Section 7.1           

Given the equation of a line, the slope of the line may be calculated from any two points on the line.

 

Problem:         Find the slope of the line  3x – 4y = 12

 

Solution:             x      y

                            0    - 3                                 Let ()  =  ( 0, - 3)   and  () =  (4, 0)

                            4      0                                 

 

 

Problem:         Find the slope of the line  3x – 4y = 12

 

Solution:             x      y

                            2     -3/2                             Let ()  =  ( 2, - 3/2)   and  () =  (8, 3)

                            8      3                                 

 

An alternative way of finding the slope of a line given it’s equation is to solve the equation for y.  The coefficient of the x-term gives the slope.

 

Problem:         Find the slope of the line  3x – 4y = 12

 

Solution:         3x – 4y = 12   

                        -4y = -3x + 12

                        y = (3/4)x – 3                        m = coefficient of the x-term = Ύ

 

***A line in this form (y = mx + b) is in slope-intercept form and will    

      be discussed further in section 7.2.

 

Objective 5: It is possible to draw the graph of a line given its slope and any point on the line.  First plot the point on the Cartesian plane.  Using the slope, find a second point. (Remember – slope is the change in y-values divided by the change in x-values.)  These two points will determine the line.

 

Problem:         Graph the lines that passes through the point (-3, -2) and has a slope of ½.

                       

Solution:         1.          Plot the point (-3, -2)

2.         From the point (-3, -2) move up 1 unit, then 2 units to the right.  Plot this new point.  (-2, 0)

3.         Draw the line determined by these two points.

Objective 6: Two distinct, nonvertical lines are parallel if their slopes are equal.

Two distinct, nonvertical lines are perpendicular if their slopes are negative reciprocals.  (You could also say that the product of their slopes is –1.)

 

Problem:         Are these lines parallel, perpendicular, or neither?

                        L through the points (-1, 2) and (3, 5)

                        M through the points (4, 7) and (8, 10)

 

Solution:         Slope of line L:    m =

                        Slope of line M:   m =

                        lines L and M are parallel (they have the same slopes.)

           

Problem:         Are these lines parallel, perpendicular, or neither?

                        y = 2x + 9     and     y = -2x – 3

 

Solution:         y = 2x + 9 has a slope of 2.

                        y = -2x – 3 has a slope of –2.

                        Since 2  -2 the lines are not parallel.

                        Since 2(-2)  -1 the lines are not perpendicular.

 

Problem:         Are these lines parallel, perpendicular, or neither?

                        3x + 5y = 6  and  5x – 3y = 2

 

Solution:         The line 3x + 5y = 6 could be written as y = (-3/5)x + (6/5)

                        The line 5x – 3y = 2 could be written as y = (5/3)x – (2/3)

                        Since (-3/5) (5/3) the lines are not parallel.

                        But since (-3/5)(5/3) = -1, the lines are perpendicular.

 

ASSIGNMENT:     PAGE 462  problems 11 – 29 (odd)

                     PAGE 464  problems 49 – 97 (odd)

           

The Midpoint Formula:  Sometimes it is necessary to find the midpoint of a line segment which connects two points.  If the endpoints of a line segment are () and (), then the midpoint M is .

CAUTION:  The midpoint is a point and must be written in the form of a point.

            For example,  2, 3  (2, 3).

Problem:         Find the midpoint of the line segment joining the points (3, -6) and (6, 3).

Solution:         Let () = (3, -6) and let () = (6, 3).

(It does not matter which point you choose as point 1 and which you choose as point 2, but you must be consistent when “pluggin-in” to the midpoint formula.)

 

   and  

 

The midpoint of the line segment is the point

 

ASSIGNMENT:     PAGE 464 problems 41 – 47 (odds)

 

Section 7.2 Review of Equations of Lines; Linear Models

Objective 1: Given the slope of a line and its y-intercept, it is possible to write an

equation of the line.

Slope-intercept form of the equation of a line: y = mx + b,

where m is the slope and b is the y-coordinate of the y-intercept.

Problem: Find an equation of the line with slope 2 and y-intercept (0, -3)

Solution: m = 2 and b = -3. Then y = 2x – 3.

Objective 2: From section 7.1, we know that we can graph an equation if we know the slope and any point that lies on the line. Given a line in slope-intercept form, we have that information – the slope and a point, in this case the y-intercept.

Problem: Graph using the slope and the y-intercept: x + 2y = -4.

Solution: Write the equation in slope-intercept form: y = -(1/2) x – 2.

              Now we know that the slope is –1/2 and the y-intercept is (0, -2).

              Plot the point, move to another point using the slope, draw the line.

Objective 3: Given a point on a line and the slope of the line, an equation of the line can be written.

Point-slope form of the equation of a line: ,

Where m is the slope and () is any point on the line.

Problem: Find an equation of the line with slope 2/5 passing through (3, -4)

Solution: m = 2/5 and = (3, -4)

          

*****Equations of lines are often written in standard form: Ax + By = C, where A >0 and

A, B, C are integers. In the last example, the equation could be written as

2x – 5y = 26. This expresses the same line, but in standard from.

Objective 4: Given any two points on a line, it is possible to find an equation of the line. This may be done by using the two points to calculate the slope and then using that slope and either of the given points in point-slope form or it may be done by using the given points directly in two-point form of the equation of a line.

Problem: Write an equation of the line which passes through the points (-2, 6) and (1, 4).

Solution (using point-slope form): Let = (-2, 6) and = (1, 4)

Then      

           

            NOTE: We could have used either of the points in point-slope form.

                   I just choose the point that had already named point 1.

Problem: Write an equation of the line which passes through the points (-2, 6) and (1, 4).

Solution (using two-point form): Let = (-2, 6) and = (1, 4)

       

The horizontal line through the point (a, b) has equation y = b.

The vertical line through the point (a, b) has the equation x = a.

Objective 5: Since there is a "slope-based" definition of parallel and perpendicular, it may be used as an aid in writing equations of lines.

            Problem: Write an equation of the line through the point (-8, 3) which is parallel to the line 2x – 3y = 10.

            Solution: Since parallel lines have equal slopes, if we can find the slope of the line 2x – 3y = 10, then that is also the slope of the line that we are trying to find. The quickest way to find the slope is to write the line in slope-intercept form.

2x – 3y = 10 m = 2/3 and = (-8, 3)

This line passes through (-8, 3) and is parallel to 2x – 3y = 10

Problem: Write an equation of the line through the point (-8, 3) which is perpendicular to the line

2x – 3y = 10.

Solution: From the last example we know that the slope of the line 2x – 3y = 10 is 2/3. Then every line perpendicular to 2x – 3y = 10 has a slope of –3/2.

m = -3/2 and = (-8, 3)

This line passes through (-8, 3) and is perpendicular to 2x – 3y = 10.

ASSIGNMENT: PAGE 447-9 problems 7 – 14 (all), 15 – 37, 41 – 67 (odd)

 

7.3 Functions

Objective 1: A relation is simply a set of ordered pairs.

Example: { (-4, 1), (-2, 1), (-2,0)} is a relation.

A function is a special type of relation in which for each value of the first component of the ordered pairs, there is exactly one value for the second component (each first component is paired with a unique second component.)

Problem: Is this a relation or a function? { (0, 3), (-1, 2), (-1, 3) }

Solution: This is a relation. In a function, no two ordered pairs may have a single first component paired with two different second components. In this case –1 is paired with both 2 and 3.

Problem: Is this a relation or a function? {(0, 6), (1, 8), (3, 10), (-8, 4), (-2, 10)}

Solution: This is a function. For each different x-value there is exactly one y-value. (3, 10) and (-2, 10) have the same second components but that is acceptable as long as no first component is paired with different second components.

We often represent functions and relations using tables and graphs. The first component of an ordered pair is the independent variable and is generally placed on the horizontal axis of the Cartesian plane. The second component of an ordered pair is the dependent variable and is placed on the vertical axis of the Cartesian plane. The dependent variable depends on the independent variable. As the independent variable changes, so does the dependent variable.

Objective 2: The domain of a relation is the set of values for which the relation is defined. When determining the domain from a set of ordered pairs, it is the set of values of the independent variable (x). The range of a relation is the set of all values of the dependent variable (y).

Problem: Give the domain and the range. Is this a function?

{(2, 5), (3, 7), (4, 9), (5, 11)}

Solution: Domain {2, 3, 4, 5}

Range {5, 7, 9, 11}

This set of ordered pairs does represent a function.

Problem: Give the domain and the range. Is this a function?

{(1, 10), (2, 15), (2, 19), (3, 19), (5, 27)}

Solution: Domain {1, 2, 3, 5}

Range {10, 15, 19, 27}

This set of ordered pairs does not represent a function.

It is also possible to determine domain and range from a graph. The domain corresponds to the interval along the x-axis (reading left to right) where the graph occurs. The range corresponds to the interval along the y-axis (reading bottom to top) where the graph occurs.