P.2

coordinates, lines and increment

Rectangular Coordinate System

The horizontal line is called the x-axis.

The vertical line is called the y-axis.

The point of intersection is the origin.

x-axis

y-axis

origin

Quadrant IQuadrant II

Quadrant III

Quadrant IV

The four regions in the x-y plane are known as quadrants, labeled as follows:

x

yQuadrant I

x > 0, y > 0

Quadrant IV

x > 0, y < 0

Quadrant III

x < 0, y < 0

Quadrant II

x < 0, y > 0

Example : Plotting Points

Plot the point (3,2). Start at the origin and move 3 units to the right. From that point,move 2 units up. Now plot your point

Plotting Points Each point in the xy-

plane corresponds to a unique rdered pair (a, b).

Plot the point (2, 4). Move 2 units right Move 4 units up

2 units

4 units

Increment and distance

When a particle moves from one points in the plane to another, the net changes in its coordinates are called increments.

1 2x x x

An increment in a variable is a net change in the that variable. If x changes from x1 to x2, the increment in x is

Definition

Theorem: Distance FormulaTheorem: Distance Formula

The distance between two points

P x y1 1 1 , and P x y

2 2 2 , , denoted

by d P P1 2, is

2 2 2 2

1 2 2 1 2 1,d P P x y x x y y

Example: Find the distance between the points (3,8) and (-1,2)

P P1 23 8 1 2 , , ,

d P P x x y y1 2 2 1

2

2 1

2,

d P P1 2

2 21 3 2 8,

d P P1 2

2 24 6,

d P P1 2 16 36, 52 2 13

Midpoint Formula The midpoint of a line segment

with endpoints (x1, y1) and (x2, y2) is

1 2 2 1, .2 2

x x y y

Midpoint Formula

Find the midpoint M of the segment with endpoints (10, 5) and (6, 4).

10 ( 6) ( 5 4) 1, 2,

2 2 2

Finding Ordered Pairs that are Solutions of Equations

For the following equation find three ordered pairs that are solutions Of the equation y = 5x+2 Let y = 3

3 = 5x + 25 = 5x1 = x (1,3)

Let x = 0y = 5(0) + 2y = 2(0, 2)

Let x = 1y = 5(1) + 2y = 7(1, 7)

Definitions: X-Intercept The x-intercept is a point on any

graph where the graph touches the x-axis.

The y-coordinate of the x-intercept is always zero.

The x-intercept is denoted by the point (x ,0), where x is any real number.

The x-intercept is also known as the zero or root of an equation.

Definitions: Y-Intercept The y-intercept is a point on

any graph where the graph touches the y-axis.

The x-coordinate of the y-intercept is always zero.

The y-intercept is denoted by the point (0, y), where y is any real number.

Procedure for Finding Intercepts

1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x.

2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y.

Example: Find the x- and y-intercepts of the graph of y = x2 + 4x – 5.

To find the x-intercepts, let y = 0 and solve for x.0 = x2 + 4x – 5 Substitute 0 for y.

0 = (x – 1)(x + 5) Factor.

x – 1 = 0 x + 5 = 0

x = 1 x = –5 Solve for x.

So, the x-intercepts are (1, 0) and (–5, 0).

Example: Find the x- and y-intercepts of the graph of y = x2 + 4x – 5.

To find the y-intercept, let x=0 and solve for y.

y = 02 + 4(0) – 5 = –5

So, the y-intercept is (0, –5).

Graph of the Functions

The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates (x,y) satisfy the equation.

Graphing The graph of an equation is found

by plotting points that are solutions of the equation.

The intercepts of the graph are good points to find first.

x-intercept is an x-value where the graph intersects the x-axis. y = 0

y-intercept is a y-value where the graph intersects the y-axis. x = 0

Graphing an Equation by Point Plotting Step 1 Find the intercepts. Step 2 Find as many additional

ordered pairs as needed. Step 3 Plot the ordered pairs

from Steps 1 and 2. Step 4 Connect the points from

Step 3 in a smooth line or curve.

Example: Sketch the graph of y = –2x + 3.

x y (x, y)

–2 7 (–2, 7)

–1 5 (–1, 5)

0 3 (0, 3)

1 1 (1, 1)

3/2 0 (3/2, 0)

Step 1: Find the intercepts.Step 2: Find as many additional ordered pairs as needed.

Example: Sketch the graph of y = –2x + 3.

3. Plot the points in the coordinate plane.

4 8

4

8

4

–4

x

y

x y (x, y)

–2 7 (–2, 7)

–1 5 (–1, 5)

0 3 (0, 3)

1 1 (1, 1)

3/2 0 (3/2, 0)

4. Connect the points with a straight line.

4 8

4

8

4

–4

x

y

Example: Sketch the graph of y = –2x + 3.

Example 2: Graphing Intercepts

Graph 4y + 5x = 20.Substitute zero for x: 4y = 20 or y =

5.Hence, the y-intercept is (0,5).Substitute zero for the y: 5x = 20 or

x = 4.Hence, the x-intercept is (4,0).

Example

Graph the equation y = 5x + 2

31

0-2/5

20

yx

Example: Sketch the graph of y = x 2.

x y (x, y)

–2 4 (–2, 4)

–1 1 (–1, 1)

0 0 (0, 0)

1 1 (1, 1)

2 4 (2, 4)

3 9 (3, 9)

4 16 (4, 16)

y

x2 4

2

6

8

–2

Example: Sketch the graph of y = | x | .

x y (x, y)

–2 2 (–2, 2)

–1 1 (–1, 1)

0 0 (0, 0)

1 1 (1, 2)

2 2 (2, 2)

y

x–2 2

2

4

Example: Is (3,5) on the graph of ?42 xy

Substitute x = 3 and y = 5 into the equation:

?435 2

495 True!

Therefore, (3,5) is on the graph of the equation.

Definition:

The standard form of an equation of a circle with radius r and center (h, k) is

x h y k r 2 2 2

Definition: A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point (h, k). The fixed distance r is called the radius, and the fixed point (h, k) is called the center of the circle.

x

y

(h, k)

r(x, y)

Standard form of an equation of a circle

where the center of the circle is at the origin (0,0) and with a radius of r.

2 2 2rx y

Example Graph x2 + y2 = 16

3

3

04

40

yx

7

7

Unit Circle equation

where the center of the circle is atthe origin (0,0) and with a radius of1, is called the unit circle.

2 2 1x y

Since the radius = 1, use the center (0,0) as a reference point and then move 1 point to the left, right, up and down.

Continued.

Continued.

Equation of Straight lines

Straight linesDefinition:

1 1 2 2

2 1

2 1

2 1

The of the line through the

distinct points ( , ) and ( , )

Change in Riseis

Change in Run

where

o e

0

sl p

.

x y x y

y y y y

x x x x

x x

2 1

2 1

yThe Slope of the line = tan = .

x

yy

x x

Possibilities for a Line’s Slope

Positive Slope

0m

Line rises from left to right.

Possibilities for a Line’s Slope

Negative Slope

0m

Line falls from left to right.

Possibilities for a Line’s Slope

Zero Slope

0m

Line is horizontal.

Possibilities for a Line’s Slope

Undefined Slope is

undefined.

m

Line is vertical.

Example: Find the slope of the line passing

through the pair of points (2,1) and (3, 4).

1 1 2 2Let ( , ) (2,1) and

Solu

( , ) (3,4).

tion

x y x y

2 1

2 1

Slopey y

mx x

4 1

3 2

3

1 3.

Example: Find the Slope

Find the slope of the line passing through

the pair of points ( 1,3) and (2,4) or state

that the slope is undefined.

Solution

1 1 2 2Let ( , ) ( 1,3) and ( , ) (2,4).x y x y

2 1

2 1

Slopey y

mx x

4 3

2 ( 1)

1

3

The slope is and

the line from l

positi

eft to

ve,

r riises ght.

Practice Exercise

Find the slope of the line passing through

the points (4, 1) and (3, 1) or state

that the slope is undefined.

Answer

The slope is zero.

Thus, the line is a horizontal line.

m

Equation of a Horizontal Line

A horizontal line

is given by an

equation of the

form

where is the

-intercept.

b

b

y

y

Y-interceptis 40m

The graph of 4y

Equation of a Vertical Line

A vertical line is

given by an

equation of the

form

where is the

-intercept.

a

x

x a

X-intercept is -5

Slope is

undefined

The graph of -5x

Example: Draw the graph of the equation x = 2.

y

x

x = 2

Example : Graphing a Horizontal Line

Graph 5 in the

rectangular coordinate system.

y

Y-intercept is 5.

Practice Exercises

Graph each equation in the rectangular

coordinate system.

1. 4

2. 0

y

x

Answers to Practice Exercises

2..1

The Equation of the line

1 2 1

1 2 1

.y y y y

mx x x x

1- With two points P1 (x1 , y1 ) and P2 (x2 , y2 )

1 1 1 2 2 2

The equation of the line passing throuhg the points

P (x , y ) and P (x , y ) is

Example

Write the point-slope of the equation

of the line passing throuhg the points

(3,5) and (8,15). Then solve the

equation for y.

Solution

1 2 1

1 2 1

5 15 5 102

3 8 3 5

y y y y

x x x x

y

x

( 5) 2( 3)y x Then solve for gives:

2 1

y

y x

The Equation of the line

1 1

The point-slope equation of a nonvertical

line of slope that passes through the

point ( , ) is

m

x y

1 1( .)y y m x x

2- Point-slope Form of the Equation of a Line

Example : Writing the Point-Slope Equation of a Line

1 1

We use the point-slope equation of a line

with

Solut

4, 1, and 3.

ion

m x y

Write the point-slope form of the equation

of the line passing through (1,3) with a slope

of 4. Then solve the equation for .y

1 14, 1, and 3m x y

1 1( )y y m x x 3 4( 1)y x 3 4 4y x

4 1y x

Practice Exercises

1. Write the point-slope form of the equation

of the line passing through (4,-1) with a slope

of 8. Then solve the equation for .y

2. Write the point-slope form of the equation

of the line passing through the points ( 2,0)

and (0,2). Then solve the equation for .y

Answers to Practice Exercises

1. 8 33

2. 2

y x

y x

3. The Slope-Intercept Form of the Equation of a Line

The slope-intercept

equation of a

nonvertical line

with slope and

-intercept is

m

y b

y mx b

(0, )b

y

x

Y-intercept is b

Slope is m

A line with slope

and -intercept .

m

y b

Example : Graphing by Using the Slope and y-Intercept

Give the slope and the -intercept of the

line 3 2. Then graph the line.

y

y x

Solution 3 2y x

The slope

is 3

The -intercept

is 2.

y

The graph of 3 2.y x

First use the -intercept 2, to

plot the point (0,2). Starting

at (0,2), move 3 units up and

1 unit to the right. This gives

us the second point of the line.

Use a straightedge to draw a

line through the tw

y

o points.

Text Example

Graph the line whose equation is y = 2/3 x + 2.

Solution:

y = 2/3 x + 2 The slope is

2/3.

The y-intercept is 2.

Text Example cont.Graph the line whose equation is y = 2/3x + 2.

We plot the second point on the line by starting at (0, 2), the first point. Then move 2 units up (the rise) and 3 units to the right (the run). This gives us a second point at (3, 4).

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

-5

Practice Exercises

Give the slope and -intercept

of each line whose equation is

given. Then graph the line.

y

1. 3 2

32. 3

4

y x

y x

Answers to Practice Exercises

1. 3, 2m b 32. , 3

4m b

General Form of the Equation of a Line

0

Every line has an equation that can

be written in the general form

where, , , and are three

real numbers, and and

are not both zero.

A B C

A B

Ax By C

Equations of Lines

1 2 1

1 2 1

1 1

1. Two points :

2. Point-slope form:

3. Slope-intercept form:

4. Horizontal line:

5. r

(

e

)

V t

y y y y

x x x x

y y m x x

y mx b

y b

ical line:

6. General form: 0

x a

Ax By C

Example : Finding the Slope and the y-Intercept

Find the slope and the -intercept of the

line whose equation is 4 6 12 0.

y

x y

SolutionFirst rewrite the equation in slope-intercept

form . We need to solve for .y mx b y

4 6 12 0x y

6 4 12y x 4 12

6 6y x

22

3y x

23

The coefficient of ,

, is the slope and

the constant term, 2,

is the -intercept.

x

y

23 , 2.m b

Practice Exercises

a. Rewrite the given equation in

slope-intercept form.

b. Give the slope and y-intercept.

c. Graph the equation.

1. 6 5 20 0

2. 4 28 0

x y

y

Answers to Practice Exercises

651. 4

6slope

5-intercept 4.

y x

m

y b

Answers to Practice Exercises

2. 7

slope 0

-intercept 7.

y

m

y b

Definitions: Parallel Lines

Two lines are said to be parallel if they do not have any points in common.

Two distinct non-vertical lines are parallel if and only if they have the same slope and have different y-intercepts.

Definitions: Perpendicular Lines

Two lines are said to be perpendicular if they intersect at a right angle.

Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

121 mm1

2

1

mm

Parallel and PerpendicularLines

Example the following lines are perpendicular

53

123

xyandxy

Text Example

Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

Text Example cont.

y – y1 = m(x – x1)

y1 = 2 x1 = -3

Solution . Notice that the line passes through the point (-3, 2). Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

-5

(-3, 2)

Rise = 2

Run = 1

y = 2x + 1

Text Example cont.

Solution Parallel lines have the same slope. Because the slope of the given line is 2, m = 2 for the new equation.

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

-5

(-3, 2)

Rise = 2

Run = 1

y = 2x + 1

y – y1 = m(x – x1)

y1 = 2 m = 2 x1 = -3

Text Example cont.Solution The point-slope form of the line’s equation is y – 2 = 2[x – (-3)]

y – 2 = 2(x + 3)

Solving for y, we obtain the slope-intercept form of the equation.

y – 2 = 2x + 6

y = 2x + 8

Text Example

Write an equation of the line passing through (-3, 2) and perpendicular to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

Text Example cont.

y – y1 = m(x – x1)

y1 = 2 x1 = -3

Solution Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.

Text Example cont.

Solution perpendicular lines have the product of their slopes is -1. Because the slope of the given line is 2, m = -1/2 for the new equation.

y – y1 = m(x – x1)

y1 = 2 m = -1/2 x1 = -3

Text Example cont.Solution The point-slope form of the line’s equation is y – 2 = -1/2[x – (-3)]

y – 2 = -1/2(x + 3)

Solving for y, we obtain the slope-intercept form of the equation.

y – 2 = -1/2x -3/2

y = -1/2x + 1/2

Find the equation of the line parallel to y = -3x + 5 passing through (1,5).

Since parallel lines have the same slope, the slope of the parallel line is m = -3.

y y m x x 1 1

y x 5 3 1

y x 5 3 3y x 3 8

Example: Find the equation of the line perpendicular to y = -3x + 5 passing through (1,5).

Slope of perpendicular line:

13

13

y y m x x 1 1

y x 513

1

y x 513

13

y x 13

143