Precalculus Study Guide
Copyright Maplesoft, a division of Waterloo Maple Inc., 2024
Chapter 2: Graphing Lines in the Cartesian Plane
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Introduction
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In plane Euclidean geometry, two points determine a straight line, as well as the line segment connecting the points.
In analytic geometry, the equation of a straight line can be given in any one of the forms shown in Table 2.0.1.
Form
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Equation
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Details
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Two-point
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where the two points are and
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Point-slope
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where is the slope of the line
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Slope-intercept
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where is the slope, and , the -intercept
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General
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Table 2.0.1 Formulas for the equation of a straight line
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Analytic geometry also provides expressions for the coordinates of the midpoint of a line segment, the length of that segment, and the equation of the perpendicular bisector of the segment.
In addition, this chapter explores how to obtain the coordinates of the intersection two lines, and how to find the distance between two parallel lines.
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Chapter Glossary
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The following terms in Chapter 2 are linked to the Maple Math Dictionary.
analytic geometry
bisector
Cartesian plane
coordinate
Cramer's Rule
Euclidean geometry
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intercept
length
line
line segment
matrix
midpoint
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negative
parallel
perpendicular
reciprocal
slope
transversal
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Typical Problems
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2.1. In the Cartesian plane, the two points and determine a line segment. For this segment, find
a) the length ;
b) the midpoint ;
c) the slope ;
d) the equation;
e) the equation of the perpendicular bisector.
2.2. Calculate the coordinates of the point of intersection of the lines .
2.3. Find the distance between the parallel lines whose equations are and .
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Maple Initializations
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Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.
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Solutions
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Problem 2.1
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2.1 - Mathematical Solution
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In the Cartesian plane, the two points and determine a line segment.
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2.1 (a) - Length of Line Segment
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The length of this line segment is computed with the distance formula
Taking and , the distance between the endpoints, namely,
= =
is the length of the line segment.
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2.1 (b) - Midpoint of Line Segment
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The midpoint of the segment is the point whose coordinates are
Again taking and , the midpoint will be
=
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2.1 (c) - Slope of Line Segment
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The slope of the line segment joining the points and is given by
Hence, the slope of the segment joining the points and is
=
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2.1 (d) - Equation of Line
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The equation of the line through the points and can be found from the two-point form of the straight line, namely,
This is equivalent to the point-slope form of the line, namely,
or even
Since the slope was calculated in Part (c), the required equation of the line joining the points and is then
or
=
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2.1 (e) - Equation of Perpendicular Bisector
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The equation of the perpendicular bisector of the line segment joining the points and is found by applying the point-slope form of the straight line to the midpoint of the segment. The midpoint was found in Part (b) to be
The slope of this segment was found in Part (c) to be . Therefore, the slope of the line perpendicular to this segment is the negative reciprocal of , namely, . Hence, the desired equation is
or
=
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2.1 - Maplet Solution
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The equation of the line through the two points and can be found with the Maple Lines Tutor, a thumbnail sketch of which can be seen in Figure 2.1.1.
In this tutor, select the Two Points radio button and enter the two given points. Then click the Display button. The Lines Tutor will then provide the equation of the line in the two forms shown in Table 2.1.1, and draw a graph of the line.
Form
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Equation
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Slope-intercept
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General
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Table 2.1.1 Equations provided by Lines Tutor
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Figure 2.1.1 Thumbnail image of the Lines Tutor
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To launch the Lines Tutor, select "Lines" from the Tools≻Tutors≻Precalculus menu. Alternatively, click the following link:
Lines Tutor
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The length, midpoint, slope, and equation of the perpendicular bisector of this line segment are obtained with
Line Tutor #2
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The coordinates of the points and are entered in the data fields for A: and B:. The buttons labeled Length, Midpoint, Slope, and Perp. Bisector yield, respectively, the length, midpoint, slope and perpendicular bisector of the segment determined by the points A and B.
To launch the Line Tutor #2, click the following link:
Line Tutor #2
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Figure 2.1.2 Thumbnail image of the Line Tutor #2
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2.1 - Interactive Solution
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Enter the given data
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Enter the first point as a Maple list.
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Context Panel: Assign to a Name≻A
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Enter the first point as a Maple list.
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Context Panel: Assign to a Name≻B
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Part (a) - Length of Line Segment
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Form the differences
by writing
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Context Panel: Assign to a Name≻
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Compute
by writing .
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Part (b) - Midpoint of Line Segment
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Obtain the midpoint by writing .
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Context Panel: Assign to a Name≻
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Part (c) - Slope of Line Segment
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Obtain the slope by writing
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Context Panel: Assign to a Name≻
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Part (d) - Equation of Line
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By referencing as , and as , implement the point-slope form of the line
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Part (e) - Equation of Perpendicular Bisector
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By referencing the - and -coordinates of the midpoint as and , respectively, implement the point-slope form of the line
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where is the slope of the line orthogonal to the line in Part (d).
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2.1 - Programmatic Solution
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Be sure to have initialized the worksheet by pressing the Initialize button in the Initialization section.
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(a) Length of the line segment
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(b) Midpoint of the segment
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(c) Slope of the segment
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(d) Equation of line through and
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(e) Equation of perpendicular bisector of segment
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Problem 2.2
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2.2 - Mathematical Solution
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Several methods are available for calculating the coordinates of the point of intersection of the lines
Figure 2.2.1 provides a graphical estimate of the coordinates of the intersection of the graphs of these lines. This is an approximate solution of the equations.
Other methods include elimination, substitution, and Cramer's rule.
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Figure 2.2.1 Graph of the lines and
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The Method of Elimination
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Solve each equation for the same variable, and equate results. For example, solving each equation for , and equating results, leads to the new equation
Multiplying this equation by both 7 and leads to
and then to
From this, so
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The Method of Substitution
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Solve one equation of one of its variables, and substitute into the other equation. For example, solving the first equation for leads to
Substitution of this into the second equation gives
from which it follows that
and
Hence, , and
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Cramer's Rule
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Table 2.2.1 Cramer's rule
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Cramer's rule gives the solution of the equations
as the ratios
where , and are the determinants, respectively, of the matrices
, ,
Since the determinant of the matrix
is
the solution of the given equations, using Cramer's rule, requires computing the determinants of the matrices
, ,
The determinants are
= 62
=
= 26
Hence, the solution of the given equations is
=
=
These calculations are summarized in Tables 2.2.1 and 2.2.2.
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Table 2.2.1 Cramer's rule
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Table 2.2.2 Cramer's rule applied to Problem 2.2
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2.2 - Maplet Solution
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Given the lines
the coordinates of the point of intersection, namely, , can be obtained with
Line Tutor #3
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Clicking this link will launch the tutor with the solution embedded as shown in Figure 2.2.2.
To use this tutor, write the given lines in the form , enter just the right-hand sides of these equations, and then press the Intersection button to obtain the desired result.
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Figure 2.2.2 Thumbnail image of Line Tutor #3
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To launch Line Tutor #3, click the following link:
Line Tutor #3
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2.2 - Interactive Solution
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Form a sequence of the two equations.
Context Panel: Solve≻Solve
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Obtain Figure 2.2.1
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Form a sequence of the two equations.
Context Panel: Plots≻Plot Builder
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2.2 - Programmatic Solution
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Enter the two equations.
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Apply the solve command.
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Obtain Figure 2.2.1.
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Problem 2.3
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2.3 - Mathematical Solution
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The distance between the parallel lines whose equations are and can be found as the length of a transversal perpendicular to both lines. This transversal can be constructed at any point on one of the lines by selecting an arbitrary point on the first line, finding the equation of the perpendicular through this point, then finding the intersection of this transversal with the second line. The length of the segment between the points on the two lines is then the distance between the parallel lines. See Figure 2.3.1.
Suppose the point is selected on the first line whose slope is . The equation of the perpendicular through is
=
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Figure 2.3.1 Graph of parallel lines cut by a perpendicular transversal
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The intersection of this line with the second given line is found by the method of substitution, which yields
= 10
or
so
and
=
The length of the segment connecting the points and is then
= =
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2.3 - Maplet Solution
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To find the distance between the two parallel lines whose equations are
and
use
Line Tutor #3
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Clicking this link will launch the tutor with the solution embedded as shown in Figure 2.3.2.
To use this tutor, write the given lines in the form , enter just the right-hand sides of these equations, using * for multiplication, and then press the Separation button to obtain the desired result, namely, , in both exact and floating-point form.
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Figure 2.3.2 Thumbnail image of Line Tutor #3
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Note that if the Intersection button is pressed, the message returned indicates that the lines are parallel, and hence, do not intersect.
To launch Line Tutor #3, click the following link:
Line Tutor #3
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2.3 - Interactive Solution
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Enter the expression for the first line.
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Context Panel: Solve≻Isolate Expression for≻
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At the point , obtain the equation of the line perpendicular to the first line.
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Implement the point-slope equation , where is the negative reciprocal of the slope of the first line.
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Type the equation of the second line, type a comma, and reference the equation of the transversal by its equation label (Control L) to form a sequence of two equations.
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Context Panel: Solve≻Solve
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Obtain the length of the line segment between the point and the intersection of the transversal with the second line.
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Tools≻Tasks≻Browse: Algebra≻Distance between Two Points
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Distance between Two Real Points
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Enter points as lists:
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Point 1:
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Point 2:
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2.3 - Programmatic Solution
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Be sure that the "Initialize" button in the Initialization section has been pressed. Then, enter the equations of the given lines.
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Obtain the common slope of these lines by writing the first line in the form .
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The coefficient of is the slope, .
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At the point on the first line, obtain the equation of the line orthogonal to any line with slope .
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Obtain the intersection of the second line with the transversal just constructed.
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Write the intersection as a "point."
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Compute the distance between on the first line with the point at the intersection of the transversal and the second line.
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Exercises - Chapter 2
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2.1. In the Cartesian plane, the two points and determine a line segment. For this segment, find
(a) the length;
(b) the midpoint;
(c) the slope;
(d) the equation;
(e) the equation of the perpendicular bisector.
2.2. Calculate the coordinates of the point of intersection of the lines .
2.3. Find the distance between the parallel lines whose equations are and .
2.4. Find the distance between the parallel lines whose equations are and .
2.5. On the line whose equation is , find the coordinates of the points that are 9 units from the point whose -coordinate is 1. Hint: Solve the equation for .
2.6. Find the equations of the lines parallel to, and 5 units from the line whose equation is . Hint: At any point on the given line, find the equation of a perpendicular; then use Exercise 2.5.
2.7. In the Cartesian plane, the three points , and are the vertices of a triangle. Find the coordinates of the intersection of the medians of this triangle. Hint: A median in a triangle is a line connecting a vertex to the midpoint of the side opposite that vertex. A theorem in plane geometry states that the medians of a triangle intersect in a single point called the centroid of the triangle.
2.8. Find the equation of the line parallel to, and midway between, the parallel lines whose equations are and .
2.9. Find the coordinates of the points that trisect the line segment whose endpoints are and .
2.10. Find the equations of the lines parallel to, and one-third and two-thirds the distance between the parallel lines whose equations are and .
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