Precalculus Study Guide
Copyright Maplesoft, a division of Waterloo Maple Inc., 2021
Chapter 2: Graphing Lines in the Cartesian Plane
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.
where the two points are ⁡x1,y1 and ⁡x2,y2
where m=y2−y1x2−x1 is the slope of the line
where m is the slope, and b, the y-intercept
Table 2.0.1 Formulas for the equation of a straight line
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.
The following terms in Chapter 2 are linked to the Maple Math Dictionary.
2.1. In the Cartesian plane, the two points ⁡1,2 and ⁡4,7 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⁢x−3⁢y=5,4⁢x+7⁢y=9.
2.3. Find the distance between the parallel lines whose equations are 3⁢x+4⁢y=7 and 3⁢x+4⁢y=10.
Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.
2.1 - Mathematical Solution
In the Cartesian plane, the two points ⁡1,2 and ⁡4,7 determine a line segment.
2.1 (a) - Length of Line Segment
The length of this line segment is computed with the distance formula
Taking ⁡x1,y1=⁡1,2 and ⁡x2,y2=⁡4,7, the distance between the endpoints, namely,
d=1−42+2−72 = −32+−52=9+25 = 34
is the length of the line segment.
2.1 (b) - Midpoint of Line Segment
The midpoint of the segment is the point whose coordinates are
Again taking ⁡x1,y1=⁡1,2 and ⁡x2,y2=⁡4,7, the midpoint will be
⁡xm,ym=⁡1+42,2+72 = ⁡52,92
2.1 (c) - Slope of Line Segment
The slope of the line segment joining the points ⁡x1,y1 and ⁡x2,y2 is given by
Hence, the slope of the segment joining the points ⁡x1,y1=⁡1,2 and ⁡x2,y2=⁡4,7 is
m=7−24−1 = 53
2.1 (d) - Equation of Line
The equation of the line through the points ⁡x1,y1 and ⁡x2,y2 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,
Since the slope m=53 was calculated in Part (c), the required equation of the line joining the points ⁡x1,y1=⁡1,2 and ⁡x2,y2=⁡4,7 is then
y=53 x−53+2 = 53 x+13
2.1 (e) - Equation of Perpendicular Bisector
The equation of the perpendicular bisector of the line segment joining the points ⁡x1,y1=⁡1,2 and ⁡x2,y2=⁡4,7 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 m=53 . Therefore, the slope of the line perpendicular to this segment is the negative reciprocal of 53, namely, −35. Hence, the desired equation is
y=−35 x+32+92 = −35 x+6
2.1 - Maplet Solution
The equation of the line through the two points ⁡1,2 and ⁡4,7 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.
Table 2.1.1 Equations provided by Lines Tutor
Figure 2.1.1 Thumbnail image of the Lines Tutor
To launch the Lines Tutor, select "Lines" from the Tools≻Tutors≻Precalculus menu. Alternatively, click the following link:
The length, midpoint, slope, and equation of the perpendicular bisector of this line segment are obtained with
Line Tutor #2
. (Clicking this link will launch the tutor with the solution embedded as shown in Figure 2.1.2.)
The coordinates of the points ⁡1,2 and ⁡4,7 are entered in the data fields for A:⁡x1,y1 and B:⁡x2,y2. 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
Figure 2.1.2 Thumbnail image of the Line Tutor #2
2.1 - Interactive Solution
Enter the given data
Enter the first point as a Maple list.
Context Panel: Assign to a Name≻A
1,2→assign to a name
Context Panel: Assign to a Name≻B
4,7→assign to a name
Part (a) - Length of Line Segment
Form the differences x2−x1,y2−y1
by writing B−A
Context Panel: Assign to a Name≻d
B−A→assign to a name
by writing d12+d22.
Press the Enter key
Part (b) - Midpoint of Line Segment
Obtain the midpoint x1+x22,y1+y22 by writing A+B2.
Context Panel: Assign to a Name≻MP
A+B2→assign to a name
Part (c) - Slope of Line Segment
Obtain the slope y2−y1x2−x1 by writing d2d1
Context Panel: Assign to a Name≻m
d2d1→assign to a name
Part (d) - Equation of Line
By referencing x1 as A1, and y1 as A2, implement the point-slope form of the line
Part (e) - Equation of Perpendicular Bisector
By referencing the x- and y-coordinates of the midpoint as MP1 and MP2, respectively, implement the point-slope form of the line
where m⊥ is the slope of the line orthogonal to the line in Part (d).
y= −1m x−MP1+MP2
2.1 - Programmatic Solution
Be sure to have initialized the worksheet by pressing the Initialize button in the Initialization section.
Enter the data.
(a) Length of the line segment AB
(b) Midpoint of the segment AB
(c) Slope of the segment AB
(d) Equation of line through A and B
(e) Equation of perpendicular bisector of segment AB
2.2 - Mathematical Solution
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.
Figure 2.2.1 Graph of the lines 2⁢x−3⁢y=5 and 4⁢x+7⁢y=9
The Method of Elimination
Solve each equation for the same variable, and equate results. For example, solving each equation for y, and equating results, leads to the new equation
5−2 x−3=9−4 x7
Multiplying this equation by both 7 and −3 leads to
35−14 x=12 x−27
and then to
From this, x=6226=3113 so y=5−2 3113−3=65−62−39= −113
The Method of Substitution
Solve one equation of one of its variables, and substitute into the other equation. For example, solving the first equation for y leads to
Substitution of this into the second equation gives
4 x+7 5−2 x−3=9
from which it follows that
−12 x+35−14 x= −27
Hence, x=6226=3113, and y=5−2 3113−3=65−62−39= −113
a11 x+a12 y=c1a21 x+a22 y=c2}⇒
Table 2.2.1 Cramer's rule
Cramer's rule gives the solution of the equations
as the ratios
where mx,my, and d are the determinants, respectively, of the matrices
Mx=c1a12c2a22, My=a11c1a21c2, D=a11a12a21a22
Since the determinant of the matrix
the solution of the given equations, using Cramer's rule, requires computing the determinants of the matrices
Mx=5−397, My=2549, D=2−347
The determinants are
mx=35−⁡−3⋅9 = 62
my=18−20 = −2
d=14−⁡−3⋅4 = 26
Hence, the solution of the given equations is
x=6226 = 3113