g(3)=10, are in function notation. To start, let's write these points as coordinate pairs. Remember that the input x is the x-coordinate and the output g(x) is the y-coordinate. g(x)=y⇔(x,y)g(5)=4⇔(5,4)g(3)=10⇔(3,10) Now we are able to write an equation for function g in slope-intercept form. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will follow two steps.We will first find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form.Finding the Slope We know that the line of function g passes through the points (5,4) and (3,10). Let's substitute these points into the slope formula and find the slope. m=x2−x1y2−y1Substitute (5,4) & (3,10)m=3−510−4Subtract termm=-26Calculate quotientm=-3 Thus, the slope of the line is -3. The original solver has also found the slope as -3. Therefore, there is not any error in this step.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (5,4) as the original solver has. y−4=-3(x−5) In this step, the original solver has applied the point-slope form. Thus, s/he has a wrong equation at the end. | |

Exercises 32 To calculate the slope, we will use the slope formula and check whether it was done correctly. m=x2−x1y2−y1Substitute (4,3) & (1,2)m=4−13−2Subtract termsm=31 The slope was determined correctly. The error must have been made when writing the equation. When substituting one of the points in the point slope formula, we can see that the first point's y-value and the second point's x-value were used. This is wrong as you have to use the x- and y-coordinates from the same point. A correct version of this would be: y−2=y−3=31(x−1) or 31(x−4) | |

Exercises 33 aFirst, we will let each step on the x-axis represent 1000 stickers so: x=1⇒1000 stickers.x=2⇒2000 stickers.x=4⇒3000 stickers.Now we can create our equation. Creating our equation The company charges $225 for the first 1000 stickers, so the function we are looking for passes through the point (1,225). After this point, each additional 1000 stickers cost $80. This means that after x=1, the function rises by 80 which gives it a slope of 80. So far we have: C=80n+b. Since we already determined that the line passes through the point (1,225), we can substitute this point in the equation and calculate b. C=80n+bn=1, C=225225=80⋅1+b Solve for b Multiply225=80+bLHS−80=RHS−80145=bRearrange equation b=145 Therefore, an equation that represents the total cost of the stickers is: C=80n+145.bTo find the total cost of 9000 stickers, we have to substitute n=9 in the function found in part A, as n is the number of of thousands of stickers. C=80n+145n=9C=80⋅9+145 Evaluate RHS MultiplyC=720+145Add terms C=865 The cost of 9000 stickers is $865. | |

Exercises 34 aLet's find the rate of change for consecutive data pairs in the table.Rate of change betweenx2−x1y2−y1m Column 4 & Column 38−6858−654102 Column 3 & Column 26−4654−450102 Column 2 & Column 14−2450−246102 We can see that the rate is constant, so the situation can be modeled by a linear equation.bTo determine the processing fee and the daily fee, let's find the linear equation which models this situation. In the previous part, we have already determined that the rate of change is 102. So far, we have: C=102t+b. To determine b, let's substitute on of the given points in the equation. We will use (2,246). C=102t+bt=2, C=246246=102⋅2+b Solve for b Multiply246=204+bLHS−204=RHS−20442=bRearrange equation b=42 The equation which models this situation is: C=102t+42. Now we know the daily fee is $102. The processing fee is $42 as this is a fixed amount that you pay only once no matter how many days you rent the beach house.cSince we can spend no more than $1200 on the beach house rental, the cost must be less than or equal to 1200. We can represent this with the following inequality: 102t+42≤1200. Let's solve it. 102t+42≤1200 Solve for n LHS−42≤RHS−42102t≤1158LHS/102≤RHS/102t≤1021158Use a calculatort≤11.35294… t≤11.4 Since t is number of days, and we cannot rent the house for a fraction of a day, we can conclude that the maximum number of days we can rent the beach house is 11. | |

Exercises 35 Here we will look at two methods to graph the line created by the equation. First, we need to recall what the different parts of the point-slope form tell us. The general form of the equation is: y−y1=m(x−x1), where m is the slope and (x1,y1) is a point on the line.First method The first method is by plotting the point we can identify from the equation and then using the slope to plot another point on the line. From the equation: y−1=23(x−4), we identify a slope of 23 and the point (4,1). We can plot this point and then use the slope, to find a second point.Drawing a line through the points we have our graph.Second method A second method is to use the equation to find a second point by simplify substituing an arbitrary y- or x-coordinate into the equation and solving for the other variable. Let's substitute x=8 and solve for y. y−1=23(x−4)x=8y−1=23(8−4) Solve for y Subtract termy−1=23⋅4ca⋅b=ca⋅by−1=212Calculate quotienty−1=6LHS+1=RHS+1 y=7 Now we can plot our two points and connect them with a straight line to graph our function. | |

Exercises 36 Although the exercise only calls for us to represent the function in two other ways, we will demonstrate three so that you can decide which methods you like the best!Equation In the exercise, we are told that the function has a slope of 52 and that it passes through the point (12,-5). Using this information, it easy to write our function as an equation in point-slope form: y−y1=m(x−x1)⇒y−(-5)=52(x−12). We are able to simplify this equation a little bit while keeping it in point-slope form by using the fact that subtraction of a negative is the same thing as addition. Therefore, our final equation becomes: y−(-5)=52(x−12)⇒y+5=52(x−12).Graph Another way to represent this function is by graphing it on a coordinate plane. We can plot the given point and use the slope to find a second point. By connecting these points with a straight line, we have our graph of the function.Table Yet another way to represent our function is by using a table of values. Since the slope is 52, each point must be found using: x1+5=x2y1+2=y2 and x2−5=x1 and y2−2=y1. If we center our table at the given point, we can add to and subtract from x and y to find the nearest integer coordinate pairs found within the function.xy(x,y) 2-9(2,-9) 7-7(7,-7) 12-5(12,-5) 17-3(17,-3) 22-1(22,-1) | |

Exercises 37 It is easier to use point-slope form since we are already given two points on the line. We need to calculate the slope either way but there is no need to make another calculation for the y-intercept. Therefore, for a simpler, better, and faster solution we should use point-slope form. | |

Exercises 38 aWhen you draw an imaginary line through these points it will cut the x-axis between 0 and 2 and again cut the y-axis at a point below the x-axis. Therefore, it appears to have a negative y-intercept.bThe first point is at x=4 and its vertical distance from x-axis is quite close to 2. We can approximate it to be 1.8, giving us the coordinates (4,1.8). Following a similar reasoning, the second point is on (8,4). Let's find the slope of the line that passes through them. m=x2−x1y2−y1Substitute (8,4) & (4,1.8)m=8−44−1.8 Simplify RHS Subtract termsm=42.2ba=b⋅10a⋅10m=4022ba=b/2a/2 m=2011 A slope of 2011 means if we take 20 steps to the right, the graph increases by 11 steps. Conversely, if we take 20 steps to the left, we would have to go down by 11 steps to get to another point. | |

Exercises 39 Let's begin by sketching their graphs. The first graph has a slope of 2 and a y-intercept of 0. To draw the second graph, we will write it in slope-intercept form. y−1=2(x+3)Distribute 2y−1=2x+6LHS+1=RHS+1y=2x+7 The second graph also has a slope of 2 but a y-intercept of 7. Let's graph this stuff!If we recall the chapter on transformations, we know that: Horizontal translation: Vertical translation: f(x−h)f(x)+k By rewriting the general point-slope form, we can relate it to these types of transformations: y−k=m(x−h)⇔y=m(x−h)+k We see that point-slope form represents a horizontal translation by h and a vertical translation by k of the function y=mx. Let's substitute k=1 and h=-3 into this general equation and simplify y=m(x−h)+kh=-3, k=1y=m(x−(-3))+1a−(-b)=a+by=m(x+3)+1 This means y−1=2(x+3) is a translation of 3 units to the left and 1 unit up of the function y=2x. Let's show this below. | |

Exercises 40 aThe amount of money each sibling is given is equal to the y-intercept of the function that models their spending behavior. This is because x=0 is the time when they receive their money. Nothing has been spent yet.Sibling A From the graph, we see that the y-intercept occurs at (0,80) which means this sibling received $80.Sibling B From the function, we see that the y-intercept occurs at (0,90) because this is the function's constant. This sibling received $90.Sibling C From the table, we can't immediately tell how much money was received as we don't have an ordered pair where x=0. However, because the siblings spend at a constant rate, the rate of change between x=0 and x=1 will be equal to the change between all other consecutive points in the table. Since two of the points are (1,100) and (2,75), we can calculate the rate of change. Slope=x2−x1y2−y1Substitute (2,75) & (1,100)Slope=2−175−100 Simplify RHS Subtract termsSlope=1-251a=a Slope=-25 The slope is -25. This means if we run time backwards and go from x=1 to x=0, the corresponding y-value is 125. Now we can summarize and determine who received the most money: Sibling A: Sibling B: Sibling C: $80 (least)$90 $125 (most)bComparing the spending rates means comparing the slopes of the spending functions. We have already determined the slope for sibling C as m=-25. For sibling B, we have a function where we instantly can recognize the slope by looking at the coefficient of x: y=-22.5x+90 The slope for sibling B is -22.5. To find the slope for sibling A, we can use the points from the graph and substitute them into the slope formula. Slope=x1−x2y1−y2Substitute (4,20) & (2,50)Slope=4−220−50Subtract termsSlope=2-30Calculate quotientSlope=-15 Sibling A spends their money at a rate of $15 per week. Let's summarize: Sibling A: Sibling B: Sibling C: $15 a week (least)$25 a week (most)$22.5 a weekcWe can determine when all of the money has been spent by finding x when: y=0 We know that sibling B has $25 left at the end of 4 weeks. Since they spend $25 a week, by the end of week 5, they will have no money left. Sibling C spends according to the function y=-22.5x+90. By substituting y=0 we can solve for x. y=-22.5x+90y=00=-22.5x+90 Solve for x LHS+22.5x=RHS+22.5x22.5x=90LHS/22.5=RHS/22.5 x=4 Sibling C runs out of money at week 4. Sibling A spends $15 a week and starts out with $80. This means we can model their spending behavior with the function y=-15x+80 Let's substitute y=0 and solve for x. y=-15x+80y=00=-15x+80 Solve for x LHS+15x=RHS+15x15x=80LHS/15=RHS/15x=5.33333… x≈5.3 Sibling A runs out of money after 5.33 weeks. Sibling A: Sibling B: Sibling C: $5.3 weeks (last)$5 weeks4 weeks (first) | |

Exercises 41 The product of any nonzero real number and its reciprocal is 1. x⋅x1=1 Therefore, the reciprocal of 5 is 51. Let's check that this is true by calculating the product. 5⋅51=?15⋅5a=a1=1 We were able to confirm that the reciprocal of 5 is 51. | |

Exercises 42 The product of any nonzero real number and its reciprocal is 1. x⋅x1=1 Therefore, the reciprocal of -8 is -81. Let's check that this is true by calculating the product. -8⋅(-81)=?1-a(-b)=a⋅b8⋅(81)=?18⋅8a=a1=1 We were able to confirm that the reciprocal of -8 is -81. | |

Exercises 43 The product of any nonzero real number and its reciprocal is 1. x⋅x1=1 For a fraction, the reciprocal is the inverse of the fraction. yx⋅xy=1 Therefore, the reciprocal of -72 is -27. Let's check that this is true by calculating the product. -72⋅(-27)=?1-a(-b)=a⋅b72⋅27=?1Multiply fractions7⋅22⋅7=?1Multiply1414=?1aa=11=1 We were able to confirm that the reciprocal of -72 is -27. | |

Exercises 44 The product of any nonzero real number and its reciprocal is 1. x⋅x1=1 For a fraction, the reciprocal is the inverse of the fraction. yx⋅xy=1 Therefore, the reciprocal of 23 is 32. Let's check that this is true by calculating the product. 23⋅32=?1Multiply fractions2⋅33⋅2=?1Multiply66=?1aa=11=1 We were able to confirm that the reciprocal of 23 is 32. |

##### Mathleaks Courses

Need more help with your math studies? Visit Writing Equations in Point-Slope Form (Algebra 1) to access Mathleaks own courses. It includes theory, exercises and tests, try it for free here: mathleaks.com/study

##### Other subchapters in Writing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Writing Equations in Slope-Intercept Form
- Writing Equations of Parallel and Perpendicular Lines
- Quiz
- Scatter Plots and Lines of Fit
- Analyzing Lines of Fit
- Arithmetic Sequences
- Piecewise Functions
- Chapter Review
- Chapter Test
- Cumulative Assessment