Embed Size (px)
Citation preview
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
