Writing Equations of Parallel and Perpendicular Lines

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Exercises 1 When two nonvertical lines have the same slope, this means that they will run side by side forever and ever and never intersect. An example is shown below.The lines shown above are considered parallel, two lines that have the same slope and will never intersect one another.
Exercises 2 Two nonvertical lines are perpendicular if their slopes are negative reciprocals. Numbers are negative reciprocals if their product is -1. We need to find the slope that, when multiplied by -75​, will give us a product of -1. -75​m=-1Put minus sign in numerator7-5​m=-1LHS⋅7=RHS⋅7-5m=-7LHS/(-5)=RHS/(-5)m=-5-7​-b-a​=ba​m=57​ The slope of the second line must be 57​ if the two lines are perpendicular.
Exercises 3 Two lines are parallel if their slopes are identical. For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line we choose, since the result will be the same. Let's start with line a, which passes through (-3,1) and (0,3). m=x2​−x1​y2​−y1​​Substitute (-3,1) & (0,3)m=0−(-3)3−1​a−(-b)=a+bm=0+33−1​Add and subtract termsm=32​ The slope of line a is 32​. We will use the same method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-3,1)(0,3)0−(-3)3−1​32​ b(0,0)(3,2)3−02−0​32​ c(-2,-3)(3,0)3−(-2)0−(-3)​53​ Now that we have identified the slope of each line, we can see that a and b have the same slope, so they are parallel.
Exercises 4 Two lines are parallel if their slopes are identical. For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (0,5) and (2,0). m=x2​−x1​y2​−y1​​Substitute (0,5) & (2,0)m=2−(0)0−5​Subtract termsm=-25​Calculate quotientm=-2.5 We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(0,5)(2,0)2−00−5​-2.5 b(3,4)(5,0)5−30−4​-2 c(4,6)(5,4)5−44−6​-2 Now that we've identified the slope of each line, we can see that b and c have the same slope, so we know that they are parallel.
Exercises 5 Two lines are parallel if their slopes are identical. For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (-1,-2) and (1,0). m=x2​−x1​y2​−y1​​Substitute (-1,-2) & (1,0)m=1−(-1)0−(-2)​ Simplify a−(-b)=a+bm=1+10+2​Add termsm=22​aa​=1 m=1 We will use the same method to calculate the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-1,-2)(1,0)1−(-1)0−(-2)​1 b(4,2)(2,-2)2−4-2−2​2 c(0,2)(-1,1)-1−01−2​1 Now that we've identified the slope of each line, we can see that a and c have the same slope, so we know that they are parallel.
Exercises 6 Two lines are parallel if their slopes are identical. For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (-1,3) and (1,9). m=x2​−x1​y2​−y1​​Substitute (-1,3) & (1,9)m=1−(-1)9−3​ Simplify a−(-b)=a+bm=1+19−3​Add and subtract termsm=26​Calculate quotient m=3 We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-1,3)(1,9)1−(-1)9−3​3 b(-2,12)(-1,14)-1−(-2)14−12​2 c(3,8)(6,10)6−310−8​32​ Now that we've identified the slope of each line, we can see that none of the lines have the same slope, so none of them are parallel.
Exercises 7 Two lines are parallel if their slopes are identical. Let's recall the slope-intercept form of a line. y=mx+b​ Here, we can identify its slope as the value of m. We will rewrite each equation in this form and highlight their slopes.LineGiven EquationSlope-intercept formSlope a4y+x=8y=-41​x+2-41​ b2y+x=4y=-21​x+2-21​ c2y=-3x+6y=-23​x+3-23​ Now that we've identified the slope of each line, we can see that none of the lines have the same slope, so none of them are parallel.
Exercises 8 Lines are parallel if their slopes are identical. For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. Let's write each equation in slope-intercept form, highlighting their slopes.LineGiven EquationSlope-intercept formSlope a3y−x=6y=31​x+231​ b3y=x+18y=31​x+631​ c3y−2x=9y=32​x+332​ Now that we've identified the slope of each line, we can see that a and b have the same slope, so we know that they are parallel.
Exercises 9 When lines are parallel, they have the same slope. Consider the given line, which is written in slope-intercept form. Slope-intercept form:Given line:​ y=mx+b y=2x+2​ In the given equation we can see that the slope of the line is 2. Therefore, we know that all the lines that are parallel to the given one have a slope of 2. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=2x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the point (-1,3). By substituting this point into the above equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=2x+bx=-1, y=33=2(-1)+b Solve for b a(-b)=-a⋅b3=-2+bLHS+2=RHS+25=bRearrange equation b=5 Now that we know that b=5, we can write the equation of the line that is parallel to y=2x+2 and passes through the point (-1,3). y=2x+5​
Exercises 10 When lines are parallel, they have the same slope. y=-5x+4​ In the given equation, we can see that the line has a slope of -5. Therefore we know that all lines that are parallel to our given line will have a slope of -5. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=-5x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (1,2). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=-5x+bx=1, y=22=-5(1)+b Solve for b (-a)b=-ab2=-5+bLHS+5=RHS+57=bRearrange equation b=7 Now that we have the y-intercept, we can write the equation for the line that is parallel to y=-5x+4 and passes through the point (1,2). y=-5x+7​
Exercises 11 When lines are parallel, they have the same slope. To help us identify the slope of this line, let's first convert it into slope-intercept form, y=mx+b, where m is the slope and (0,b) is the y-intercept. 3y−x=-12 Solve for y LHS+x=RHS+x3y=x−12LHS/3=RHS/3y=3x−12​Split into factorsy=3x​−312​Calculate quotient y=31​x−4 With this, we can more easily identify the slope m and y-intercept b. y=31​x−4​ Now that we've identified the slope, we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=31​x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (18,2). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=31​x+bx=18, y=22=31​(18)+b Solve for b Multiply2=6+bLHS−6=RHS−6-4=bRearrange equation b=-4 As we can see we got exactly the same y-intercept as in the given equation. So we can write the equation for the line that is parallel to 3y−x=-12 and passes through (18,2). y=31​x−4​
Exercises 12 When lines are parallel, they have the same slope. To help us identify the slope of this line, let's first convert it into slope-intercept form, y=mx+b, where m is the slope and (0,b) is the y-intercept. 2y=3x+10 Solve for y LHS/2=RHS/2y=23x+10​Split into factorsy=23x​+210​Calculate quotient y=23​x+5 With this, we can more easily identify the slope m and y-intercept b. y=23​x+5​ This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=23​x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (2,-5). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=23​x+bx=2, y=-5-5=23​(2)+b Solve for b 2a​⋅2=a-5=3+bLHS−-3=RHS−-3-8=bRearrange equation b=-8 Now that we have the y-intercept, we can write the equation of the line that is parallel to y=23​x+5 and passes through (2,-5). y=23​x+(-8) ⇒y=23​x−8​
Exercises 13 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Parallel Lines For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (-3,-1) and (-5,-4). m=x2​−x1​y2​−y1​​Substitute (-3,-1) & (-5,-4)m=-5−(-3)-4−(-1)​ Simplify a−(-b)=a+bm=-5+3-4+1​Add termsm=-2-3​-b-a​=ba​ m=23​ We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-3,-1)(-5,-4)-5−(-3)-4−(-1)​23​ b(-6,-4)(-2,-6)-2−(-6)-6−(-4)​-21​ c(-3,-6)(0,-1)0−(-3)-1−(-6)​35​ The lines have all different slopes. Therefore, none of them are parallel.Perpendicular Lines Lines with different slopes are not parallel. To determine whether they are perpendicular, we calculate the product of their slopes. If the product equals -1, then the lines are perpendicular. Let's start by calculating the product of the slopes of lines a and b. m1​⋅m2​m1​=23​, m2​=-21​23​(-21​)a(-b)=-a⋅b-23​(21​)Multiply fractions-43​ The product of the slopes of lines a and b is not -1. Therefore, lines a and b are not perpendicular. We will use a similar method to check whether lines a and c, or b and c are perpendicular.LinesSlope 1Slope 2Product a & b23​-21​-43​ × a & c23​35​25​ × b & c-21​35​-65​ × Neither product equals -1. This means that none of the lines are perpendicular.
Exercises 14 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Parallel Lines For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (-1,1) and (2,0). m=x2​−x1​y2​−y1​​Substitute (-1,1) & (2,0)m=2−(-1)0−1​ Simplify a−(-b)=a+bm=2+10−1​Add and subtract termsm=3-1​Put minus sign in front of fraction m=-31​ We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-1,1)(2,0)2−(-1)0−1​-31​ b(0,5)(3,4)3−04−5​-31​ c(0,0)(2,5)2−05−0​25​ Lines a and b have the same slope. Therefore, they are parallel.Perpendicular Lines Lines with different slopes are not parallel. To determine whether they are perpendicular, we calculate the product of their slopes. If the product equals -1, then the lines are perpendicular. Let's start by calculating the product of the slopes of lines a and c. m1​⋅m2​m1​=-31​, m2​=25​-31​⋅(25​)Multiply fractions-65​ The product of the slopes of lines a and c is not -1. Therefore, lines a and c are not perpendicular. We will use a similar method to check whether lines a and b, or b and c are perpendicular.LinesSlope 1Slope 2Product a & b-31​-31​91​ × a & c-31​25​-65​ × b & c-31​25​-65​ × Neither product equals -1. This means that none of the lines are perpendicular.
Exercises 15 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Parallel Lines For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (-2,1) and (0,3). m=x2​−x1​y2​−y1​​Substitute (-2,1) & (0,3)m=0−(-2)3−1​ Simplify a−(-b)=a+bm=0+23−1​Add and subtract termsm=22​aa​=1 m=1 We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(-2,1)(0,3)0−(-2)3−1​1 b(4,1)(6,4)6−44−1​23​ c(1,3)(4,1)4−11−3​-32​ The lines have all different slopes. Therefore, none of them are parallel.Perpendicular Lines Lines with different slopes are not parallel. To determine whether they are perpendicular, we calculate the product of their slopes. If the product equals -1, then the lines are perpendicular. Let's start by calculating the product of the slopes of lines a and b. m1​⋅m2​m1​=1, m2​=23​1⋅23​Identity Property of Multiplication23​ The product of the slopes of lines a and b is not -1. Therefore, lines a and b are not perpendicular. We will use a similar method to check whether lines a and c, or b and c are perpendicular.LinesSlope 1Slope 2Product a & b123​23​ × a & c1-32​-32​ × b & c23​-32​-1 ✓ We have found that lines b and c are perpendicular to each another.
Exercises 16 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they parallel? For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (2,10) and (4,13). m=x2​−x1​y2​−y1​​Substitute (2,10) & (4,13)m=4−213−10​Subtract termsm=23​ We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(2,10)(4,13)4−213−10​23​ b(4,9)(6,12)6−412−9​23​ c(2,10)(4,9)4−29−10​-21​ Now that we've identified the slope of each line, we can see that a and b have the same slope, so they are parallel.Are they perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and therefore perpendicular. Let's start with checking lines a and c. m1​⋅m2​=?-1m1​=23​, m2​=-21​23​⋅(-21​)=?-1Multiply fractions-43​​=-1 Therefore, lines a and c are neither parallel nor perpendicular. We will use a similar method to check if lines a and b or b and c are perpendicular.LinesSlope 1Slope 2Product a & c23​-21​-43​ a & b23​23​49​ b & c23​-21​-43​ None of the products is equal to -1, so none of the lines are perpendicular. Notice that when two lines are parallel they cannot be perpendicular at the same time.
Exercises 17 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they parallel? To start with let's write each equation in slope-intercept form, highlighting their slopes.LineGiven EquationSlope-intercept formSlope a4x−3y=2y=34​x−32​34​ by=34​x+2y=34​x+234​ c4y+3x=4y=-43​x+1-43​ Now that we've identified the slope of each line, we can see that a and b have the same slope, so they are parallel.Are they perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and are therefore perpendicular. Let's start with checking lines a and c. m1​⋅m2​=?-1m1​=34​, m2​=-43​34​⋅(-43​)=?-1Multiply fractions-1212​=?-1aa​=1-1=-1 Therefore, lines a and c are perpendicular. We will use a similar method to check if lines a and b or b and c are perpendicular.LinesSlope 1Slope 2Product a & b34​34​916​ a & c34​-43​-1 b & c34​-43​-1 We have found that line c is perpendicular to lines a and b. Notice that when two lines are parallel they cannot be perpendicular at the same time.
Exercises 18 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they parallel? To start, let's write each equation in slope-intercept form, highlighting their slopes.LineGiven EquationSlope-intercept formSlope ay=6x−2y=6x−26 b6y=-xy=-61​x-61​ cy+6x=1y=-6x+1-6 Now that we've identified the slope of each line, we can see that none of the lines have the same slope, so they are not parallel.Are they perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and are therefore perpendicular. Let's start with checking lines a and b. m1​⋅m2​=?-1m1​=6, m2​=-61​6⋅(-61​)=?-1Multiply-1=-1 Therefore, lines a and b are perpendicular. We will use a similar method to check if lines a and c or b and c are perpendicular.LinesSlope 1Slope 2Product a & b6-61​-1 a & c6-6-36 b & c-61​-61 We have found that lines a & b are perpendicular to one another.
Exercises 19 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. After that, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Slope of the Perpendicular Line Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of the slopes of perpendicular lines is -1. m1​⋅m2​=-1​ Note that the given equation is written in slope-intercept form. y=21​x−9​ In the above formula we can see that the slope is 21​. By substituting this value for m1​ into our negative reciprocal equation, we can solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=21​21​⋅m2​=-1LHS⋅2=RHS⋅2m2​=-2 Any perpendicular line to the given equation will have a slope of -2.Writing the Perpendicular Line's Equation Let's write a partial equation in slope-intercept form for a perpendicular line to the given equation. y=-2x+b​ By substituting the given point (7,10) into this equation for x and y, respectively, we can solve for the y-intercept b of the perpendicular line. y=-2x+bx=7, y=1010=-2(7)+b Solve for b (-a)b=-ab10=-14+bLHS+14=RHS+1424=bRearrange equation b=24 Now that we have the y-intercept, we can write the equation of the perpendicular line to y=21​x−9 through the point (7,10). y=-2x+24​
Exercises 20 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. After that, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Perpendicular Line's Slope Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m1​⋅m2​=-1​ For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. y=34​x+6​ In the given equation we can see that its slope is 34​. By substituting this value into our negative reciprocal equation for m1​, we can solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=34​34​⋅m2​=-1 Solve for m2​ LHS⋅3=RHS⋅34m2​=-3LHS/4=RHS/4m2​=4-3​Put minus sign in front of fraction m2​=-43​ With this, we can identify that any line perpendicular to the given equation will have a slope of -43​.Writing the Perpendicular Line's Equation With the slope m2​=-43​, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=-43​x+b​ By substituting the given point (-4,-1) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line. y=-43​x+bx=-4, y=-1-1=-43​(-4)+b Solve for b Put minus sign in denominator-1=-43​(-4)+b-4a​⋅-4=a-1=3+bLHS−3=RHS−3-4=bRearrange equation b=-4 Now that we have the y-intercept, we can write the equation for the line that is both perpendicular to y=34​x+6 and passes through the point (-4,-1). y=-43​x−4​
Exercises 21 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. After that, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Perpendicular Line's Slope Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m1​⋅m2​=-1​ For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. Since the given equation is not written in slope-intercept form, we'll have to rewrite it before identifying the slope. 2y=8x−6LHS/2=RHS/2y=4x−3 Now we can identify the slope and the y-intercept. y=4x−3​ We can see that the slope is 4. Let's substitute this value into our negative reciprocal equation for m1​, and solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=44⋅m2​=-1 Solve for m2​ LHS/4=RHS/4m2​=4-1​Put minus sign in front of fraction m2​=-41​ With this, we can identify that any line perpendicular to the given equation will have a slope of -41​.Writing the Perpendicular Line's Equation With the slope m2​=-41​, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=-41​x+b​ By substituting the given point (-3,3) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line. y=-41​x+bx=-3, y=33=-41​(-3)+b Solve for b -a(-b)=a⋅b3=43​+bLHS−43​=RHS−43​3−43​=bWrite as a fraction412​−43​=bSubtract fractions49​=bRearrange equation b=49​ Now that we have the y-intercept, we can write the equation of the line that is both perpendicular to 2y=8x−6 and passes through the point (-3,3). y=-41​x+49​​
Exercises 22 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. After that, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Perpendicular Line's Slope Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m1​⋅m2​=-1​ For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. Since the given equation is not written in slope-intercept form, we'll have to rewrite it before identifying the slope. 2y+4x=12LHS−4x=RHS−4x2y=-4x+12LHS/2=RHS/2y=-2x+6 Now we are able to identify the slope and the y-intercept. y=-2x+6​ We can see that the slope is -2. By substituting this value into our negative reciprocal equation for m1​, we can solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=-2-2⋅m2​=-1 Solve for m2​ LHS/-2=RHS/-2m2​=-2-1​-b-a​=ba​ m2​=21​ With this, we can identify that any line perpendicular to the given equation will have a slope of 21​.Writing the Perpendicular Line's Equation With the slope m2​=21​, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=21​x+b​ By substituting the given point (8,1) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line. y=21​x+bx=8, y=11=21​(8)+b Solve for b Multiply1=4+bLHS−4=RHS−4-3=bRearrange equation b=-3 Now that we have the y-intercept, we can write the equation of the line that is both perpendicular to 2y+4x=12 and passes through the point (8,1). y=21​x−3​
Exercises 23
Exercises 24
Exercises 25 Parallel lines have the same slope. Therefore, the slope of a parallel line to y=41​x+2 is 41​, and not -4. This is the mistake they have made. The line whose equation we want to write passes through the point (1,3). We can substitute this information in the point-slope form to write its equation. y−y1​=m(x−x1​)↓y−3=41​(x−1)​ Finally, let's isolate the y-variable to obtain the slope-intercept form of the line. y−3=41​(x−1) Solve for y Distribute 41​y−3=41​x−41​LHS+3=RHS+3y=41​x−41​+3a=44⋅a​y=41​x−41​+412​Add fractions y=41​x+411​
Exercises 26 Two lines are perpendicular if and only if their slopes are negative reciprocals. Consider the given line. y=31​x+5​ Here, the slope is 31​. This means that any perpendicular line will have a slope of -3. In the solution provided, they wrote a slope of 3. This is the error. Let's correct it. y−y1​=m(x−x1​)Substitute valuesy−(-5)=-3(x−4)a−(-b)=a+by+5=-3(x−4)Distribute -3y+5=-3x+12LHS−5=RHS−5y=-3x+7
Exercises 27 To write the equation of the line which represents the new pipe, we need to know its slope. Since it is perpendicular to the old one, the slopes are negative reciprocals, which means their product is -1. We will calculate the slope of the existing pipe by substituting two of its points, (-3,-1) and (0,3), into the Slope Formula. m1​=x2​−x1​y2​−y1​​Substitute (-3,-1) & (0,3)m1​=0−(-3)3−(-1)​a−(-b)=a+bm1​=0+33+1​Add termsm1​=34​ The slope of the line which represents the old pipe is 34​. We will use it to find m2​, the slope of the line which represents the new pipe. 34​⋅m2​=-1⇔m2​=-43​​ The slope of the perpendicular line is -43​. Since this line passes through (2,0), we can substitute these values into the point-slope form and find the equation of this line. y−y1​=m(x−x1​)Substitute valuesy−0=-43​(x−2) Solve for y Subtract termy=-43​(x−2)Distribute -43​y=-43​x+43​⋅2ca​⋅b=ca⋅b​y=-43​x+46​ba​=b/2a/2​ y=-43​x+23​
Exercises 28 We are told the bike path will be parallel to the railroad tracks. This means they will have the same slope. Let's calculate the slope of the railroad tracks by substituting the points we see in the graph, (8,0) and (11,4), into the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (8,0) & (11,4)m=11−84−0​Subtract termsm=34​ The slope of the railroad tracks and, consequently, of the bike path is 34​. By looking at the graph shown, we see that the bike path will pass through the point (4,5). We can substitute this information into the point-slope form of a line to create our equation. x−x1​=m(y−y1​)↓x−4=34​(y−5)​ The above is a possible equation for the bike path. If we want to obtain the slope-intercept form, we need to isolate y. y−5=34​(x−4) Solve for y Distribute 34​y−5=34​x−34​(4)ca​⋅b=ca⋅b​y−5=34​x−316​LHS+5=RHS+5y=34​x−316​+5a=33⋅a​y=34​x−316​+315​Add fractions y=34​x−31​
Exercises 29
Exercises 30 In order to determine the value of a, we need to know the slopes of the equations. Thus, we will first isolate the y-variables in both equations so that they will be in slope-intercept form and we will be able to determine their slopes. Given Form​6y=-2x+4 2y=ax−5 ​Slope-Intercept Form​⇒ y=-31​x+32​⇒ y=2a​x−25​​ As a result, the slope of the first line is -31​ and the slope of the second line is 2a​. Now that we know the slopes of the lines, we can start with finding the value of a such that the lines are parallel.Parallel Lines We learned that nonvertical lines are parallel if and only if they have the same slope. ℓ1​∥ℓ2​⟺m1​=m2​​ According to the statement above, let's find the value of a. m1​=m2​m1​=-31​, m2​=2a​-31​=2a​LHS⋅3=RHS⋅3-1=23a​LHS⋅2=RHS⋅2-2=3aLHS/3=RHS/3-32​=aRearrange equationa=-32​ As a result, the lines will be parallel when the value of a is -32​.Perpendicular Lines We also know that nonvertical lines are perpendicular if and only if their slopes are negative reciprocals. ℓ1​⊥ℓ2​⟺-m1​1​=m2​​ Using the statement above, we can find the value of a. -m1​1​=m2​m1​=-31​, m2​=2a​-1/-31​=2a​1/ba​=ab​-1(-3)=2a​-a(-b)=a⋅b3=2a​LHS⋅2=RHS⋅26=aRearrange equationa=6 Thus, the lines will be perpendicular if the value of a is 6.
Exercises 31 If our friend is correct and the path of the puck forms a right angle, then the two lines that form the path are perpendicular. Slopes of perpendicular lines are negative reciprocals, meaning their product is -1. To check this, we first have to find the slopes.First Line From the graph we can tell that the line passes through the points (-4,0) and (0,8). Let's substitute these points into the Slope Formula. m1​=x2​−x1​y2​−y1​​Substitute (-4,0) & (0,8)m1​=0−(-4)8−0​ Simplify right-hand side a−(-b)=a+bm1​=0+48−0​Add and subtract termsm1​=48​Calculate quotient m1​=2 The slope of the first line is 2.Second Line The second line passes through the points (0,8) and (-4,16). Again, we will substitute these two points into the Slope Formula to find the slope of the second line that makes the path. m2​=x2​−x1​y2​−y1​​Substitute (0,8) & (-4,16)m2​=-4−016−8​ Simplify right-hand side Subtract termsm2​=-48​Put minus sign in front of fractionm2​=-48​Calculate quotient m2​=-2 The slope of the second line is -2.Product Let's multiply our slopes and check if the product equals -1. m1​⋅m2​=?-1m1​=2, m2​=-22(-2)=?-1a(-b)=-a⋅b-2⋅2=?-1Multiply-4​=-1 × Since the product of the slopes is not -1, the lines are not perpendicular. Consequently, our friend was wrong.
Exercises 32
Exercises 33 If lines are perpendicular, the product of their slopes is always -1. The slopes of perpendicular lines are negative reciprocals. Since the question states that both lines have positive slopes, the product of them can never equal anything other than a positive number. Therefore, the statement is never true.
Exercises 34 A vertical line can be modeled as x=a where a is any arbitrary number. If we draw a couple of vertical lines we can easily see that each of them is parallel to the y-axis.However, what about when the vertical line has the equation x=0?When the vertical line is the same line as the y-axis they are no longer distinct, they are the same line. They will intersect an infinite number of times. Therefore, the answer is sometimes.
Exercises 35 Let's draw two arbitrary lines that have the same y-intercept.These lines both intercept the y-axis at 2, but they are not perpendicular. They don't form a right angle. Notice, we could have drawn the lines in such a way that they formed a right angle. Therefore, the statement is sometimes true.
Exercises 36 Imagine that you are lucky enough to have this beautiful masterpiece of art as your school math club logo.Since vertical lines are undefined, let's ignore those when considering parallel and perpendicular lines. Let's consider the other parallel and perpendicular lines present in the graph.Parallel lines Parallel lines are lines that have the same slope and different y-intercepts. All of the horizontal lines in our logo have a slope of 0 and, therefore, are parallel. The equations for these lines are: ​y=14y=12y=10y=4y=2y=0.​Perpendicular lines Perpendicular lines have negative reciprocal slopes. This means that: m1​×m2​=-1. The perpendicular lines in our logo are the diagonal lines running from corner to corner. The equations for these lines are: ​y=xy=-x+14.​ We can check that they are perpendicular by multiplying their slopes: 1⋅(-1)=-1.
Exercises 37 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. (3,6),(4,8),(5,10),(6,9),(7,14)​ Examining the ordered pairs, we can see that there is no x-value with multiple y-values. Therefore, the relation is a function.
Exercises 38 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. (-1,6),(1,4),(-1,2),(1,6),(-1,5)​ Examining the ordered pairs, we can see that there is at least one x-value with multiple y-values. Therefore, the relation is not a function.
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