X – 2– 1 012 y = 2x x – 2– 1 012 y – 4– 2 024 Olympic College - Topic 7 Graphing a Linear Equation Topic 7 Graphing a Linear Equation 1. The Linear Equation

x – 2 – 1 0 1 2

y = 2x

x – 2 – 1 0 1 2

y – 4 – 2 0 2 4

Olympic College - Topic 7 Graphing a Linear Equation

Topic 7 Graphing a Linear Equation1. The Linear Equation

Definition: A linear equation is one of the form y = mx + b, for example y = 2x – 5, where xrepresents the x coordinate and y the y coordinate. The set of all points whichsatisfy a linear equation make a straight line.

One of the skills that are useful is to be able to draw any straight line by using its equation.There are three main techniques used in sketching a linear equation.

A table of values. Use the x and y-intercepts. Using the slope and y-intercept.

A.Using a Table of values to sketch a Linear Equation

Example 1: Sketch the graph of the linear equation y = 2x using the given table of values.

Solution: We are given a table of values with x-coordinates varying from – 2, to 2For each value of x we find the corresponding value of y.

So for x = – 2 y = 2x = 2(– 2) = – 4This information means that the point

(– 2, – 4) lies on the line.This process is continued for each valueof x which results in the table below.

-5 -4 -3 -2 -1 0 1 2 3 4 5x

y

5

4

3

2

1

-1

-2

-3

-4

-5

We only really need two points in order to draw a line but it is important that weuse at least three as this will enable us to detect any errors in our computations.

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y = 2x

x – 1 0 1 2

y=3x – 2

x – 1 0 1 2

y=3x – 2 – 5 – 2 1 4


Example 2: Sketch the graph of the linear equation y = 3x – 2 using the given table of values.

Solution: We are given a table of values with x-coordinates varying from – 1, to 2For each value of x we find the corresponding value of y.

So for x = – 1 y = 3x – 2 = 3(– 1) – 2 = – 5This information means that the point

(– 1, – 5) lies on the line.

This process is continued for each valueof x which results in the table below.

using a table of values.

Solution: We are not given the values of x to use in this situation. We can choose to use anyx-values we like but it makes sense to choose values that give whole numbers for

the corresponding y – values. So we use x = – 4, – 2, 0 and 2.

So for x = – 4 y = =

This information means that the point(– 4, – 1) lies on the line.

This process is continued for each valueof x which results in the table below.

-5 -4 -3 -2 -1 0-1

1 2 3 4 5x

y5

4

3

2

1

-2

-3

-4

-5

y = 3x – 2

Example 3: Sketch the graph of the linear equation y =

x

y=

-1

-2

-3

4

3

2

1

-5 -4 -3 -2 -1 0 1 2 3

y

5

-4

-5

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y = 3x – 2

x4 5

x

y=

– 4

– 1

– 2

0

0

1

2

2

x – 3 0 3 6 9

y 4 2 0 – 2 – 4

x

y


Example 3: Sketch the graph of the linear equation 2x + 3y = 6 using a table of values.

Solution version 1:

We are not given the values of x to use in this situation so we can choose to use any x-values welike but it makes sense to choose values that give whole numbers for the corresponding

y – values.

In these circumstances we use x = – 3, 0, 3, 6 and 9 as these are all multiples of 3 and they willwork well with the 3y and the 6 that are also multiples of 3.

So for x = 6 2x + 3y = 6Replace the x with 6

Subtract 12 from both sides

Divide both sides by 3

2(6) + 3y12 + 3y

3y

y

===

=

=

66– 6

– 2

-5

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-1

-2

-5 -4 -3 -2 -1 0 1 2 3 4x

This information means that the point (6, – 2) lies on the line.This process is continued for each value of x which results in the table below.

These points are now plotted on the Cartesian grid below to give the graph of the linear equation.

y5

4

3

2

1

-3

-4

5 6 7 8 9

2x+3y = 6

x – 3 0 3 6 9

y 4 2 0 – 2 – 4

x

y

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Example 3: Sketch the graph of the linear equation 2x + 3y = 6 using a table of values.

Solution version 2:

In this question the equation of the line is given in its standard form 2x + 3y = 6 we cantransform this into its equivalent slope intercept form by using algebra.

2x + 3y = 6– 2x + 63y =

=

Subtract 2x from both sides

divide both sides by 3

=y

We can now sketch y = in the usual way. We are not given the values of x to use in this

situation so we can choose to use any x-values we like but it makes sense to choose values thatgive whole numbers for the corresponding y – values. In this situation we use x = – 3, 0, 3, 6 and

9 as these are all multiples of 3 and they will work well with the .

So for x = 6 y= =

-1

-2

-5 -4 -3 -2 -1 0 1 2 3 4x

This information means that the point (6, – 2) lies on the line.This process is continued for each value of x which results in the table below.

These points are now plotted on the Cartesian grid below to give the graph of the linear equation.

y5

4

3

2

1

-3

-4

-5

5 6 7 8 9

2x+3y = 6

x – 6 – 3 0 3 6

y

x

y

x – 2 – 1 0 1 2

y

x

y

x – 10

y

x

y

x – 2 – 1 0 1 2

y

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Exercise 1A:

1.

2.

3.

4.

5.

Draw the line has with the equation y =





using the table of values below.





6.

7.

Draw the line has with the equation 2x – 3y = 12 using the table of values below.

Draw the line has with the equation 4x + 2y = 8 using the table of values below.

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In a similar fashion the y-intercept of a line is thepoint where it cuts the y-axis and can be found by

Solution: We find the y-intercept by putting x = 0 and finding the corresponding y-value.

y = 4x + 12 = 4(0) + 12 = 12

The y-intercept is the point (0,12)

We find the x-intercept by putting y = 0 and finding the corresponding x-value.

y0

==

4x + 124x + 12

subtract 12 from both sides

divide both sides 4

– 12

– 3

=

=

=

4x

x


B. Using x and y-intercepts to sketch a Linear Equation.y

The x-intercept of a line is the point where it cutsthe x-axis, it can be found by putting y = 0 into the

equation of a line and solving for x.

y-interceptx

putting x = 0 into the equation of a line and solving for y.

x-intercept

Example 1: Sketch the graph of the linear equation y = 4x + 12 by finding its x and y-intercepts

x6

4

2

– 3 – 2 – 1– 4

The x-intercept is the point (– 3,0)

We now plot the x and y –intercepts and by joining these points we have the required line.y

12

10

8


Example 2: Sketch the graph of the linear equation 2x + 3y = 6 by finding its x andy-intercepts.

Solution: We find the y-intercept by putting x = 0and finding the corresponding y-value.

We find the x-intercept by putting y=0and finding the corresponding x-value.

2x + 3y2(0) + 3y

3yy

====

6662

2x + 3y2x + 3(0)

2xx

====

6663

The y-intercept is the point (0,2) The x-intercept is the point (3,0)

Note: When you use the x and y intercepts to draw a line you no longer have the advantage ofbeing able to easily detect errors in your calculations that would appear if you plot 3 ormore points so you must be extra careful when you use this method.

This method also does not always yield easy to use points as complex fractions canarise.

For example, 2x + 7y = 5 will give an x –intercept of and a y–intercept of

These points do not lend themselves to plotting accurate positions and so the resultingline can be inaccurate as a result.

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We now plot the x and y –intercepts and by joining these points we have therequired line.

y

x

2x + 3y = 6

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Exercise 1B:

Draw each of the following lines by first finding their x and y-intercepts.

1.

2.

y = 2x – 7

y = 7x + 21

3.

4.

5.

y=

y=

y=

+6

+9

– 9

6.

7.

8.

9.

2x + 4y = 8

10x + 5y = – 5

2x – 4y = 12

18x + 3y = 9

10. 4x – =


C. Using slope and y-intercepts to sketch a Linear Equation.

The slope of a line can be found by usingthe formula.

Slope =

in a geometric way the slope of a line canbe seen as starting at one point and movingit up or down in the Rise movement first andthen moving it to the right in the Run movement.

We then join the starting point to the finishing point.To draw the line with the required slope.

In order to draw a line using the slope and y-intercept we use the y-intercept as the starting pointand then from this position we use the slope to create the line.

Example 1: Sketch the line y = 3x – 2 using its slope and y-intercept.

Solution: This line has a slope of m = 3and a y-intercept of (0,– 2).Since the slope is 3 then

this means that the Rise = 3and the Run = 1.

We use the y-intercept of (0,– 2)as the starting point and then

x

Startingpoint

Finishingpoint

y

Run

Rise

x

y

pointmove up 3 (the Rise) and thenmove to the right 1 place (Run)

to the right to reach the finishing point at (1,1).

We then joint these two points and continue the line to sketch the line y = 3x – 2

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Finishingpoint

Slope =Run

Rise

Starting


Example 2: Sketch the line y = – 4x + 3 using its slope and y-intercept.

Solution: This line has a slope of m = – 4and a y-intercept of (0,3).Since the slope is – 4 then

this means that the Rise = –4and the Run = 1.

We use the y-intercept of (0,3)as the starting point and then

move down 4 places (the Rise) and thenmove to the right 1 place (Run)to the right to reach the finishing point at (1, – 1).

We then joint these two points and continue the line to sketch the line y = – 4x + 3

Example 3: Sketch the line y = using its slope and y-intercept.

Solution: This line has a slope of m =

and a y-intercept of (0,– 4).Since the slope is 3 then

Slope =

this means that the Rise = 3and Run = 4.

We use the y-intercept of (0,– 4)as the starting point and then

move up 3 places (the Rise) and thenmove to the right 4 places (the Run)to reach the finishing point at (4,– 1).

We then joint these two points and continue the line to sketch the line y =

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x

y

Startingpoint

Slope =

FinishingpointRun

Rise

x

y

Finishingpoint

Run

Rise

Startingpoint


Example 4: Sketch the line y = using its slope and y-intercept.

Solution: This line has a slope of m =

and a y-intercept of (0,3).Since the slope is 3 then

Slope =

this means that the Rise = – 2and Run = 5.

We use the y-intercept of (0,3)as the starting point and then

move down 2 places (the Rise) and thenmove to the right 5 place (the Run)to reach the finishing point at (5,1).

We then joint these two points and continue the line to sketch the line y =

Example 5: Sketch the line 2x – 4y = 6 using its slope and y-intercept.

Solution: We begin by rearranging the equation of this line from its general form of2x – 4y = 6 into its slope intercept form.

2x – 4y = 12– 4y = –2x + 12

=

=

This line has a slope of m =

and a y-intercept of (0, 3).

Since the slope is then

Slope =

this means that the Rise = 1 and Run = 2

We use the y-intercept of (0, 3) as the starting point and then move up one place(the Rise) and then move to the right 2 places (the Run) to reach the finishing pointat (2, 2).

We then joint these two points and continue the line to sketch the line 2x – 4y = 6

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x

y

Startingpoint

Finishingpoint

Run

Rise

y=

x

y

Finishingpoint

Run

Startingpoint

Rise

2x – 4y = 6


Page | 12

Example 6: Sketch the line y = 4.

Solution:

This is a special situation as lines

which have this form which containno x term have a slope of 0.

This line will have a y-intercept at (0,4)and a slope of zero.

A slope of zero has a horizontal slopeAnd will pass through the y-intercept (0,4).

Example 7: Sketch the line x = – 3

Solution:

This is a special situation as lines

1.

4.

y=7

x= – 4

2.

5.

y= – 5

y = 2x

3.

6.

x=3

y = – 3x

7. y = 2x – 4 8. y = 3x + 2 9. y = 4x – 7

10. y = 0 – 3x – 2 11. y = +5 12. y = +1

13. y = – 4 14. 3x + 4y = 12 15. x + 5y = – 10

16. 2x – 4y = 12 17. 18x + 8y = 16 18. x – =

y =4

x

y

y

which have this form do not havea y-intercept and a vertical slope.

xTo draw lines of this form we draw aa vertical line that passes through thex-axis at – 3.

x = – 3Exercise 1C:

Draw each of the following lines by using their slopes and y-intercepts.


2 Finding the Equation of a Line

There are two main methods for finding the equation of a line, they are.

Using the slope and y-intercept.

Using the slope and the coordinates of any point on the line.

Method 1: The equation of the line that has a y-intercept of (0,b) and a slope of m is

y = mx + b

Example 1: Find the equation of the line with a slope of 4 and a y-intercept of (0,3)

Solution: Slope = m = 4 and y-intercept b = 3 we get y = 4x + 3

We get the equation of the line is y = 4x + 3

Example 2: Find the equation of the line with a slope of – 5 and a y-intercept of (0,2)

Solution: Slope = m = – 5 and y-intercept b = 2

We get the equation of the line is y = – 5x + 2

Example 3: Find the equation of the line with a slope of and passing through the origin.

Solution: Slope = m = and y-intercept is (0,0) and so b = 0

We get the equation of the line is y = x

Example 4: Find the equation of a horizontal line that has a y-intercept of (0,5)

Solution: Slope = m = and y-intercept is (0,5) and so b = 5

We get the equation of the line is y = 0x + 5 which is y = 5

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Method 2: The equation of a line that passes through the point (x1,y1) with a slope of m can befound by using the formula.

y – y1 = m(x – x1)Example 5: Find the equation of the line with a slope of 3 passing through the point (2,7)

In this situation slope = m = 3 and the point (x1,y1) = (2,7)

y – y1

y – 7y – 7

y

====

m(x – x1)3(x – 2)3x – 63x + 1

Replace the values for m and x1 and y1

Use the distributive lawAdd 7 to both sides

Example 6: Find the equation of the line with a slope of – ½ passing through the point (– 4 ,9)

In this situation slope = m = – ½ and the point (x1,y1) = (– 4 ,9)

y – y1

y – 9y – 9

y

====

m(x – x1)– ½ (x + 4)– ½x – 2– ½x + 7



Example 7: Find the equation of the line with a slope of 0 passing through the point (– 1,– 4)

In this situation slope = m = 0 and the point (x1,y1) = (– 1,– 4)

y – y1 = m(x – x1)y+4y+4

y

===

0(x + 1)0– 4


Use the distributive lawSubtract 4 from both sides

Example 8: Find the equation of the line passing through the points (2,4) and (– 3 ,14)

To find the slope of the line that passes through the points (x1,y1) and (x2,y2)we use the formula slope = m = = = = – 2

In this situation slope = m = – 2 and the point (x1,y1) = (2 ,4)

y – y1

y – 4y – 4

y

====

m(x – x1)– 2(x – 4)– 2x + 8– 2x + 12




Example 9: Find the equation of the line passing through the points (– 3, – 4) and (– 5 , – 9)

To find the slope of the line that passes through the points (x1,y1) and (x2,y2)

we use the formula slope = m = = = = =

In this situation slope = m = and the point (x1,y1) = (– 3, – 4)

y – y1 = m(x – x1)

y+4

y+4

y

=

=

=

(x + 3)

x+

x+


Use the distributive law

Subtract 4 = from both sides

Example 10: Find the equation of the line passing through the points (– 3, 0) and (– 2 , 5)

To find the slope of the line that passes through the points (x1,y1) and (x2,y2)we use the formula slope = m = = = = =5

In this situation slope = m = 5 and the point (x1,y1) = (– 3, 0)

y – y1 = m(x – x1)

y+0

y

=

=

5 (x + 3)

5x + 15



Example 11: Find the equation of the line passing through the points (5, 4) and (5 , 9)

To find the slope of the line that passes through the points (x1,y1) and (x2,y2)we use the formula slope = m = = = = undefined

Since the slope is undefined we are looking at a vertical line that passes throughthe point (5,4).

This is a special situation and all lines which are vertical have the equationx = a number in this case x = 5

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Example 12: Find the equation of the lines drawn below.

we use the formula slope = m = = = =

In this situation slope = m = and the point (x1,y1) = (– 1, – 2)

y – y1 = m(x – x1)y+2

y+2

=

=

(x + 1)

x+



y = x – subtract 2 = from both sides

Line 3: Since line is vertical we can state its equation directly in this case it will be x = – 6

Line 4: Two points on this line are (2, – 4) and (–2 , 2)


we use the formula slope = m = = = = =

In this situation slope = m = and the point (x1,y1) = (2, – 4)

y – y1 = m(x – x1)y+4

y+4

y

=

=

=

(x – 2)

x+3

x – 1



subtract 4 from both sides

1

-5 -4 -3 -2 -1 0-1

x

y5432

-2

-3-4-5

Solution: In order to find the equation of a sloping line we need to find the coordinates of two pointson the line (any two points will do)

Line 1: Since line is horizontal we can state its equation directly in this case it will be y = 4.

Line 2: Two points on this line are (– 1, – 2) and (2 , 0)


1 2 3 4 5 6 7 8 9

y

x

12

3 4


Page | 17

Method 3: There is another method that can be used to find the equation of a line when youknow its slope and one point on the line.The examples below show this method in action.

Example 13: Find the equation of the line with a slope of 4 that passes through the point (3, – 8)

Solution: The method that we use is to start with the assumption that the equation of this linewill eventually become y = mx + b where m is the slope and (0,b) is the y-intercept.

Next, since the line has a slope = m = 4 we can replace this value into the equationof the line y = mx + b to get y = 4x + b.

Next we use the fact that the point (x,y) = (3, – 8) lies on the line and bysubstituting the x and y coordinates into the equation y = 4x + b and solving for bwe can get the final equation.

y = 4x + b– 8 =– 8 =– 20 =

4(3) + b12 + bb

So the equation of the above line is y = 4x – 20

Example 14: Find the equation of the line that passes through the points (4, – 6) and (– 2, – 3)

Solution: We start by finding the slope

First we find the slope of the line that passes through the points (x1,y1) and (x2,y2)

And to do this we use the formula slope = m = = = = =

Next, since the line has a slope = m =

equation of the line y = mx + b to get y =

we can replace this value into the

x + b.

Next we use the fact that the point (x,y) = (4, – 6) lies on the line and by

x + b and solvingsubstituting the x and y coordinates into the equation y =

for b we can get the final equation.

y = x+b

– 6 =

– 6 =– 4 =

+b

– 2 +bb

So the equation of the above line is y = x – 4

Find the equation of each line with a Slope = , passing through the point (5,3)

Page | 18


Exercise 2:

1.

2.

1

5

Find the equation of each line with a Slope = – 1, passing through the point (0,3)

3. Find the equation of each line with a Slope =2

3, passing through the point (0, – 5)

4.

5.

6.

7.

8.

9.

Find the equation of each line with a Slope = –3, passing through the point (–2,2).

Find the equation of each line with a Slope = 0, passing through the point (–7,3).

Find the equation of each line with a Slope = 10, passing through the point (–2,0).

Find the equation of each line passing through the points (–1,2) and (1,4)

Find the equation of each line passing through the points (0,4) and (4,2)

Find the equation of each line passing through the points (–3,3) and (0, –10)

10. Find the equation of each line passing through the points (2, –2) and (–1, –7)

11. Find the equation of each line passing through the points (–2, –3) and (–4, –5)

12. Find the equation of the lines drawn below.

12

5

4

-5

-1

-2

1

-5 -4 -3 -2 -1 0x

y5

4

3

2

-3

-4

1 2 3 4 5 6 7 8 9

y

x

3

6

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3 Puzzle questions involving the Equation of a Line

Example 1: Find the equation of the line that is parallel to the line 2x – 3y = 9 and passesthrough the point (3,6).

Solution: To find the slope of the line 2x – 3y = 9 we rearrange it into its slope interceptform.

2x – 3y = 9–2x + 9– 3y

y

=

=

=

subtract 2x from both sides

divide both sides by – 3

This line has a slope of

We are now going to find the slope of the line passing through the point (3,6)

In this situation slope = m = and the point (x1,y1) = (3,6)

y – y1 = m(x – x1)y – 6

y – 6

y

=

=

=

(x – 3)

x – 2

x+4



add 6 to both sides

Example 2: Find the equation of the line that is perpendicular to the line 2x – 3y = 9 and passesthrough the point (3,6).

Solution: To find the slope of the line 2x – 3y = 9 we rearrange it into its slope intercept form

using Example 1 we get y =

To find the slope of the perpendicular line we use the property that m1 m2 = – 1m1 m2 = – 1

= – 1( ) m2

m2 =

We are now going to find the slope of the line passing through the point (3,6)

In this situation slope = m = and the point (x1,y1) = (3,6)

y – y1 = m(x – x1)y – 6

y – 6

=

=

(x – 3)

x+



y = x+ add 6 = to both sides


Example 3: Are the lines with equation y = 2x – 2 and 2x + 4y = 9 parallel, perpendicular orneither.

Solution: In order to answer this question we find the slopes of both lines.

Line 1: y = 2x – 2 has a slope of 2

Line 2: 2x + 4y = 9

– 2x + 94y =

=

y = has a slope of

The two lines have slopes 2 and these lines are perpendicular since

m1 m2 =

Example 4: Are the lines with equation y = – 4x + 6 and 12x + 3y = 8 parallel, perpendicular orneither.

Solution: In order to answer this question we find the slopes of both lines.

Line 1: y = – 4x + 6 has a slope of – 4

8

– 12x + 8

Line 2: 12x + 3y =

3y =

=

y = has a slope of

The two lines have slopes and these lines are parallel since their slopes areidentical.

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Exercise 3:

1.

2.

3.

4.

5.

6.

7.

8.

9.

Find the equation of the line that is parallel to the line 3x – 2y = 8 and passes through thepoint (1, – 7).

Find the equation of the line that is perpendicular to the line 3x – 2y = 10 and passes throughthe point (2,0).

Find the equation of the line that is parallel to the line 2x + 3y = 0 and passes through thepoint (0,2).

Find the equation of the line that is perpendicular to the line 7x – 2y = – 8 and passesthrough the point (5,9).

Find the equation of the line that is parallel to the line 7x – y = – 8 and passes through thepoint (2, –3).

Find the equation of the line that is perpendicular to the line 7x – y = – 8 and passes throughthe point (2, –3).

Are the lines with equation y = 2x – 2 and 2x + 4y = 9 parallel, perpendicular or neither.

Are the lines with equation y = 4x + 3 and 3x + 2y = 9 parallel, perpendicular or neither.

Are the lines with equation 2x – 2y = 0 and 2x + 4y = 0 parallel, perpendicular or neither.

10. Are the lines with equation 4x – 3y = 7 and 8x – 6y = 7 parallel, perpendicular or neither.

11. The equation of the line that has the same slope as 4x – 10y = 24 and the same y-intercept as12y + 72 = 15x.

A. y = B. y = C. y = D. y =

12. The equation of the line that has the same slope as 4x – 10y = 24 and the same y-intercept as12y + 72 = 15x.

A. y = B. y = C. y = D. y =

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13. Match the equation with its graph. –7x + 7y = –49

A. C.

B. D.

2 4 6 8 10–4–6–10 –8 x

10

8

6

4

2

–2–2

–4

–6

–8

–10

y

2 4 6 8 10–4–6–10 –8 x

10

8

6

4

2

–2–2

–4

–6

–8

–10

y

2 4 6 8 10–4–6–10 –8 x

10

8

6

4

2

–2–2

–4

–6

–8

–10

y

2 4 6 8 10–4–6–10 –8 x

10

8

6

4

2

–2–2

–4

–6

–8

–10

y

x – 2 – 1 0 1 2

y = 2x-1 – 5 – 3 – 1 1 3

x – 2 – 1 0 1 2

y= – 1 1 3 5 7

x – 6 – 3 0 3 6

y – 8 – 6 – 4 – 2 0

x – 4 – 2 0 2 4

y 12 8 4 0 – 4

Page | 23


Solutions:

Exercise 1A:

1.

2.

3.

4.

5.

6.

7.

Draw the line has with the equation 2x – 3y = 12 using the table of values below.

Draw the line has with the equation 4x + 2y = 8 using the table of values below.

x

y=

– 6

– 3

– 3

– 2

0

– 1

3

0

6

1

x

y=

– 10

– 5

– 5

– 3

0

– 1

5

1

10

3

x

y=

– 6

– 5

– 3

– 3

0

– 1

3

1

6

3

Page | 24


Exercise 1B:

Draw each of the following lines by first finding their x and y-intercepts.

1.

2.

y = 2x – 7

y = 7x + 21

x-intercept =

x-intercept = – 3

y-intercept = – 7

y-intercept = 21

3.

4.

y=

y=

+6

+9

x-intercept = – 3

x-intercept =

y-intercept = 6

y-intercept = 9

5. y= – 9 x-intercept = 12 y-intercept = – 9

6.

7.

8.

9.

2x + 4y = 8

10x + 5y = – 5

2x – 4y = 12

18x + 3y = 9

x-intercept = 4

x-intercept =

x-intercept = 6

x-intercept =

y-intercept

y-intercept

y-intercept

y-intercept

=

=

=

=

2

– 1

– 3

3

10. 4x – = x-intercept = y-intercept = – 8

Exercise 1C:

Draw each of the following lines by using their slopes and y-intercepts.

1.2.3.4.

y=7y= – 5x=3x= – 4

Slope = 0 (horizontal)Slope = 0 (horizontal)Slope = undefined (vertical)Slope = undefined (vertical)

y-intercept = 7y-intercept = – 5y-intercept = does not existy-intercept = does not exist

5.6.7.8.9.10.

y = 2xy = – 3xy = 2x – 4y = 3x + 2y = 4x – 7y = – 3x – 2

Slope = 2Slope = – 3Slope = 2Slope = 3Slope = 4Slope = – 3

y-intercept = 0y-intercept = 0y-intercept = – 4y-intercept = 2y-intercept = – 7y-intercept = – 2

11. y= +5 Slope = y-intercept = 5

12. y =

13. y =

+1

– 4

Slope =

Slope =

y-intercept = 1

y-intercept = – 4

14. 3x + 4y = 12

15. x + 5y = – 10

16. 2x – 4y = 12

17. 18x + 8y = 16

Slope =

Slope =

Slope =

Slope =

y-intercept = 3

y-intercept = – 2

y-intercept = – 3

y-intercept = 2

18. x – = Slope = y-intercept =

Page | 25


Exercise 2:

1. y= 2. y = 3. y = 4. y=

5.7.

8.

9.

10.11.

y=Slope = m = 1

Slope = m =

Slope = m =

Slope = m = 3Slope = m = 1

6. y =equation of line y =

equation of line y =


equation of line y =equation of line y =

12. Line 1:Line 2:Line 3:Line 4:

Line 5:

Line 6:

Slope = m =Slope = m = – 2Slope = m = undefinedSlope = m =

Slope = m =

Slope = m =

equation of line y =equation of line y =equation of line x = 8equation of line y =



Exercise 3:

1.

2.

3.

4.

slope = m =

slope = m =

slope = m =

slope = m =

point (1, – 7)

point (2, 0)

point (0, 2)

point (5, 9)





5.6.

7.

8.

9.

slope = m = 7 point (2, –3).slope = m = 7 point (2, –3).

slope = m1 = 2 slope = m2 =



equation of line y =equation of line y =

Lines are perpendicular

Lines are neither

Lines are neither

10. slope = m1 = slope = m2 = Lines are parallel

11. A 12. A 13. A

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X – 2– 1 012 y = 2x x – 2– 1 012 y – 4– 2 024 Olympic College - Topic 7 Graphing a Linear Equation Topic 7 Graphing a Linear Equation 1. The Linear Equation