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Precalculus Study Guide

Chapter 2: Graphing Lines in the Cartesian Plane

Introduction

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	Equation	Details
Two-point	$\frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$	where the two points are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
Point-slope	$y - y_{1} = m (x - x_{1})$	where $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is the slope of the line
Slope-intercept	$y = m x + b$	where $m$ is the slope, and $b$ , the $y$ -intercept
General	$A x + B y = C$
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.

Chapter Glossary

The following terms in Chapter 2 are linked to the Maple Math Dictionary.

analytic geometry

bisector

Cartesian plane

coordinate

Cramer's Rule

Euclidean geometry

intercept

length

line

line segment

matrix

midpoint

negative

parallel

perpendicular

reciprocal

slope

transversal

Typical Problems

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 $3 x + 4 y = 10$ .

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 2.1

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 $d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$ Taking $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 7)$ , the distance between the endpoints, namely, $d = \sqrt{{(1 - 4)}^{2} + {(2 - 7)}^{2}}$ = $\sqrt{{(- 3)}^{2} + {(- 5)}^{2}} = \sqrt{9 + 25}$ = $\sqrt{34}$ 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 $(x_{m}, y_{m}) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ Again taking $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 7)$ , the midpoint will be $(x_{m}, y_{m}) = (\frac{1 + 4}{2}, \frac{2 + 7}{2})$ = $(\frac{5}{2}, \frac{9}{2})$

	2.1 (c) - Slope of Line Segment
	The slope of the line segment joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ Hence, the slope of the segment joining the points $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 7)$ is $m = \frac{7 - 2}{4 - 1}$ = $\frac{5}{3}$

2.1 (d) - Equation of Line

The equation of the line through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found from the two-point form of the straight line, namely,

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $(x - x_{1})$

This is equivalent to the point-slope form of the line, namely,

$y - y_{1} = m (x - x_{1})$

or even

$y = m (x - x_{1}) + y_{1}$

Since the slope $m = \frac{5}{3}$ was calculated in Part (c), the required equation of the line joining the points $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 7)$ is then

$y = \frac{5}{3}$ $(x - 1) + 2$

$y = \frac{5}{3}$ $x - \frac{5}{3} + 2$ = $\frac{5}{3}$ $x + \frac{1}{3}$

2.1 (e) - Equation of Perpendicular Bisector

The equation of the perpendicular bisector of the line segment joining the points $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 7)$ 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

$(x_{m}, y_{m}) = (\frac{5}{2}, \frac{9}{2})$

The slope of this segment was found in Part (c) to be $m = \frac{5}{3}$ . Therefore, the slope of the line perpendicular to this segment is the negative reciprocal of $\frac{5}{3}$ , namely, $- \frac{3}{5}$ . Hence, the desired equation is

$y = - \frac{3}{5}$ $(x - \frac{5}{2}) + \frac{9}{2}$

$y = - \frac{3}{5}$ $x + \frac{3}{2} + \frac{9}{2}$ = $- \frac{3}{5}$ $x + 6$

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.

Form	Equation
Slope-intercept	$y = \frac{5}{3}$ $x + \frac{1}{3}$
General	$- 5 x + 3 y = 1$
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: Lines Tutor .

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: $(x_{1}, y_{1})$ and B: $(x_{2}, y_{2})$ . 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

$[1, 2]$ $\overset{assign to a name}{\to}$

•	Enter the first point as a Maple list.

•	Context Panel: Assign to a Name≻B

$[4, 7]$ $\overset{assign to a name}{\to}$

Part (a) - Length of Line Segment

•	Form the differences $[x_{2} - x_{1}, y_{2} - y_{1}]$ by writing $B - A$

•	Context Panel: Assign to a Name≻ $d$

$B - A$ $\overset{assign to a name}{\to}$

•	Compute $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ by writing $\sqrt{d_{1}^{2} + d_{2}^{2}}$ .

•	Press the Enter key

$\sqrt{d_{1}^{2} + d_{2}^{2}}$

Part (b) - Midpoint of Line Segment

•	Obtain the midpoint $[\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}]$ by writing $\frac{(A + B)}{2}$ .

•	Context Panel: Assign to a Name≻ $MP$

$\frac{(A + B)}{2}$ $\overset{assign to a name}{\to}$

Part (c) - Slope of Line Segment

•	Obtain the slope $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ by writing $\frac{d_{2}}{d_{1}}$

•	Context Panel: Assign to a Name≻ $m$

$\frac{d_{2}}{d_{1}}$ $\overset{assign to a name}{\to}$

Part (d) - Equation of Line

•	By referencing $x_{1}$ as $A_{1}$ , and $y_{1}$ as $A_{2}$ , implement the point-slope form of the line

$y = m (x - x_{1}) + y_{1}$

$y = m (x - A_{1}) + A_{2}$

Part (e) - Equation of Perpendicular Bisector

•	By referencing the $x$ - and $y$ -coordinates of the midpoint as ${MP}_{1}$ and ${MP}_{2}$ , respectively, implement the point-slope form of the line

$y = m_{⊥} (x - x_{1}) + y_{1}$

where $m_{⊥}$ is the slope of the line orthogonal to the line in Part (d).

$y = - \frac{1}{m} (x - {MP}_{1}) + {MP}_{2}$

2.1 - Programmatic Solution

•	Be sure to have initialized the worksheet by pressing the Initialize button in the Initialization section.

•	Enter the data.

>	$A, B ≔ [1, 2], [4, 7]$

(a) Length of the line segment $AB$

>	$Distance (A, B)$

(b) Midpoint of the segment $AB$

>	$MP ≔ Midpoint (A, B)$

>	$m ≔ Slope (A, B)$

(d) Equation of line through $A$ and $B$

>	$Line (A, B) [1]$

(e) Equation of perpendicular bisector of segment $AB$

>	$Line (- \frac{1}{m}, MP) [1]$

Problem 2.2

2.2 - Mathematical Solution

Several methods are available for calculating the coordinates of the point of intersection of the lines

$2 x - 3 y = 5, 4 x + 7 y = 9$

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

$\frac{5 - 2 x}{- 3} = \frac{9 - 4 x}{7}$

Multiplying this equation by both 7 and $- 3$ leads to

$35 - 14 x = 12 x - 27$

and then to

$62 = 26 x$

From this, $x = \frac{62}{26} = \frac{31}{13}$ so $y = \frac{5 - 2 (\frac{31}{13})}{- 3} = \frac{65 - 62}{- 39} = - \frac{1}{13}$

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

$y = \frac{5 - 2 x}{- 3}$

Substitution of this into the second equation gives

$4 x + 7 (\frac{5 - 2 x}{- 3}) = 9$

from which it follows that

$- 12 x + 35 - 14 x = - 27$

and

$62 = 26 x$

Hence, $x = \frac{62}{26} = \frac{31}{13}$ , and $y = \frac{5 - 2 (\frac{31}{13})}{- 3} = \frac{65 - 62}{- 39} = - \frac{1}{13}$

Cramer's Rule

$\begin{array}{c} a_{11} x + a_{12} y = c_{1} \\ a_{21} x + a_{22} y = c_{2} \end{array}} \Rightarrow$	$x = \frac{\|\begin{array}{c} c_{1} & a_{12} \\ c_{2} & a_{22} \end{array}\|}{\|\begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\|}, y = \frac{\|\begin{array}{c} a_{11} & c_{1} \\ a_{21} & c_{2} \end{array}\|}{\|\begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\|}$
Table 2.2.1 Cramer's rule

Cramer's rule gives the solution of the equations

$a_{11} x + a_{12} y = c_{1}$

$a_{21} x + a_{22} y = c_{2}$

as the ratios

$x = \frac{m_{x}}{d}$

$y = \frac{m_{y}}{d}$

where $m_{x}, m_{y}$ , and $d$ are the determinants, respectively, of the matrices

$M_{x} = [\begin{array}{c} c_{1} & a_{12} \\ c_{2} & a_{22} \end{array}]$ , $M_{y} = [\begin{array}{c} a_{11} & c_{1} \\ a_{21} & c_{2} \end{array}]$ , $D = [\begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}]$

Since the determinant of the matrix

$[\begin{array}{c} P & Q \\ R & S \end{array}]$

$P S - Q R$

the solution of the given equations, using Cramer's rule, requires computing the determinants of the matrices

$M_{x} = [\begin{array}{c} 5 & - 3 \\ 9 & 7 \end{array}]$ , $M_{y} = [\begin{array}{c} 2 & 5 \\ 4 & 9 \end{array}]$ , $D = [\begin{array}{c} 2 & - 3 \\ 4 & 7 \end{array}]$

The determinants are

$m_{x} = 35 - (- 3) \cdot 9$ = 62

$m_{y} = 18 - 20$ = $- 2$

$d = 14 - (- 3) \cdot 4$ = 26

Hence, the solution of the given equations is

$x = \frac{62}{26}$ = $\frac{31}{13}$

$y = - \frac{2}{26}$ = $- \frac{1}{13}$

These calculations are summarized in Tables 2.2.1 and 2.2.2.

$\begin{array}{c} a_{11} x + a_{12} y = c_{1} \\ a_{21} x + a_{22} y = c_{2} \end{array}} \Rightarrow$	$x = \frac{\|\begin{array}{c} c_{1} & a_{12} \\ c_{2} & a_{22} \end{array}\|}{\|\begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\|}, y = \frac{\|\begin{array}{c} a_{11} & c_{1} \\ a_{21} & c_{2} \end{array}\|}{\|\begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\|}$
Table 2.2.1 Cramer's rule

$x = \frac{\|\begin{array}{c} 5 & - 3 \\ 9 & 7 \end{array}\|}{\| \begin{array}{c} 2 & - 3 \\ 4 & 7 \end{array} \|} = \frac{62}{26} = \frac{31}{13}$	$y = \frac{\|\begin{array}{c} 2 & 5 \\ 4 & 9 \end{array}\|}{\| \begin{array}{c} 2 & - 3 \\ 4 & 7 \end{array} \|} = \frac{- 2}{26} = - \frac{1}{13}$
Table 2.2.2 Cramer's rule applied to Problem 2.2

2.2 - Maplet Solution

Given the lines

$2 x - 3 y = 5$

$4 x + 7 y = 9$

the coordinates of the point of intersection, namely, $(\frac{31}{13}, - \frac{1}{13})$ , can be obtained with Line Tutor #3 .

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 $y = ...$ , enter just the right-hand sides of these equations, and then press the Intersection button to obtain the desired result.

Figure 2.2.2 Thumbnail image of Line Tutor #3

To launch Line Tutor #3, click the following link: Line Tutor #3 .

2.2 - Interactive Solution

Form a sequence of the two equations.

Context Panel: Solve≻Solve

Obtain Figure 2.2.1

Form a sequence of the two equations.

Context Panel: Plots≻Plot Builder

2.2 - Programmatic Solution

Enter the two equations.

>	$E_{1} ≔ 2 x - 3 y = 5 &semi; E_{2} ≔ 4 x + 7 y = 9$

Apply the solve command.

>	$solve (\{E_{1}, E_{2}\}, \{x, y\})$

Obtain Figure 2.2.1.

>	$implicitplot ([E_{1}, E_{2}], x = 0 .. 5, y = - 5 .. 5)$

Problem 2.3

2.3 - Mathematical Solution

The distance between the parallel lines whose equations are $3 x + 4 y = 7$ and $3 x + 4 y = 10$ 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 $(1, 1)$ is selected on the first line whose slope is $m = - \frac{3}{4}$ . The equation of the perpendicular through $(1, 1)$ is

$y = \frac{4}{3}$ $(x - 1) + 1$ = $\frac{4}{3}$ $x - \frac{1}{3}$

Figure 2.3.1 Graph of parallel lines cut by a perpendicular transversal

The intersection of this line with the second given line is found by the method of substitution, which yields

$3 x + \frac{4 (4 x - 1)}{3}$ = 10

$\frac{25}{3}$ $x = \frac{34}{3}$

$x = \frac{34}{25}$

and

$y = \frac{4}{3}$ $\frac{34}{25} - \frac{1}{3}$ = $\frac{37}{25}$

The length of the segment connecting the points $(1, 1)$ and $(\frac{34}{25}, \frac{37}{25})$ is then

$\sqrt{{(1 - \frac{34}{25})}^{2} + {(1 - \frac{37}{25})}^{2}}$ = $\frac{\sqrt{81 + 144}}{25} = \frac{\sqrt{225}}{25}$ = $\frac{15}{25} = \frac{3}{5}$

2.3 - Maplet Solution

To find the distance between the two parallel lines whose equations are

$3 x + 4 y = 7$ and $3 x + 4 y = 10$

use Line Tutor #3 .

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 $y = ...$ , enter just the right-hand sides of these equations, using * for multiplication, and then press the Separation button to obtain the desired result, namely, $\frac{3}{5}$ , in both exact and floating-point form.

Figure 2.3.2 Thumbnail image of Line Tutor #3

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

2.3 - Interactive Solution

•	Enter the expression for the first line.

•	Context Panel: Solve≻Isolate Expression for≻ $y$

At the point $(x_{1}, y_{1}) = (1, 1)$ , obtain the equation of the line perpendicular to the first line.

•	Implement the point-slope equation $y = m (x - x_{1}) + y_{1}$ , where $m$ is the negative reciprocal of the slope of the first line.

•	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.

•	Context Panel: Solve≻Solve

Obtain the length of the line segment between the point $(1, 1)$ and the intersection of the transversal with the second line.

Tools≻Tasks≻Browse: Algebra≻Distance between Two Points

Distance between Two Real Points

Enter points as lists:

Point 1:

Point 2:

2.3 - Programmatic Solution

Be sure that the "Initialize" button in the Initialization section has been pressed. Then, enter the equations of the given lines.

>	$L_{1} ≔ 3 x + 4 y = 7 &semi; L_{2} ≔ 3 x + 4 y = 10$

Obtain the common slope of these lines by writing the first line in the form $y = m x + b$ .

>	$Y_{1} ≔ solve (L_{1}, y)$

The coefficient of $x$ is the slope, $m$ .

>	$m ≔ coeff (Y_{1}, x)$

At the point $(1, 1)$ on the first line, obtain the equation of the line orthogonal to any line with slope $m$ .

>	$L_{3} ≔ Line (- \frac{1}{m}, [1, 1]) [1]$

Obtain the intersection of the second line with the transversal just constructed.

>	$Q ≔ solve (\{L_{2}, L_{3}\})$

Write the intersection as a "point."

>	$P ≔ eval ([x, y], Q)$

Compute the distance between $(1, 1)$ on the first line with the point $P$ at the intersection of the transversal and the second line.

>	$Distance (P, [1, 1])$

Exercises - Chapter 2

2.1. In the Cartesian plane, the two points $(6, - 5)$ and $(- 3, 8)$ determine a line segment. For this segment, find

(a) the length;
(b) the midpoint;

(d) the equation;

(e) the equation of the perpendicular bisector.

2.2. Calculate the coordinates of the point of intersection of the lines $6 x + 11 y = 13, 5 x - 9 y = 1$ .

2.3. Find the distance between the parallel lines whose equations are $7 x - 11 y = 2$ and $7 x - 11 y = 12$ .

2.4. Find the distance between the parallel lines whose equations are $2 x + 3 y = 5$ and $6 y = 21 - 4 x$ .

2.5. On the line whose equation is $5 x + 7 y = 11$ , find the coordinates of the points that are 9 units from the point whose $x$ -coordinate is 1. Hint: Solve the equation $9 = \sqrt{{(a - 1)}^{2} + {(y (a) - y (1))}^{2}}$ for $a$ .

2.6. Find the equations of the lines parallel to, and 5 units from the line whose equation is $5 x - 3 y = 4$ . 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 $(1, 2), (7, 5)$ , and $(3, 10)$ 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 $x + y = 1$ and $x + y = 9$ .

2.9. Find the coordinates of the points that trisect the line segment whose endpoints are $(3, - 4)$ and $(13, 12)$ .

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 $x + y = 1$ and $x + y = 9$ .

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