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1) Nasirah Binti Che Daud D20081032296
2) Wan Masturah Binti Wan Mad Mohtar D20081032356
3) Ayuni Amalina Binti Mukhtar D20081032369
4) Nur Ain Bt Ahmad Fikri D20081032311
5) Nur Syazwani Bt Wan Aziz D20061026742
Coordinate Plane
• A basic concept for coordinate geometry.
• It describes a two-dimensional plane in terms of two perpendicular axes: x and y.
• The x-axis-horizontal direction
• the y-axis-vertical direction of the plane.
• Points are indicated by their positions along the x and y-axes in the form (a,b)
• L coordinates is (–3, 1.5)
Equation Of A Line
• An equation of a line can be written
y = mx + b
where
m is the slope
b is the y-intercept
• Slant of a line is called the slope/gradient.
• Slope is the ratio of the change in the y-
value over the change in the x-value.
Slopes = Change in y value
Change in x value
- The rate at which line rises (or falls) vertically for every unit across to the right.
y
x0
m
Q(x2,y2)
P(x1, y1)(y2-y1)
(x2-x1)
Gradient line
P(x1,y1) ,Q (x2,y2):
12
12
x-x
y-y m
Where.. 12 xx
y
x0
P(2,2)
Q(6,5)
If θ < 90, m is positive
θ
Gradient line of PQ :
12
12
x-x
y-y mPQ
26
25
4
3m PQ
Example 1:• Given two points, P = (0, –1) and Q = (4,1), on the
line we can calculate the slope of the line.• Slopes= Change in y value
Change in x value
= 1-(-1)
4 - 0
= 2
4
= 1
2
y
0x
M (2,7)
N (5,2)
If θ > 90, m is negative
θ
Gradient line of MN :
52
27mMN
3
5
3
5mMN
• Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line?
Slopes = Change in y value
Change in x value
= -1-3
0-(-2)
= -4
2
= -2
Example 2:
y
X0
P Q
-If the line PQ is parallel with the x-axis,
θ = 0° OR θ = 180° ….Hence, m = 0
m = 0
How about parallel & perpendicular line ??What are their gradient ??....
1.) Two lines are parallel if and only if both have the same gradient ; m1=m2
2.) Two lines with m1 and m2 gradient
perpendicular if and only if m2m1= -1
Slopes Of Parallel Lines
y
x0
P(2,2)
Q(6,5)
Gradient line of PQ :
12
12
x-x
y-y mPQ
26
25
4
3m PQ
R (4,2)
S (8,5)
Gradient line of RS :
12
12
x-x
y-y mRS
48
25
4
3m RS
mPQ // mRS
Question…
1.) Does the straight-line AB and CD below parallel??....
A(1,9) , B(5,8) , C(5,2) ,D(1,3)
12
12
ABmxx
yy
12
12
CDmxx
yy
51
23
15
98
4
1
4
1
mAB = mCD…….hence, AB // CD
Slopes Of Perpendicular Lines
• Two lines are perpendicular if the product of their slopes (m) is –1
• The line y= ½ x-1
perpendicular to
y= -2x-1
Because:
½ x (-2) =-1
y
x0
P(2,5)
Gradient line of PQ :
12
12
x-x
y-y mPQ
42
25
2
3m PQ
Q (6,3)
R (9,5)
Gradient line of RS :
12
12
x-x
y-y mQR
69
35
3
2m RS
mPQ mRS
Y-intercept
• The y-intercept is where the line intercepts (meets) the y-axis.
• The midpoint of a segment divides the segment into two segments of equal length.
• The midpoint between the two points (x1,y1) and (x2,y2) is
2,
2
2121 yyxx
Example:
• The midpoint of the points A(1,4) and B(5,6) is
)5,3(2
10,
2
6
2
64,
2
51
Distance Formula
• The distance between the two points (x1,y1) and (x2,y2) is
Dividing Point with Ratio
Formula for inside point :
nm
nymy
12
nm
nxmx
12
x = y =
Inside dividing ratio
11, yxA
),( yxP
),( 22 yxB
Given points A(1,2) and B(19,23). If P (x,y) dividing inside AB with ratio 1:2, find the value of x and y.
A(1,2)
y
xA(1,2)
P(x,y)
B(19,23)
m
n
Solution:
y=
=
=
=
nm
nymy
12
21
)2)(2()23)(1(
3
27
9
x =nm
nxmx
12
3
219
21
)1)(2()19)(1(
3
21 7
)9,7(P
Outside dividing ratioy
x),( 11 yxA
),( 22 yxB
),( yxP
Outside dividing point
nm
nymy
12
nm
nxmx
12
x=
Formula :
y =
Given that points A(-5,-6) and B(-1,-2). Get the coordinates that
dividing outside AB with the ratio of 5:3
1
5
2
1
x
x2
6
2
1
y
y
3
5
n
m
nm
nymy
nm
nxmxP
1212
,
35
)6(3)2(5,
35
)5(3)1(5P
2
1810,
2
155P )4,5(
EQUATION TYPE
• Gradient Type
• Interception Type
• General Type
GRADIENT TYPE
y = mx + c
y
x
P(x,y)
m = gradient
c = y-interception
c
INTERCEPTION TYPE
x + ya b = 1
a = x-interception
b = y-interception
x
y
P(0,b)
Q(a,0)
GENERAL TYPE
ax + by + c = 0
a, b, c are constant
METHOD TO FIND STRAIGHT-LINE EQUATION
•Give Two Points
or
•One Point, One Gradient
GIVEN TWO POINTS
P(1,-4)
Q(3,4)
First, find gradient (m):
m =
m = =
m = 4
Then, find c
Substitute q(3,4) in equation y = mx + c
(4) = (4)(3) + c
4 = 12 + c
C= -8
the equation
y = 4x-8
y2 – y1
x2 – x1
4-(-4)3-1
82
y = 4x - 8
ONE POINT, ONE GRADIENT
P(2,8)
Substitute m and point p(2,8)
In equation y = mx + c
(8) = (3)(2) + C
8 = 6 + C
Hence c = 2
So the equation=
y = 3x + 2 y = 3x + 2
m = 3
y = 2x + 10Change this equation into General Type
y = 2x + 10
y – 2x – 10 = 0 or 2x – y + 10 = 0
Then, change 2x – y + 10 = 0 into Interception type
2x – y + 10 = 0
2x – y = -10
2x – y = -10
-10 -10 -10
2x + y = 1 or x + y = 1
10 10 5 10
Brain-Storming Corner
Given a line with two given point. Find the equation of the straight line
(-3, 2)
(2, 7)
y
x
Gradient ;
7 – 2 = 5
2 – (-3) 5
= 1
Find c by substitute (2,7) in equation y = mx + c
7 = (1)(2) + c
c = 5
Equation
y = mx + c
y = x + 5
Change the equation into general type and interception type
y = x + 5
y – x – 5 = 0 or x – y + 5 = 0
Then, change x – y + 5 = 0 into Interception type
x – y + 5 = 0
x – y = -5
x – y = -5
-5 -5 -5
x + y = 1
5 5
Subtopics….
• The nearest point to the straight line
• The distance between two straight lines that parallel to each other
• Intersection of the straight line• Area of rectangle• Area of triangle
P
Q
ax + by +c = 0
d
(h,k)
The nearest point to straight line
Perpendicular distance :
-The shortest distance betweenthem / the length of a perpendicularline segment from the line to thepoint
y
x
The nearest point to straight line
22d
ba
cbkah
d
P
Q
ax + by +c = 0
(h,k)
The nearest distance from point (h,k) to straight line ax +by +c =0 is ;
0 if22
ba
cbkah Hence, the point is on the other side.
Example :• Find the point distance and location of (2,1) and (-3,2)
towards straight line 2y-3x-1=0
Solution :
From straight line 2y-3x-1= 0 ,
a = -3 b = 2 c = -1
Point (2,1), hence h=2, k=1
d1 22 )2()3(
1)1(2)2(3
13
5
22 )2()3(
1)2(2)3(3
13
12
Point (-3,2), hence, h=-3, k=2
d2
d1 d2 13
5
13
12
2y-3x-1=0
y
x
(2,1)(-3,2)
d1d2
0
The distance between 2 straight-lines that parallel
to each other
L1
L2
METHOD
1) Find the coordinate on one of the line2) Find the point perpendicular distance from the other line
Example :• Find the distance between the parallel lines 5x+12y+1=0 and
5x+12y+8=0
Solution:
Take 5x+12y +1=0
When x=0, y=
Coordinate is (0, )
12
1
12
1
The distance to line 5x+12y+8=0 is :
a= 5 b= 12 c=8 h= 0 k=
13
7
14425
8121
120 d
12
1,0
22d
ba
cbkah
12
1
Intersection of straight-line
The coordinate for two straight lines intersection can be found by
solving both equation stimultaneously
7
4 y
Solution :
2x-3y=6 …………(1)
4x+y =16 …………(2)
(1) x 2
4x+y =16 …………(2)
4x-6y=12 …………(3)
(2) - (3)
7y= 4
7
27 x
2x-3y=6
4x+y=16
y
x0
P
QR
S
Choose A Quizzes
SELECT THE DIFFICULTY LEVEL
EASY
EXPERT
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6C D
F
B
E
A
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
3 UNITS LEFT,
5 UNITS UP
C D
F
B
E
A
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
C D
F
B
E
A
2 UNITS RIGHT,
4 UNITS DOWN
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
C D
F
B
E
A
5 UNITS RIGHT,
0 UNITS UP/DOWN
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
C D
F
B
E
A
5 UNITS LEFT,
3 UNITS DOWN
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
C D
F
B
E
A
0 UNITS LEFT/RIGHT
4 UNITS UP
y
5
4
3
2
1 x
-5 -4 -3 -2 -1 1-1
2 3 4 5
-2
-3
-4
-5
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6C D
F
B
E
A
0 UNITS RIGHT/LEFT
0 UNITS UP/DOWN
Positive slopeFor example: Given two points, P = (0, –1) and Q = (4,1), on the line we can calculate the slope of the line.
• y-intercept
• Gradient
• Equation
1/2 4
Y = 1/2x - 1
Y = 2x + 4
-1 2
Q4Q1 Q3Q2 Q5 Q6
Negative slopeFor example: Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line?
• y-intercept
• Gradient
• Equation
-2 3
Y = 3x - 2
Y = -2x - 1
-1 -2
Q4Q1 Q3Q2 Q5 Q6
• In coordinate geometry, two lines are parallel if their slopes (m) are equal.
• For example: The line y=1/2x+1 is parallel to the line y=1/2x-1. Their slopes are both the same.
Slopes Of Parallel Lines
Q4Q1 Q3Q2 Q5 Q6
• In the coordinate plane, two lines are perpendicular if the product of their slopes (m) is –1.
• For example: The line Y=1/2X-1 is perpendicular to the line y = –2x– 1. The product of the two slopes is 1/2 x (-2) = -1
Slopes Of Perpendicular Lines
Q4Q1 Q3Q2 Q5 Q6
• To find a point that is halfway between two given points, get the average of the x-values and the average of the y-values.
• The midpoint between the two points (x1,y1) and (x2,y2) is:
• For example:
The midpoint of the points A(1,4) and B(5,6) is
Mid Point Formula
Q4Q1 Q3Q2 Q5 Q6
• For example: To find the distance between A(1,1) and B(3,4), we form a right angled triangle with AB as the hypotenuse. The length of AC = 3 – 1 = 2. The length of BC = 4 – 1 = 3.
Applying Pythagorean Theorem:
• AB2 = 22 + 32
AB2 = 13AB = /13
Distance Formula
Q4Q1 Q3Q2 Q5 Q6
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