What is wrong with the following “solution” of
5x  5x  1  0
2
5  52  4(5)(1)
 b  b 2  4ac
x
;x 
2a
2(5)
5 5
1
x
; x  5
10
2
1
1
x1   5 and x2   5
2
2
Learning Outcomes
At the end of this chapter, you should be able to:
• appreciate the importance of coordinate geometry
• plot points in a coordinate system
• determine the distance between two points
• find the midpoint of a line segment
• find the slope of a line
• find the equation of a line
• slope-intercept form
• point-slope form
• explain the meaning of slope and y-intercept in the real world context
• enhance your critical thinking and self-management skills
Prior knowledge: Plane, Points, Lines, Linear equations, Solution of Linear Equation
Why do we need to study Coordinate Geometry?
4.1 The Rectangular Coordinate System
Definition 4.1 Coordinate Geometry is a system where the position
of points on the plane is described using an ordered pair of numbers.
Let us recall that a plane is a flat surface that goes on forever in both directions..
If we were to place a point on the plane, coordinate geometry gives us a way to
describe exactly where it is by using two numbers.
Consider this grid:
What is the exact
location of “x”?
x is at (5, C)
The Rectangular Coordinate Plane or the Cartesian Plane
y
5
4
Quadrant II
(-,+)
3
Quadrant I
(+,+)
2
1
-5
-4 -3
-2
-1
1
2
3
4
-1
Quadrant III
(-,-)
-2
-3
Quadrant I V
(+,-)
-4
-5
Figure 4.1: The Coordinate Plane
5
x
The Rectangular Coordinate Plane or the Cartesian Plane
y
Quadrant II
(-,+)
Quadrant I
(+,+)
5
4
Plot the ff.
points,
label them.
F(-1, 5)
G(-3, - 4)
H(0, - 5.5)
J(5, - 1)
B( , )
3
A( , )
2
1
D( , )
-5
-4 -3
E( , )
-2
-1
1
2
3
4
5
x
-1
-2
-3
C( , )
-4
-5
Quadrant III
(-,-)
Figure 4.1: The Coordinate Plane
Quadrant I V
(+,-)
The Rectangular Coordinate Plane or the Cartesian Plane
y
Quadrant II
(-,+)
Quadrant I
(+,+)
5
4
3
2
1
-5
-4 -3
-2
-1
1
2
3
4
5
x
-1
-2
-3
-4
-5
Quadrant III
(-,-)
Figure 4.1: The Coordinate Plane
Quadrant I V
(+,-)
Name: ___________________ Group: _____
y
Point
x
y
Coordinates
Quadrant/
Axis
5
P1
P2
P3
P2
3
2
P3
-5
1
-4 -3 -2 -1
P6
1
2
3
-1
P4
-2
P5
P6
P1
4
-3
P4
-4
-5
Figure 3.4: Plotting of Points
P5
4
5
x
The X & Y-intercepts
y
line a
5
Self test:
Identify the x & y
Intercepts of
line b:
4
3
x-intercept
(0.5, 0)
2
X-int: _______
Y-int: _______
x
1
-5
-4 -3
-2
-1
-1
1
2
3
4
5
line b
-2
-3
-4
y-intercept
(0, -1)
-5
an x-intercept is a point on the graph where y is zero,
4.3the graph where x is zero.
and a y-intercept is a Figure
point on
The X & Y-intercepts
Without the graph, we can also find the intercepts as long as the equation is given.
For example in 3x + 2y = 12; the x intercept is solved by setting y = 0
So, setting y = 0, we have: 3x + 2(0) = 12; Solving for x,
we have 3x = 12; therefore x = 4, the x-intercept.
In the same manner, the y-intercept is solved by setting x = 0
Therefore, setting x = 0, we have: 3(0) + 2y = 12; Solving for y,
we have 2y = 12; So, y = 6, the y-intercept.
Self Test: Find the x and y intercepts in the linear
equation 6x – 10y = 30
y-intercept: _________
x-intercept: _________
To enhance the concepts of intercepts, you can do Exercise 4.1
Board Work # 5 page 210 before proceeding to the next topic.
Distance and
Midpoint Formulas Page 200
y
Find the Distance & Midpoint
of points A & B.
5
4
3
Midpoint
M(0, -1)
The distance between
two points:
B(2, 3)
x1, y1  & x2 , y2 
2
2
d  x2  x1    y2  y1 
2
1
:
-5
-4 -3
-2
-1
:
x
1
-1
2
.
5
x1, y1  & x2 , y2 
-3
-4
-5
4
The midpoint between
two points:
-2
A(-2, -5)
3
C
 x  x y  y2 
MP   1 2 , 1

2
2


Figure 4.4: Distance & Midpoint Between Two Points
Distance and Midpoint Formulas
Page 201
Self Test: Quadrilateral ABCD in Figure 4.5 is
drawn in aycoordinate plane below.
a. Use the distance formula to show
10
that line segment AB and DC
9
are equal in length.
8
d
7
B
6
C
x2  x1 2   y2  y1 2
5
4
:
3
:
2
1
0
A1
2
3
4
5
D
.
6
7
x
8
Figure 3.7: Exercises on Distance & Midpoint Between Two Points
Figure 4.5:
Challenge!
What are diagonals?
a.Use the Midpoint formula to show
that the midpoints of the diagonals
BD and AC
have the same coordinates.
 x  x y  y2 
MP   1 2 , 1

2
2


To enhance the concepts “distance bet. 2 points” and “midpoint”, you can do
Exercise 4.1 Board Work #4 & C applications on pages 209-210
before proceeding to the next topic.
The bus breaks down on your way to school. The conductor calls the garage for a tow truck.
There is a foot bridge halfway between the garage and the bus.
a) How far is the garage from the bus?
b) Give the exact location of the foot bridge. Draw the foot bridge on the map.
c) How far is the foot bridge from the school?
In the map
on the right,
each unit
represents
one mile.
Slope of a Line: Page 202
Slope of a line is a number that measures its "steepness",
usually denoted by letter “m”.
What do you know about slope?
Consider the following definitions of slope.
• On the coordinate plane, the slant of a line is called the slope.
• It is the change in y for a unit change in x along the line.
• Slope is the ratio of the change in the y-value over the change
in the x-value.
• Slope of a line is the ratio of its rise to run.
Slope of a Line: Page 202
“NO slope”
m = “0”
Slope of a Line: Page 202
Slope of a Line: Page 202
Slope of a line is a number that measures its "steepness",
usually denoted by letter “m”.
Why do we use “m” for slope?
Slope" was once called the "modulus of slope",
the word "modulus" being used in its sense of
"number used to measure"
It is originated from the Arabic word
MOMAS means tangent.
http://wiki.answers.com/Q/Why_is_slope_represented_by_the_letter_m
Slope of a Line: Page 202
Slope of a line is a number that measures its "steepness",
usually denoted by letter “m”.
…but…what is slope in the real world?
In the real world,
• slope of a line tells us how something changes
over time.
• If we find the slope we can find the rate of
change over that period.
- Carpenters use the terms rise and run to
describe the steepness of a stairway or
a roofline.
- We can use rise and run to describe the
steepness of a hill.
Challenge!
Why do you think we are using “m” instead of “s” to represent slope?
How do we find the slope?
change in y
Slope (m) 
change in x
rise
 or
run
Step1: Choose two exact points
on the line & then connect
them by a straight line.
Step2: Draw a right triangle
using the two points you have
selected as the vertices of the
two acute angles.
Step3: To get the rise, count the
# of units of the vertical leg;
To get the run,, count the #
of units of the horizontal leg.
change in y
Slope (m) 
change in x
y
rise
 or
run
5
rise 4
slope (m) 
 2
run 2
4
3
run = 2 units
B(2, 3)
2
1
-5
-4 -3
-2 -1
D(0,-1) -1
-2
rise = 4 units
1
-3
-4
A(-2,-5)
-5
run = 3 units
Figure 4.6
2
3
4
5
rise = 6 units
Step1: Choose exact points
on the line & then connect
them by a straight line.
Step2: Draw a right triangle
using the two points you have
x
selected as the vertices of the
two acute angles.
Step3: To get the rise, count the
# of units of the vertical leg;
To get the run,, count the #
of units of the horizontal leg.
rise 6
slope (m) 
 2
run 3
Slope (m) 
y
rise
 or
run
5
4
3
B(2, 3)
Use points A & B to
get the slope using
the ratio rise to the
run.
2
1
-5
-4 -3
-2 -1
D(0,-1) -1
-2
-3
-4
A(-2,-5)
-5
Figure 4.6
change in y
change in x
x
1
2
3
4
5
What conclusion
can you draw
about the slope
of the line using
any 2 points on it?
Finding the slope (m) by the formula
change in y y2  y1
y1  y2
m

or
change in x x2  x1
x1  x2
where : x2  x1 or x1  x2  0.
Using the formula, let us find the slope of the
Line in Fig. 4.6 using:
A] Points A(-2, -5) and B(2, 3).
y 2  y1 3  (5) 3  5 8
slope (m) 


 2
x2  x1 2  (2) 2  2 4
B] Points A(-2, -5) and D(0, -1).
line “b” falls from left
to right; slope is negative.
m = _________
y
Direction of the Line and its Slope
line a
line b
5
line “a” rises from left
to right; slope is positive
m = _________
4
3
2
1
-5
-4 -3
-2
x
-1
-1
line c
1
-2
-3
line “c” is horizontal;
no rise or rise = 0
m = _________
-4
-5
Figure 4.7
2
3
4
5
line “d” is vertical;
no run or run = 0
m = ________
line d
INTERPRETING SLOPE (m) using lines in Fig. 4.7
5
ma 
2
Slope (m) is POSITIVE: This means, for every 5-unit increase in y,
x increases by 2 units. Line “a” rises from left to right.
mb  1
mc  0
Slope (m) is NEGATIVE: This means, for every 1-unit decrease in y,
x increases by 1unit. Line “b” falls from left to right.
Slope (m) is ZERO:
; This ;means, y does not change as x increases;
line c is horizontal; therefore the slope of horizontal line is “zero”.
md  undefined / does not exist
Slope (m) is UNDEFINED or does not exist: This means,
as y increases, x does not change; thus, the two x coordinates
are the same, so the difference is zero. In short, vertical line
has NO defined slope.
Page 204
Slopes of
parallel lines
y
line “b”
5
line “a”
Page 204
4
3
What do you know
about the slopes
of parallel lines?
2
1
-5
-4 -3
-2
x
-1
-1
10
5
m 
4
2
1
2
3
-2
-3
m
-4
-5
Figure 4.8: Slopes of Parallel Lines
4
5
2
5
Slopes of
Perpendicular
lines
y
5
Page 205
4
Please read the
word of caution
on page 205!
line “a”
3
What do you know
about the slopes
of perpendicular
lines?
5
m
4
2
1
-5
-4 -3
-2
-1
-1
1
2
3
-2
4
5
x
line “b”
m
4
5
-3
-4
-5
Figure 4.9: Perpendicular lines
Who can interpret
the slope of:
- line a
- line b
Slope &
steepness
of a line
y
10
line a : m  3
9
Page 206
line b : m  2
8
7
Which between lines “a” and “b” is steeper? Justify.
6
Which between lines “c” and “d” is steeper? Justify.
5
4
Which lines have the same steepness? Justify.
3
line d : m  1
2
line c : m  2
1
0
x
1
2
3
4
5
6
7
8
Arrange the lines in order from the steepest to the least
Figure 4.10:
Are lines “b” and “c” perpendicular to each other? Justify
Page 206
Slope &
steepness
of a line
Complete the statements below:
 The bigger the absolute value of the slope, the _______________
is the line.
___________________ line have the same slopes.
Two lines are perpendicular to each other if ___________________
______________________________________________________.
 Horizontal lines have a slope equal to ____________.
 ___________________ lines have no slopes.
 A line that rises from left to right has ______________ slope.
 A line that ___________________________________
has a negative slope.
y
10
line a
1. Find the slope of
line “a” only!
line b
9
8
line c
7
6
line d
5
line e
4
3
3. Explain how did
you do it?
2
1
0
2. Approximate the
slopes of
- line b
- line c
- line d
- line e
1
Figure 4.11:
2
3
4
5
6
7
8
4. What concepts did
x you use in this
exercise?
y
10
9
8
Do the Self-Test on Pages 206 - 207!
7
B
6
C
5
4
3
2
1
0
A1
2
3
4
5
D
x
6
7
8
Do Page 216 - #2 (a to i)
Figure 4.11
y
Graphing of Line
given its Slope &
a Point
5
4
3
2
(1, 3)
1
-5
-4 -3 -2
(-2, -1)
-1
1
-1
-2
-3
-4
-5
(-5, -5)
Figure 4.12:
2
Page 207
Draw a line passing
through the point (-2, -1)
4
with
m
3
Step 1:Plot (-2, -1)
Step 2: Decide on the direction.
Since m is +, the line goes up
to the right.
Step 3: Find another point
x is
m = 4/3; that
Do the
3 using
4 5
4 units up from (-2, -1)
self-test
& 3 units to the right.
on
That point is ((1, 3). Or you
Page 208!
can go 4 units down from
(-2, -1) & 3 units left. That
point is (-5, -5).
Step 4: Connect the points by a
straight line.
Do the Mathematical Investigations on Page 210
before proceeding to the next lesson!
y
REVIEW
1. Meaning of slope & y-intercept in the real world
Example: Taxi fare: y = 7.50x + 40
rise
2. Finding the slope given a line: m 
rune
Finding the slope given 2 points: (0, 6) & (8, -10)
3. Finding the slope & the y-intercept given the equation
y = mx + b
Example: y = 2x – 5;
m = 2 & b = -5
2y = -6x + 10 m = - 3 & b = 5
4.
Page 208
y
5
• Draw a line passing
through the point (-1, 2)
4
3
with m 
2
3
3
, & m
2
2
1
-5
-4 -3
-2
-1
1
-1
-2
-3
2
3
4
5
x
• Draw a line with a slope
of 3 and
- a y-int of 2
- an x-int of 2
-4
-5
Figure 4.12:
Go to Page 216, do Letter B (3-6)
In a graphing paper.
y
4.2 Equation of a line
A straight line is defined by a linear equation whose general
form is Ax + By + C = 0, where A, B are not both 0.
Form
Equation
1. Slope Intercept
y = mx + b where:
slope
y-intercept
Remark
Use this form when
you know the slope
“m”and the yintercept “b”.
Example1.
Find the equation of a line with a slope of 3
and a y-intercept of 5. Equation: y = 3x + 5
Your turn: Give the equation of the line with m = 6 & b = -1
Equation: y = 6x - 1
y
4.2 Equation of a line
A straight line is defined by a linear equation whose general
form is Ax + By + C = 0, where A, B are not both 0.
Form
Equation
2. Point Slope
Form
Your turn: Give the equation of the line with m = 4 &
and passing through the point (0, 3). y = 4x + 3
Remark
Use this form when
you know a point on
the line and the
slope (or can
determine the
slope).
Example2.
Find the equation of the line with a slope of -3
and the line passes through the point (2, 4)
Equation: y = -3x + 10
10
What is the y-intercept “b”of this line? __________
y
4.2 Equation of a line
A straight line is defined by a linear equation whose general
form is Ax + By + C = 0, where A, B are not both 0.
Form
3. Horizontal Line
Equation
y=b
Remark
This equation also
describes what is
happening to the ycoordinates on the
line. In this case, it is
always “b”.
Example3.
Find the equation of the horizontal line
with a y intercept of -3. y = - 3
y
4.2 Equation of a line
A straight line is defined by a linear equation whose general
form is Ax + By + C = 0, where A, B are not both 0.
Form
4. Vertical Line
Equation
x=k
Remark
This equation also
describes what is
happening to the xcoordinates on the
line. In this case, it is
always “k”.
Example3.
Find the equation of the line which is parallel to
y–axis and passing through (5, 0). x = 5
y
4.2 Equation of a line
A straight line is defined by a linear equation whose general
form is Ax + By + C = 0, where A, B are not both 0.
Example 3: Find the equation of the line whose slope is 4
and passing through the point (0, -3).
Example 4: Find the equation of the line that passes
through the points (-3, 5) and (-5, -8).
First, find the slope:
Example 5: Find the slope and y-intercept for the
equation: 6x + 3y = 9.
Do the Self-Test on page 214.
Do Exercise 4.2 pages 215-216.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 1: Physical Fitness
The membership fee (f), in pesos, at a gym is based on
the number of people (p) that a new member recruits to join
the gym at the time of registration.
The membership fee formula is given by the equation:
f = – 300p + 5000, where:
f = membership fee, and
p = the number of recruits at the
time of registration
Relating the equation f = – 300p + 5000, to y = mx + b,
we have:
y = -300x + 5000
What is the slope of the membership formula?
What does the slope of this equation mean in the context of the problem?
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 1: Physical Fitness
f = – 300p + 5000
y = -300x + 5000
• What is the slope of the membership formula?
What does the slope of this equation mean in the context of the problem?
Answer: The slope (m) is – 300; Since the slope is negative, this means
that a membership fee is decreased by Php300 for every person that
a new member recruits at the time of registration.
• What is the y-intercept of the membership formula?
What does the y-intercept of this equation mean in the context of the problem?
Answer: The y-intercept (b) is 5,000; This means the new member
will pay a membership fee of Php5,000 if he has no recruit at the
time of the registration (because x = 0); that is f = –(300)(0)+ 5,000;
f = 0 + 5,000; therefore f = Php5,000.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 2: Food Handling y = 200x + 1550
• What is the slope of the bacteria formula?
What does the slope of this equation mean in the context of the problem?
Answer: The slope (m) is 200; since the slope is positive, this means that
the number of bacteria increases by 200 for every hour that the meat is in
the warm room.
• What is the y-intercept of the bacteria formula?
What does the y-intercept of this equation mean in the context of the problem?
Answer:The y-intercept (b) 1,550; This means that, the meat has
already 1,550 bacteria when it was placed in the warm room;
that is x = 0, so y = 200(0) + 1,550; y = 1,550.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 2: Food Handling y = 200x + 1550
 The Scientist-Chef left the piece of meat in the warm room
at 9:00 AM. What number of bacteria does the meat will have
at 1:30 p.m. of the same day? Justify your answer by showing
the solution.
Answer: The number of hours “x” that elapsed from 9:00AM to 1:30 PM
is 4.5.
So, since y = 200x + 1,550;
y = 200(4.5) + 1,550;
y = 900 + 1,550;
y = 2,450, this is the number of bacteria which is in the meat at 1:30 P.M.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Activity Worksheet 4.3b page 227
A tire company wants to determine how quickly the tread
on its tires wears down with average use.
- Let x represents the number of months the tire was used.
y represents the thickness of the tire thread, in millimeters.
An equation for a line that describes this relationship is
5
y   x  20
8
Explain the meaning of slope & y-intercept in the context
of the problem.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 3: Demand & Price
BBZ company noticed that a linear relationship exits between the price of
The school bags and the number of bags sold. At Php1000 the company
sold 2,000 pieces. When the company raised the price to Php1,200 the
Company was able to sell only 1,500 pieces. Find an equation that
relates the price of school bags to the number of bags sold.
Show your solution
Let x = the number of bags sold
y = the price of bag
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 3: Demand & Price
1] Find an equation that relates the price of school bags to the number of bags sold.
Remember that if we have two points we can determine the
equation of a line. In this case, we will use “ point-slope form”:
y  y1  m( x  x1 )
• First, solve for the ;slope using the two points: (2000, Php1000) & (1500, Php1200)
y 2  y1 1,200  1000
200
2 Use either point; say let
m
;
x2  x1 1,500  2000   500 ; m   5 us use (2000, 1000);
y  y1  m( x  x1 )
we have: y  1000  
2
( x  2000)
5
Therefore, the equation that relates the price to the number of bags demanded is:
y
2
x  1,800
5
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 3: Demand & Price
Therefore, the equation that relates the price to the number of bags demanded is:
y
2
x  1,800
5
2] Explain the meaning of slope & y-intercept in the context of this
problem.
The slope 
2
means “for every 2 pesos decrease in the price
5
of bag, an additional 5 pieces is sold”.
The y-intercept 1,800 means, “if the price is Php1,800, no bag is sold”.
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 3: Demand & Price
Therefore, the equation that relates the price to the number of bags demanded is:
y
2
x  1,800
5
3] What is the unit price if the number of bags
demanded/sold is 1000?
4] How many bags are demanded/sold if the
unit price is Php500?
y
4.3 Applications a. Interpreting slope & y-intercept in the
real world context.
Example 3: Demand & Price
1. SMS department store sell 2000 bags when the unit price is Php450.
It was determined that it can sell 300 bags more with each Php100
reduction in the unit price.
a. Find the demand equation: y = mx+b
b. Explain the meaning of slope & y-intercept in the context
of this problem.
c. How many bags are demanded if the unit price is Php200?
d. What is the unit price if 3000 bags are demanded?