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== 10.12#h(0.6em)Properties of Lines


    
      
=== Horizontal and Vertical Lines


      Two special cases of linear equations are worth noting.  First, an equation such as #math.equation(block: false, alt: "y equals 4")[$y = 4$] can be thought of as an equation in two variables,


      #math.equation(block: true, alt: "0 x plus y equals 4")[$0 x + y = 4$]
      For each value of #math.equation(block: false, alt: "x")[$x$], this equation assigns the value #math.equation(block: false, alt: "4")[$4$] to #math.equation(block: false, alt: "y")[$y$].  Thus, any ordered pair of the form #math.equation(block: false, alt: "open parenthesis x , 4 close parenthesis")[$( x , 4 )$] is a solution of the equation.  For example,


      #math.equation(block: true, alt: "open parenthesis minus 1 , 4 close parenthesis , open parenthesis 2 , 4 close parenthesis and open parenthesis 4 , 4 close parenthesis")[$( − 1 , 4 ) , " " " " ( 2 , 4 ) " " " " "and" " " " " ( 4 , 4 )$]
      are all solutions of the equation.  If we draw a straight line through these points, we obtain the #strong[horizontal] line shown at left below.


      #figure(figph[horizontal line], alt: "horizontal line", caption: none)


      #figure(figph[vertical line], alt: "vertical line", caption: none)


      The other special case of a linear equation is of the type #math.equation(block: false, alt: "x equals 3")[$x = 3$],  or


      #math.equation(block: true, alt: "x plus 0 y equals 3")[$x + 0 y = 3$]
      Here, only one value is permissible for #math.equation(block: false, alt: "x")[$x$], namely #math.equation(block: false, alt: "3")[$3$], while any value may be assigned to #math.equation(block: false, alt: "y")[$y$].  Any ordered pair of the form #math.equation(block: false, alt: "open parenthesis 3 , y close parenthesis")[$( 3 , y )$] is a solution of this equation.  If we choose two solutions, say #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] and #math.equation(block: false, alt: "open parenthesis 3 , 3 close parenthesis")[$( 3 , 3 )$], and draw a straight line through these two points, we have the #strong[vertical] line shown at right above.  In general, we have the following results.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The graph of #math.equation(block: false, alt: "x equals k")[$" " " " x = k " " " "$] (#math.equation(block: false, alt: "k")[$k$] a constant) is a vertical line.


        The graph of #math.equation(block: false, alt: "y equals k")[$" " " " y = k " " " "$] (#math.equation(block: false, alt: "k")[$k$] a constant) is a horizontal line.
]


      #examplebox("Example 1")[][
+ Graph #math.equation(block: false, alt: "y equals 2")[$" " y = 2$].
+ Graph #math.equation(block: false, alt: "x equals minus 3")[$" " x = − 3$].


        
        #solutionbox[
+ #figure(figph[horline], alt: "horline", caption: none)
+ #figure(figph[verline], alt: "verline", caption: none)
]
]


      Now let's compute the slopes of the two lines in the previous example.  Choose two points on the graph of #math.equation(block: false, alt: "y equals 2")[$y = 2$], say #math.equation(block: false, alt: "open parenthesis minus 5 , 2 close parenthesis")[$( − 5 , 2 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 2 close parenthesis")[$( 4 , 2 )$].  Use these points to compute the slope.


      #math.equation(block: true, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction 2 minus 2 over 4 minus open parenthesis minus 5 close parenthesis equals the fraction 0 over 9 equals 0")[$m = display(frac(y_(2) − y_(1), x_(2) − x_(1))) = display(frac(2 − 2, 4 − ( − 5 ))) = display(frac(0, 9)) = 0$]
      The slope of the horizontal line #math.equation(block: false, alt: "y equals 2")[$y = 2$] is zero.  In fact, the slope of any horizontal line is zero, because the #math.equation(block: false, alt: "y")[$y$] -coordinates of all the points on the line are equal.  Thus


      #math.equation(block: true, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction 0 over x sub 2 minus x sub 1 equals 0")[$m = display(frac(y_(2) − y_(1), x_(2) − x_(1))) = display(frac(0, x_(2) − x_(1))) = 0$]
      On a vertical line, the #math.equation(block: false, alt: "x")[$x$]-coordinates of all the points are equal.  For example, two points on the line #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] are  #math.equation(block: false, alt: "open parenthesis minus 3 , 1 close parenthesis")[$( − 3 , 1 )$] and #math.equation(block: false, alt: "open parenthesis minus 3 , 6 close parenthesis")[$( − 3 , 6 )$].  Using these points to compute the slope, we find


      #math.equation(block: true, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction 6 minus 1 over minus 3 minus open parenthesis minus 3 close parenthesis equals the fraction 5 over 0")[$m = display(frac(y_(2) − y_(1), x_(2) − x_(1))) = display(frac(6 − 1, − 3 − ( − 3 ))) = display(frac(5, 0))$]
      which is undefined.  The slope of any vertical line is undefined because the expression #math.equation(block: false, alt: "x sub 2 minus x sub 1")[$x_(2) − x_(1)$] equals zero.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The slope of a horizontal line is zero.


        The slope of a vertical line is undefined.
]


    
    
      
=== Parallel and Perpendicular Lines


      Consider the graphs of the equations


      #math.equation(block: true, alt: "y equals the fraction 2 over 3 x minus 4; y equals the fraction 2 over 3 x plus 2")[$y = display(frac(2, 3)) x − 4 \ y = display(frac(2, 3)) x + 2$]
      shown below.


      #figure(figph[parallel lines], alt: "parallel lines", caption: none)


      The lines have the same slope, #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$],  but different #math.equation(block: false, alt: "y")[$y$]-intercepts.  Because slope measures the steepness or inclination of a line, lines with the same slope are #strong[parallel].


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Two lines with slopes #math.equation(block: false, alt: "m sub 1")[$m_(1)$] and #math.equation(block: false, alt: "m sub 2")[$m_(2)$] are #strong[parallel] if and only if #math.equation(block: false, alt: "m sub 1 equals m sub 2")[$m_(1) = m_(2)$].
]


      #examplebox("Example 2")[][
Are the graphs of the equations #math.equation(block: false, alt: "3 x plus 6 y equals 6")[$3 x + 6 y = 6$] and #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x plus 5")[$y = − display(frac(1, 2)) x + 5$] parallel?


        
        #solutionbox[
The lines are parallel if their slopes are equal.  We can find the slope of the first line by putting its equation into slope-intercept form.  We solve for #math.equation(block: false, alt: "y")[$y$]:


          #math.equation(block: true, alt: "3 x plus 6 y, equals 6, Subtract 3 x from both sides.; 6 y, equals minus 3 x plus 6, Divide both sides by 6.; y, equals the fraction minus 3 x over 6 plus the fraction 6 over 6, Simplify.; y equals, equals minus the fraction 1 over 2 x plus 1")[$3 x + 6 y & = 6 &  & "Subtract" " " 3 bold(italic(x)) " " "from both sides." \ 6 y & = − 3 x + 6 &  & "Divide both sides by" " " 6. \ y & = display(frac(− 3 x, 6)) + display(frac(6, 6)) &  & "Simplify." \ y = & = − display(frac(1, 2)) x + 1$]
          The slope of the first line is #math.equation(block: false, alt: "m sub 1 equals minus the fraction 1 over 2")[$m_(1) = − display(frac(1, 2))$].  The equation of the second line is already in slope-intercept form, and its slope is #math.equation(block: false, alt: "m sub 2 equals minus the fraction 1 over 2")[$m_(2) = − display(frac(1, 2))$].  Thus, #math.equation(block: false, alt: "m sub 1 equals m sub 2")[$m_(1) = m_(2)$],  so the lines are parallel.
]
]


      Now consider the graphs of the equations


      #math.equation(block: true, alt: "y equals the fraction 2 over 3 x minus 2; y equals minus the fraction 3 over 2 x plus 3")[$y = display(frac(2, 3)) x − 2 \ y = − display(frac(3, 2)) x + 3$]
      shown below.


      #figure(figph[perpendicular lines], alt: "perpendicular lines", caption: none)


      The lines appear to be #strong[perpendicular].  The relationship between the slopes of perpendicular lines is not as easy to see as the relationship for parallel lines.  However, for this example, #math.equation(block: false, alt: "m sub 1 equals the fraction 2 over 3")[$m_(1) = display(frac(2, 3))$] and #math.equation(block: false, alt: "m sub 2 equals minus the fraction 3 over 2")[$m_(2) = − display(frac(3, 2))$].  Note that


      #math.equation(block: true, alt: "m sub 2 equals minus the fraction 3 over 2 equals the fraction minus 1 over the fraction 2 over 3 equals the fraction minus 1 over m sub 1")[$m_(2) = − display(frac(3, 2)) = display(frac(− 1, display(frac(2, 3)))) = display(frac(− 1, m_(1)))$]
      This relationship holds for any two perpendicular lines with slopes #math.equation(block: false, alt: "m sub 1")[$m_(1)$] and #math.equation(block: false, alt: "m sub 2")[$m_(2)$], as long as #math.equation(block: false, alt: "m sub 1 not equal to 0")[$m_(1) ≠ 0$] and #math.equation(block: false, alt: "m sub 2 not equal to 0")[$m_(2) ≠ 0$].


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Two lines with slopes #math.equation(block: false, alt: "m sub 1")[$m_(1)$] and #math.equation(block: false, alt: "m sub 2")[$m_(2)$] are #strong[perpendicular] if


        #math.equation(block: true, alt: "m sub 2 equals the fraction minus 1 over m sub 1")[$m_(2) = display(frac(− 1, m_(1)))$]
        .
]


      We say that #math.equation(block: false, alt: "m sub 2")[$m_(2)$] is the #strong[negative reciprocal] of #math.equation(block: false, alt: "m sub 1")[$m_(1)$].


      #examplebox("Example 3")[][
Are the graphs of #math.equation(block: false, alt: "3 x minus 5 y equals 5")[$3 x − 5 y = 5$] and #math.equation(block: false, alt: "2 y equals the fraction 10 over 3 x plus 3")[$2 y = display(frac(10, 3)) x + 3$] perpendicular?


        
        #solutionbox[
We find the slope of each line by putting the equations into slope-intercept form.  For the first line,


          #math.equation(block: true, alt: "5 y, equals 3 x minus 5, Divide both sides by 5.; y, equals the fraction 3 over 5 x minus 1")[$5 y & = 3 x − 5 &  & "Divide both sides by" " " 5. \ y & = display(frac(3, 5)) x − 1$]
          so #math.equation(block: false, alt: "m sub 1 equals the fraction 3 over 5")[$m_(1) = display(frac(3, 5))$].  For the second line, #math.equation(block: false, alt: "y equals the fraction 5 over 3 x plus the fraction 3 over 2")[$y = display(frac(5, 3)) x + display(frac(3, 2))$], so #math.equation(block: false, alt: "m sub 2 equals the fraction 5 over 3")[$m_(2) = display(frac(5, 3))$].  Now, the negative reciprocal of #math.equation(block: false, alt: "m sub 1")[$m_(1)$]  is


          #math.equation(block: true, alt: "the fraction minus 1 over m sub 1 equals the fraction minus 1 over the fraction 3 over 5 equals the fraction minus 5 over 3")[$display(frac(− 1, m_(1))) = display(frac(− 1, display(frac(3, 5)))) = display(frac(− 5, 3))$]
          but #math.equation(block: false, alt: "m sub 2 equals the fraction 5 over 3")[$m_(2) = display(frac(5, 3))$].  Thus, #math.equation(block: false, alt: "m sub 2 not equal to the fraction minus 1 over m sub 1")[$m_(2) ≠ display(frac(− 1, m_(1)))$],  so the lines are not perpendicular.
]
]


    
    
      
=== Applications to Geometry


      These relationships for the slopes of parallel and perpendicular lines can help us solve numerous geometric problems.


      #examplebox("Example 4")[][
Show that the triangle with vertices #math.equation(block: false, alt: "A open parenthesis 0 , 8 close parenthesis , B open parenthesis 6 , 2 close parenthesis ,")[$A ( 0 , 8 ) , " " B ( 6 , 2 ) , " "$] and #math.equation(block: false, alt: "C open parenthesis minus 4 , 4 close parenthesis")[$C ( − 4 , 4 )$] shown below is a right triangle.


          #figure(figph[right triangle], alt: "right triangle", caption: none)


        
        #solutionbox[
We will show that two of the sides of the triangle are perpendicular.  The line segment #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] has slope


          #math.equation(block: true, alt: "m sub 1 equals the fraction 2 minus 8 over 6 minus 0 equals the fraction minus 6 over 6 equals minus 1")[$m_(1) = display(frac(2 − 8, 6 − 0)) = display(frac(− 6, 6)) = − 1$]
          and the line segment #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$] has slope


          #math.equation(block: true, alt: "m sub 2 equals the fraction 4 minus 8 over minus 4 minus 0 equals the fraction minus 4 over minus 4 equals 1")[$m_(2) = display(frac(4 − 8, − 4 − 0)) = display(frac(− 4, − 4)) = 1$]
          Because


          #math.equation(block: true, alt: "the fraction minus 1 over m sub 1 equals the fraction minus 1 over minus 1 equals 1 equals m sub 2 ,")[$display(frac(− 1, m_(1))) = display(frac(− 1, − 1)) = 1 = m_(2) ,$]
          the sides #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] and #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$] are perpendicular, and the triangle is a right triangle.
]
]


      Consider the graph of #math.equation(block: false, alt: "4 x minus 2 y equals 6")[$4 x − 2 y = 6$] shown below.


      #figure(figph[parallel lines], alt: "parallel lines", caption: none)


      Can we find the equation of the line that is parallel to this line, but passes through the point #math.equation(block: false, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$]?  If we can find the slope of the desired line, we can use the slope-intercept formula to find its equation.


      Now because the line we want is parallel to the given line, they must have the same slope.  To find the slope of the given line, we write its equation in slope-intercept form:


      #math.equation(block: true, alt: "4 x minus 2 y, equals 6, Subtract 4 x from both sides.; minus 2 y, equals minus 4 x plus 6, Divide both sides by minus 2.; y, equals 2 x minus 3")[$4 x − 2 y & = 6 &  & "Subtract" " " 4 bold(italic(x)) " " "from both sides." \ − 2 y & = − 4 x + 6 &  & " " "Divide both sides by" " " − 2. \ y & = 2 x − 3$]
      The slope of the given line is #math.equation(block: false, alt: "2")[$2$]. Because the unknown line is parallel to this line, its slope is also #math.equation(block: false, alt: "2")[$2$].  Now we know the slope of the desired line, #math.equation(block: false, alt: "m equals 2")[$m = 2$],  and one point on the line, #math.equation(block: false, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$].  Substituting these values into the point-slope formula will give us the equation.


      #math.equation(block: true, alt: "y minus y sub 1, equals m open parenthesis x minus x sub 1 close parenthesis; y minus 4, equals 2 open parenthesis x minus 1 close parenthesis, Apply the distributive law.; y minus 4, equals 2 x minus 2, Add 4 to both sides.; y, equals 2 x plus 2")[$y − y_(1) & = m ( x − x_(1) ) \ y − 4 & = 2 ( x − 1 ) &  & "Apply the distributive law." \ y − 4 & = 2 x − 2 &  & "Add 4 to both sides." \ y & = 2 x + 2$]
      #examplebox("Example 5")[][
Find an equation for the line that passes through the point #math.equation(block: false, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$] and is perpendicular to the line #math.equation(block: false, alt: "4 x minus 2 y equals 6")[$4 x − 2 y = 6$].


        
        #solutionbox[
We follow the same strategy as in the discussion above:  First find the slope of the desired line, then use the point-slope formula to write its equation.


          The line we want is perpendicular to the given line, so its slope is the negative reciprocal of #math.equation(block: false, alt: "m sub 1 equals 2")[$m_(1) = 2$], the slope of the given line.  Thus


          #math.equation(block: true, alt: "m sub 2 equals the fraction minus 1 over m sub 1 equals the fraction minus 1 over 2")[$m_(2) = display(frac(− 1, m_(1))) = display(frac(− 1, 2))$]
          Now we use the point-slope formula with #math.equation(block: false, alt: "m sub 2 equals the fraction minus 1 over 2")[$m_(2) = display(frac(− 1, 2))$]  and #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis equals open parenthesis 1 , 4 close parenthesis")[$( x_(1) , y_(1) ) = ( 1 , 4 )$].


          #math.equation(block: true, alt: "y minus y sub 1, equals m open parenthesis x minus x sub 1 close parenthesis; y minus 4, equals the fraction minus 1 over 2 open parenthesis x minus 1 close parenthesis, Apply the distributive law.; y minus 4, equals the fraction minus 1 over 2 x plus the fraction 1 over 2, Add 4 to both sides.; y, equals the fraction minus 1 over 2 x plus the fraction 9 over 2, the fraction 1 over 2 plus 4 equals the fraction 1 over 2 plus the fraction 8 over 2 equals the fraction 9 over 2")[$y − y_(1) & = m ( x − x_(1) ) \ y − 4 & = display(frac(− 1, 2)) ( x − 1 ) &  & "Apply the distributive law." \ y − 4 & = display(frac(− 1, 2)) x + display(frac(1, 2)) &  & "Add 4 to both sides." \ y & = display(frac(− 1, 2)) x + display(frac(9, 2)) &  & display(frac(1, 2)) + 4 = display(frac(1, 2)) + display(frac(8, 2)) = display(frac(9, 2))$]
          The given line and the perpendicular line are shown below.


          #figure(figph[perpendicular lines], alt: "perpendicular lines", caption: none)
]
]


    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Horizontal
- Vertical
- Parallel
- Perpendicular


      
      
        
==== SKILLS


        Practice each skill in the exercises listed.


        + Sketch horizontal and vertical lines: \#1–6
+ Find an equation for a horizontal or vertical line: \#7–12
+ Identify parallel or perpendicular lines: \#15–24
+ Find equations for parallel or perpendicular lines: \#25–36


      
      
        
==== Reading Questions


        + Give an example of an equation for a vertical line, and for a horizontal line.
+ Why is the slope of a vertical line undefined?
+ What is the best way to determine whether two lines are parallel?
+ Suppose you know the equation of a certain line. Explain how to find the slope of a second line perpendicular to the first line.


      
    
    
      
=== Exercises A.12


      For Problems 1–6,


      + Sketch a rough graph of each equation, and label its intercept.
+ State the slope of each line.


      
        #math.equation(block: true, alt: "y equals minus 3")[$y = − 3$]


        
          + #figure(figph[line], alt: "line", caption: none)
+ #math.equation(block: false, alt: "m equals 0")[$m = 0$]


        
      
      
        #math.equation(block: true, alt: "x equals minus 2")[$x = − 2$]


      
      
        #math.equation(block: true, alt: "2 x equals 8")[$2 x = 8$]


        
          + #figure(figph[line], alt: "line", caption: none)
+ #math.equation(block: false, alt: "m")[$m " "$] is undefined


        
      
      
        #math.equation(block: true, alt: "3 y equals 15")[$3 y = 15$]


      
      
        #math.equation(block: true, alt: "x equals 0")[$x = 0$]


        
          + #figure(figph[line], alt: "line", caption: none)
+ #math.equation(block: false, alt: "m")[$m " "$] is undefined


        
      
      
        #math.equation(block: true, alt: "y equals 0")[$y = 0$]


      
      For Problems 7–12, find the equation of the line described.


      
        A vertical line through the point #math.equation(block: false, alt: "open parenthesis minus 5 , 8 close parenthesis")[$( − 5 , 8 )$]


        
          #math.equation(block: true, alt: "x equals minus 5")[$x = − 5$]


        
      
      
        A horizontal line through the point #math.equation(block: false, alt: "open parenthesis 2 , minus 4 close parenthesis")[$( 2 , − 4 )$]


      
      
        The #math.equation(block: false, alt: "x")[$x$]-axis


        
          #math.equation(block: true, alt: "y equals 0")[$y = 0$]


        
      
      
        The #math.equation(block: false, alt: "y")[$y$]-axis


      
      
        Perpendicular to #math.equation(block: false, alt: "x equals 3")[$x = 3$] and intersecting it at #math.equation(block: false, alt: "open parenthesis 3 , 9 close parenthesis")[$( 3 , 9 )$]


        
          #math.equation(block: true, alt: "y equals 9")[$y = 9$]


        
      
      
        Parallel to the #math.equation(block: false, alt: "y")[$y$]-axis and including the point #math.equation(block: false, alt: "open parenthesis minus 1 , minus 2 close parenthesis")[$( − 1 , − 2 )$]


      
      For Problems 13 and 14,


      + Determine whether the slope of each line is positive, negative, zero, or undefined.
+ List the lines in order of increasing slope.


      
        #figure(figph[lines], alt: "lines", caption: none)


        
          + #math.equation(block: false, alt: "l sub 1")[$l_(1) " "$] negative, #math.equation(block: false, alt: "l sub 2")[$l_(2) " "$] negative, #math.equation(block: false, alt: "l sub 3")[$l_(3) " "$] positive, #math.equation(block: false, alt: "l sub 4")[$l_(4) " "$] zero
+ #math.equation(block: false, alt: "l sub 1 , l sub 2 , l sub 4 , l sub 3")[$l_(1) , " " l_(2) , " " l_(4) , " " l_(3)$]


        
      
      
        #figure(figph[lines], alt: "lines", caption: none)


      
      
        + Use your calculator to graph the equations #math.equation(block: false, alt: "y equals 3 x plus 8")[$y = 3 x + 8$] and #math.equation(block: false, alt: "y equals 3.1 x plus 6")[$y = 3.1 x + 6$] together in the standard window.  Do you think the lines are parallel?
+ Find the slope of each line in part (a).  Are the lines parallel?
+ Find the #math.equation(block: false, alt: "y")[$y$]-value for each equation when #math.equation(block: false, alt: "x equals 20")[$x = 20$]. What do your answers tell you about the two lines?


        
          + #figure(figph[lines], alt: "lines", caption: none)
+ #math.equation(block: false, alt: "m equals 3 , m equals 3.1 ,")[$m = 3 , " " m = 3.1 , " "$] No
+ #math.equation(block: false, alt: "y equals 68")[$y = 68$] for both lines. The lines intersect at #math.equation(block: false, alt: "open parenthesis 20 , 68 close parenthesis")[$( 20 , 68 )$].


        
      
      
        + Use your calculator to graph the equation #math.equation(block: false, alt: "y equals 0.001 x plus 4")[$y = 0.001 x + 4$] in the standard window.  Do you think the line is horizontal?
+ Find the slope and the #math.equation(block: false, alt: "x")[$x$]-intercept of the line in part (a). Is the line horizontal?
+ Graph the equation in part (a) in the window #linebreak() #math.equation(block: false, alt: "Xmin equals minus 5000 ; Xmax equals 5000")[$"Xmin" = − 5000 "; Xmax" = 5000$] #linebreak() #math.equation(block: false, alt: "Ymin equals minus 10 ; Ymax equals 10")[$"Ymin" = − 10 "; Ymax" = 10$] #linebreak() Find the coordinates of two convenient points on the line, and compute its slope using the slope formula.


      
      
        The slopes of several lines are given below.  Which of the lines are parallel to the graph of #math.equation(block: false, alt: "y equals 0.75 x plus 2")[$y = 0.75 x + 2$], and which are perpendicular to it?


        + #math.equation(block: false, alt: "m equals the fraction 3 over 4")[$m = display(frac(3, 4))$]
+ #math.equation(block: false, alt: "m equals the fraction 8 over 6")[$m = display(frac(8, 6))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 20 over 15")[$m = display(frac(− 20, 15))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 39 over 52")[$m = display(frac(− 39, 52))$]
+ #math.equation(block: false, alt: "m equals the fraction 4 over 3")[$m = display(frac(4, 3))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 16 over 12")[$m = display(frac(− 16, 12))$]
+ #math.equation(block: false, alt: "m equals the fraction 36 over 48")[$m = display(frac(36, 48))$]
+ #math.equation(block: false, alt: "m equals the fraction 9 over 12")[$m = display(frac(9, 12))$]


        
          parallel:  a, g, h;  perpendicular:  c, f


        
      
      
        The slopes of several lines are given below.  Which of the lines are parallel to the graph of #math.equation(block: false, alt: "y equals 2.5 x minus 3")[$y = 2.5 x − 3$], and which are perpendicular to it?


        + #math.equation(block: false, alt: "m equals the fraction 2 over 5")[$m = display(frac(2, 5))$]
+ #math.equation(block: false, alt: "m equals the fraction 25 over 10")[$m = display(frac(25, 10))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 8 over 20")[$m = display(frac(− 8, 20))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 45 over 18")[$m = display(frac(− 45, 18))$]
+ #math.equation(block: false, alt: "m equals the fraction 40 over 16")[$m = display(frac(40, 16))$]
+ #math.equation(block: false, alt: "m equals 25")[$m = 25$]
+ #math.equation(block: false, alt: "m equals the fraction minus 1 over 25")[$m = display(frac(− 1, 25))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 5 over 10")[$m = display(frac(− 5, 10))$]


      
      
        In each part, determine whether the two lines are parallel, perpendicular, or neither.


        + #math.equation(block: false, alt: "y equals the fraction 3 over 5 x minus 7 , 3 x minus 5 y equals 2")[$y = display(frac(3, 5)) x − 7 , " " " " 3 x − 5 y = 2$]
+ #math.equation(block: false, alt: "y equals 4 x plus 3 , y equals the fraction 1 over 4 x minus 3")[$y = 4 x + 3 , " " " " y = display(frac(1, 4)) x − 3$]
+ #math.equation(block: false, alt: "6 x plus 2 y equals 1 , x equals 1 minus 3 y")[$6 x + 2 y = 1 , " " " " x = 1 − 3 y$]
+ #math.equation(block: false, alt: "2 y equals 5 , 5 y equals minus 2")[$2 y = 5 , " " " " 5 y = − 2$]


        
          + parallel
+ neither
+ neither
+ parallel


        
      
      
        In each part, determine whether the two lines are parallel, perpendicular, or neither.


        + #math.equation(block: false, alt: "2 x minus 7 y equals 14 , 7 x minus 2 y equals 14")[$2 x − 7 y = 14 , " " " " 7 x − 2 y = 14$]
+ #math.equation(block: false, alt: "x plus y equals 6 , x minus y equals 6")[$x + y = 6 , " " " " x − y = 6$]
+ #math.equation(block: false, alt: "x equals minus 3 , 3 y equals 5")[$x = − 3 , " " " " 3 y = 5$]
+ #math.equation(block: false, alt: "the fraction 1 over 4 x minus the fraction 3 over 4 y equals the fraction 2 over 3 , the fraction 1 over 6 x plus the fraction 1 over 2 y equals the fraction 1 over 3")[$display(frac(1, 4)) x − display(frac(3, 4)) y = display(frac(2, 3)) , " " " " display(frac(1, 6)) x + display(frac(1, 2)) y = display(frac(1, 3))$]


      
      
        + Sketch the triangle with vertices #math.equation(block: false, alt: "A open parenthesis 2 , 5 close parenthesis , B open parenthesis 5 , 2 close parenthesis ,")[$A ( 2 , 5 ) , " " B ( 5 , 2 ) ,$] and  #math.equation(block: false, alt: "C open parenthesis 10 , 7 close parenthesis")[$C ( 10 , 7 )$].
+ Show that the triangle is a right triangle.  (Hint:  What should be true about the 
        slopes of the two sides that form the right angle?)


        
          Slope #math.equation(block: false, alt: "A B bar equals minus 1 ,")[$limits(A B)^(―) = − 1 , " "$] slope #math.equation(block: false, alt: "B C bar equals 1 ,")[$limits(B C)^(―) = 1 , " "$] slope #math.equation(block: false, alt: "A C bar equals the fraction 1 over 4.")[$limits(A C)^(―) = display(frac(1, 4)) . " "$] Hence #math.equation(block: false, alt: "A B bar ⊥ A B bar ,")[$limits(A B)^(―) ⊥ limits(A B)^(―) ,$] so the triangle is a right triangle.


        
      
      
        + Sketch the triangle with vertices #math.equation(block: false, alt: "P open parenthesis minus 1 , 3 close parenthesis , Q open parenthesis minus 3 , 8 close parenthesis ,")[$P ( − 1 , 3 ) , " " Q ( − 3 , 8 ) ,$] and #math.equation(block: false, alt: "R open parenthesis 4 , 5 close parenthesis")[$R ( 4 , 5 )$].
+ Show that the triangle is a right triangle.  (See the hint for Problem 21.)


      
      
        + Sketch the quadrilateral with vertices #math.equation(block: false, alt: "P open parenthesis 2 , 4 close parenthesis , Q open parenthesis 3 , 8 close parenthesis , R open parenthesis 5 , 1 close parenthesis ,")[$P ( 2 , 4 ) , " " Q ( 3 , 8 ) , " " R ( 5 , 1 ) ,$] and #math.equation(block: false, alt: "S open parenthesis 4 , minus 3 close parenthesis")[$S ( 4 , − 3 )$].
+ Show that the quadrilateral is a parallelogram.  (Hint:  What should be true about the 
        slopes of the opposite sides of the parallelogram?)


        
          Slope #math.equation(block: false, alt: "P Q bar equals minus 4 ,")[$limits(P Q)^(―) = − 4 , " "$]  slope #math.equation(block: false, alt: "Q R bar equals the fraction minus 7 over 2 ,")[$limits(Q R)^(―) = display(frac(− 7, 2)) , " "$] slope #math.equation(block: false, alt: "R S bar equals 4 ,")[$limits(R S)^(―) = 4 , " "$] slope #math.equation(block: false, alt: "S P bar equals the fraction minus 7 over 2.")[$limits(S P)^(―) = display(frac(− 7, 2)) . " "$]  Hence #math.equation(block: false, alt: "P Q bar ∥ R S bar")[$" " limits(P Q)^(―) ∥ limits(R S)^(―) " "$] and #math.equation(block: false, alt: "Q R bar ∥ S P bar")[$" " limits(Q R)^(―) ∥ limits(S P)^(―) " "$],  so the points are the vertices of a parallelogram.


        
      
      
        + Sketch the quadrilateral with vertices #math.equation(block: false, alt: "A open parenthesis minus 5 , 4 close parenthesis , B open parenthesis 7 , minus 11 close parenthesis , C open parenthesis 12 , 25 close parenthesis ,")[$A ( − 5 , 4 ) , " " B ( 7 , − 11 ) , " " C ( 12 , 25 ) ,$] and #math.equation(block: false, alt: "D open parenthesis 0 , 40 close parenthesis")[$D ( 0 , 40 )$].
+ Show that the quadrilateral is a parallelogram.  (See the hint for Problem 23.)


      
      
        Show that the line passing through the points #math.equation(block: false, alt: "A open parenthesis 0 , minus 3 close parenthesis")[$A ( 0 , − 3 )$] and  #math.equation(block: false, alt: "B open parenthesis 3 , the fraction 1 over 2 close parenthesis")[$B ( 3 , display(frac(1, 2)) )$]  also passes through the point #math.equation(block: false, alt: "C open parenthesis minus 6 , minus 10 close parenthesis")[$C ( − 6 , − 10 )$].


        
          Slope #math.equation(block: false, alt: "A B bar equals the fraction 7 over 6 equals")[$limits(A B)^(―) = display(frac(7, 6)) =$] slope #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$], so #math.equation(block: false, alt: "A , B ,")[$A , B ,$] and #math.equation(block: false, alt: "C")[$C$] lie on the same line.


        
      
      
        Do the points #math.equation(block: false, alt: "P open parenthesis minus 5 , minus 3 the fraction 1 over 2 close parenthesis , Q open parenthesis 4 , minus 2 close parenthesis")[$P ( − 5 , − 3 display(frac(1, 2)) ) , " " Q ( 4 , − 2 )$] and #math.equation(block: false, alt: "R open parenthesis 9 the fraction 1 over 2 , minus 1 close parenthesis")[$R ( 9 display(frac(1, 2)) , − 1 )$] lie on the same line?  Why or why not?


      
      Use graph paper for Problems 27-30.


      
        + Put the equation #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] into slope-intercept form, and graph the equation.
+ What is the slope of any line that is parallel to #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$]?
+ On your graph for part (a), sketch by hand a line that is parallel to #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$].
+ Use the point-slope formula to write an equation for the line that is parallel to the graph of #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$].


        
          + #math.equation(block: false, alt: "y equals the fraction 1 over 2 x minus the fraction 5 over 2")[$y = display(frac(1, 2)) x − display(frac(5, 2))$]
+ #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]
+ (sketch)
+ #math.equation(block: false, alt: "y equals the fraction 1 over 2 x minus 2")[$y = display(frac(1, 2)) x − 2$]


        
      
      
        + Put the equation #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] into slope-intercept form, and graph the equation.
+ What is the slope of any line that is parallel to #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$]?
+ On your graph for part (a), sketch by hand a line that is parallel to #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis minus 3 , 2 close parenthesis")[$( − 3 , 2 )$].
+ Use the point-slope formula to write an equation for the line that is parallel to the graph of #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis minus 3 , 2 close parenthesis")[$( − 3 , 2 )$].


      
      
        + Put the equation #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] into slope-intercept form, and graph the equation.
+ What is the slope of any line that is perpendicular to #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$]?
+ On your graph for part (a), sketch by hand a line that is perpendicular to #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$].
+ Use the point-slope formula to write an equation for the line that is perpendicular to the graph of #math.equation(block: false, alt: "2 y minus 3 x equals 5")[$2 y − 3 x = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$].


        
          + #math.equation(block: false, alt: "y equals the fraction 3 over 2 x plus the fraction 5 over 2")[$y = display(frac(3, 2)) x + display(frac(5, 2))$]
+ #math.equation(block: false, alt: "the fraction minus 2 over 3")[$display(frac(− 2, 3))$]
+ (sketch)
+ #math.equation(block: false, alt: "y equals the fraction minus 2 over 3 x plus the fraction 14 over 3")[$y = display(frac(− 2, 3)) x + display(frac(14, 3))$]


        
      
      
        + Put the equation #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] into slope-intercept form, and graph the equation.
+ What is the slope of any line that is perpendicular to #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$]?
+ On your graph for part (a), sketch by hand a line that is perpendicular to #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 4 , minus 3 close parenthesis")[$( 4 , − 3 )$].
+ Use the point-slope formula to write an equation for the line that is perpendicular to the graph of #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] and passes through the point #math.equation(block: false, alt: "open parenthesis 4 , minus 3 close parenthesis")[$( 4 , − 3 )$].


      
      
        Two of the vertices of rectangle #math.equation(block: false, alt: "A B C D")[$A B C D$] are #math.equation(block: false, alt: "A open parenthesis minus 5 , 2 close parenthesis")[$A ( − 5 , 2 )$] and #math.equation(block: false, alt: "B open parenthesis minus 2 , 4 close parenthesis")[$B ( − 2 , 4 )$].


        + Find an equation for the line that includes side #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$].
+ Find an equation for the line that includes side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$] .


        
          + #math.equation(block: false, alt: "y equals minus 2 x minus 8")[$y = − 2 x − 8$]
+ #math.equation(block: false, alt: "y equals the fraction 1 over 2 x minus 3")[$y = display(frac(1, 2)) x − 3$]


        
      
      
        Two of the vertices of rectangle #math.equation(block: false, alt: "P Q R S")[$P Q R S$] are #math.equation(block: false, alt: "P open parenthesis minus 2 , minus 6 close parenthesis")[$P ( − 2 , − 6 )$] and #math.equation(block: false, alt: "Q open parenthesis 4 , minus 4 close parenthesis")[$Q ( 4 , − 4 )$].


        + Find an equation for the line that includes side #math.equation(block: false, alt: "P Q bar")[$limits(P Q)^(―)$].
+ Find an equation for the line that includes side #math.equation(block: false, alt: "Q R bar")[$limits(Q R)^(―)$].


      
      For Problems 33 and 33, recall from geometry that the altitude from one vertex of a triangle to the opposite side is perpendicular to that side.


      
        + Sketch the triangle with vertices #math.equation(block: false, alt: "A open parenthesis minus 6 , minus 3 close parenthesis , B open parenthesis minus 6 , 3 close parenthesis")[$A ( − 6 , − 3 ) , " " B ( − 6 , 3 )$] and #math.equation(block: false, alt: "C open parenthesis 4 , 5 close parenthesis")[$C ( 4 , 5 )$].
+ Find the slope of the side #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$].
+ Find the slope of the altitude from point #math.equation(block: false, alt: "B")[$B$] to side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$].
+ Find an equation for the line that includes the altitude from point #math.equation(block: false, alt: "B")[$B$] to side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$].


        
          + (sketch)
+ #math.equation(block: false, alt: "m equals the fraction 4 over 5")[$m = display(frac(4, 5))$]
+ #math.equation(block: false, alt: "m equals the fraction minus 5 over 4")[$m = display(frac(− 5, 4))$]
+ #math.equation(block: false, alt: "y equals the fraction minus 5 over 4 x minus the fraction 9 over 2")[$y = display(frac(− 5, 4)) x − display(frac(9, 2))$]


        
      
      
        + Sketch the triangle with vertices #math.equation(block: false, alt: "A open parenthesis minus 5 , 12 close parenthesis , B open parenthesis 4 , minus 2 close parenthesis")[$A ( − 5 , 12 ) , " " B ( 4 , − 2 )$] and #math.equation(block: false, alt: "C open parenthesis 1 , minus 6 close parenthesis")[$C ( 1 , − 6 )$].
+ Find the slope of the side #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$].
+ Find the slope of the altitude from point #math.equation(block: false, alt: "B")[$B$] to side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$].
+ Find an equation for the line that includes the altitude from point #math.equation(block: false, alt: "B")[$B$] to side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$].


      
      For Problems 35 and 36, recall from geometry that the tangent line to a circle is perpendicular to the radius to the point of tangency.


      
        The center of a circle is the point #math.equation(block: false, alt: "C open parenthesis 2 , 4 close parenthesis")[$C ( 2 , 4 )$], and #math.equation(block: false, alt: "P open parenthesis minus 1 , 6 close parenthesis")[$P ( − 1 , 6 )$] is a point on the circle, as shown below.  Find the equation of the line tangent to the circle at the point #math.equation(block: false, alt: "P")[$P$].


        #figure(figph[circle], alt: "circle", caption: none)


        
          #math.equation(block: true, alt: "y equals the fraction 3 over 2 x plus the fraction 15 over 2")[$y = display(frac(3, 2)) x + display(frac(15, 2))$]


        
      
      
        The center of a circle is the point #math.equation(block: false, alt: "D open parenthesis minus 2 , 1 close parenthesis")[$D ( − 2 , 1 )$], and #math.equation(block: false, alt: "Q open parenthesis 1 , minus 3 close parenthesis")[$Q ( 1 , − 3 )$] is a point on the circle, as shown below.  Find the equation of the line tangent to the circle at the point #math.equation(block: false, alt: "Q")[$Q$].


        #figure(figph[circle], alt: "circle", caption: none)


      
      
        In this exercise we will show that parallel lines have the same slope.  In the figure below, #math.equation(block: false, alt: "l sub 1")[$l_(1)$] and #math.equation(block: false, alt: "l sub 2")[$l_(2)$] are two parallel lines that are neither horizontal nor vertical.  Their #math.equation(block: false, alt: "y")[$y$]-intercepts are #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. The segments #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$] and #math.equation(block: false, alt: "C D bar")[$limits(C D)^(―)$] are constructed parallel to the #math.equation(block: false, alt: "x")[$x$]-and #math.equation(block: false, alt: "y")[$y$]-axes, respectively.


        #figure(figph[parallel lines], alt: "parallel lines", caption: none)


        Explain why each of the following statements is true.


        + Angle #math.equation(block: false, alt: "A C D")[$A C D$] equals angle #math.equation(block: false, alt: "C A B")[$C A B$].
+ Angle #math.equation(block: false, alt: "D A C")[$D A C$] equals angle #math.equation(block: false, alt: "A C B")[$A C B$].
+ Triangle #math.equation(block: false, alt: "A C D")[$A C D$] is similar to triangle #math.equation(block: false, alt: "C A B")[$C A B$].
+ #math.equation(block: false, alt: "m sub 1 equals the fraction C D over A C ; m sub 2 equals the fraction A B over A C")[$m_(1) = display(frac(C D, A C)) ; " " " " m_(2) = display(frac(A B, A C))$]
+ #math.equation(block: false, alt: "m sub 1 equals m sub 2")[$m_(1) = m_(2)$]


        
          + Right angles are equal.
+ Alternate interior angles are equal.
+ Two angles of one triangle equal two angles of the other.
+ Definition of slope.
+ Corresponding sides of similar triangles are proportional.


        
      
      
        In this exercise we will show that if two lines with slopes #math.equation(block: false, alt: "m sub 1")[$m_(1)$] and #math.equation(block: false, alt: "m sub 2")[$m_(2)$] (where neither line is vertical) are perpendicular, then #math.equation(block: false, alt: "m sub 2")[$m_(2)$] is the negative reciprocal of #math.equation(block: false, alt: "m sub 1")[$m_(1)$].  In the figure below, lines #math.equation(block: false, alt: "l sub 1")[$l_(1)$] and #math.equation(block: false, alt: "l sub 2")[$l_(2)$] are perpendicular.  Their #math.equation(block: false, alt: "y")[$y$]-intercepts are #math.equation(block: false, alt: "B")[$B$] and #math.equation(block: false, alt: "C")[$C$].  The segment #math.equation(block: false, alt: "A P bar")[$limits(A P)^(―)$] is constructed through the point of intersection of #math.equation(block: false, alt: "l sub 1")[$l_(1)$] and #math.equation(block: false, alt: "l sub 2")[$l_(2)$] parallel to the #math.equation(block: false, alt: "x")[$x$]-axis.


        #figure(figph[circle], alt: "circle", caption: none)


        Explain why each of the following statements is true.


        + Angles #math.equation(block: false, alt: "A B C")[$A B C$] and #math.equation(block: false, alt: "A C B")[$A C B$] are complementary.
+ Angles #math.equation(block: false, alt: "A B C")[$A B C$] and #math.equation(block: false, alt: "B A P")[$B A P$]  are complementary.
+ Angle #math.equation(block: false, alt: "B A P")[$B A P$] equals angle #math.equation(block: false, alt: "A C B")[$A C B$].
+ Angles #math.equation(block: false, alt: "C A P")[$C A P$] and #math.equation(block: false, alt: "A C B")[$A C B$] are complementary.
+ Angle #math.equation(block: false, alt: "C A P")[$C A P$] equals angle #math.equation(block: false, alt: "A B C")[$A B C$].
+ Triangle #math.equation(block: false, alt: "A B P")[$A B P$] is similar to triangle #math.equation(block: false, alt: "C A P")[$C A P$].
+ #math.equation(block: false, alt: "m sub 1 equals the fraction B P over A P ; m sub 2 equals minus the fraction C P over A P")[$m_(1) = display(frac(B P, A P)) ; " " " " m_(2) = − display(frac(C P, A P))$]
+ #math.equation(block: false, alt: "m sub 2 equals the fraction minus 1 over m sub 1")[$m_(2) = display(frac(− 1, m_(1)))$]