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COORDINATE
GEOMETRY
7.2 Parallel Lines &
Perpendicular Lines
Prepared By : Muslimah Binti Mahbob
Objective for today’s lesson:
Make and verify conjectures about
gradient of parallel lines and
perpendicular lines, and hence, make
generalisation.
Solve problems involving equations of
parallel and perpendicular lines.
01
02
EXPLORATION
TIME
LET EXPLORE
PARALLEL
LINES
What is it about parallel lines?
Example 1
Since the gradients for both lines are equal,
thus, both lines are parallel to each other.
Since the gradients for both lines
are not equal, both lines are not
parallel lines.
Determine whether the following pairs of straight lines are
parallel lines.
(a) (b)
9
5
1
6





x
y
x
y
3
4
6
4
3
2




y
x
x
y
2
1 m
m 
5
6
5
,
9
5
6
,
1
6
)
(
2
1
2
1











m
m
m
x
y
m
x
y
a
2
3
2
3
2
3
4
2
4
3
2
)
(
1 






m
x
y
x
y
x
y
b
2
3
4
3
2
3
3
6
4
3
4
6
2 






m
x
y
x
y
y
x
2
1 m
m 
Example 2
Find the equation of the straight line by
using the coordinate (1, 4) with the
gradient of -2.
Find the gradient
of the straight line.
Find the equation of straight line which passes through (1, 4)
and is parallel to line 5
4
2 
 x
y
2
2
2
5
4
5
2
5
4
2
1 







m
x
y
x
y
x
y
 
 
6
2
4
2
2
1
2
4
1
1














x
y
x
y
x
y
x
x
m
y
y
 
1
1 x
x
m
y
y 


PERPENDICULAR
LINES
What is it about perpendicular lines?
Gradient of Parallel &
Perpendicular Lines by graphing
Proof of Gradient of
Perpendicular Lines
Example 3
Since the product of the gradients for both
lines is -1, thus, both lines are perpendicular
to each other.
Determine whether the following pairs of straight lines are
perpendicular lines.
0
4
3
0
2
3






x
y
x
y
1
2
1 

 m
m
3
1
3
2
3
1
2
3
0
2
3
1 







m
x
y
x
y
x
y
3
4
3
0
4
3
2 







m
x
y
x
y
  1
3
3
1




1
2
1 

 m
m
Example 4
Find the equation of the normal line by
using the coordinate (1, 4) with the
gradient of 0.5.
Find the gradient
of the straight line.
Find the equation of straight line which passes through (1, 4)
and is perpendicular to line 6
2 

 x
y
2
6
2
1 




m
x
y
 
 
2
7
2
1
4
2
1
2
1
1
2
1
4
1
2
1











x
y
x
y
x
y
x
x
m
y
y
 
1
1 x
x
m
y
y 


Find the gradient
normal of the
straight line.
2
1
2
1
1
1
2
1
2
2
1









m
m
m
m
m
Example 5
3
1
5
3
1
15
3
15
3
)
(







mCD
x
y
x
y
x
y
a  
 
 
3
4
3
1
2
3
2
3
1
2
3
1
2
3
1
,
2
,
1
1
1












x
y
x
y
x
y
x
x
m
y
y
mAB
A
 
 
 
15
3
6
9
3
3
3
6
6
,
3
,
3
3
1
1
1
1
2
1




















x
y
x
y
x
y
x
x
m
y
y
D
mAB
mDE
Example 5
4
40
10
5
45
9
45
9
5
15
3
3
5
3
1
15
3
:
,
3
5
3
1
:
)
(


















x
x
x
x
x
x
x
x
x
y
DE
x
y
AB
b
 
 
3
,
4
3
15
4
3
?
,
4
15
3
:










E
y
y
y
x
when
x
y
AB
 
 
4
,
7
4
7
6
2
8
1
3
2
2
,
4
2
1
2
2
,
2
1
3
,
4
2
,
2
2
2
2
2
2
2
2
2
2
1
2
1
















 







 


B
y
x
y
x
y
x
y
x
y
y
x
x
M
SOLVING
PROBLEMS
Example 6
Example 6
Example 7
2
3
24
2
3
48
3
2
48
2
3
1 









m
x
y
x
y
y
x
3
2
2
3
1
1
2
2
1
2
2







m
m
m
m
mAB
m
 
 
 
3
20
3
2
12
3
16
3
2
8
3
2
12
12
,
8
,
3
2
1
2
1












x
y
x
y
x
y
x
x
m
y
y
B
mAB
Example 8
 
5
,
5
2
10
,
2
10
2
2
8
,
2
7
3
2
,
2
)
( 2
1
2
1













 







 


M
M
M
y
y
x
x
M
a    
   
   
606
.
3
13
4
9
5
7
5
8
7
,
8
,
5
,
5
,
?
)
(
2
2
2
1
2
2
1
2













MP
MP
MP
y
y
x
x
MP
P
M
MP
b
EVALUATION
TIME
TRY
DISCUSS THE ANSWERS WITH YOUR TEACHER
HOMEWORK
Self Practice 7.4, page 187,
Q1-Q4
Intensive Practice 7.2, page 190,
Q1–Q6 & Q10–Q11
Intensive Practice 7.2, page 190,
Q1–Q7 & Q10-Q11
A problem can be real,
but it is your choice
whether you want to stay
in it or you get out of it.
GOOD LUCK & ALL THE BEST

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COORDINATE GEOMETRY (7.2 Parallel & Perpendicular Lines).pptx

  • 1. COORDINATE GEOMETRY 7.2 Parallel Lines & Perpendicular Lines Prepared By : Muslimah Binti Mahbob
  • 2. Objective for today’s lesson: Make and verify conjectures about gradient of parallel lines and perpendicular lines, and hence, make generalisation. Solve problems involving equations of parallel and perpendicular lines. 01 02
  • 6. What is it about parallel lines?
  • 7. Example 1 Since the gradients for both lines are equal, thus, both lines are parallel to each other. Since the gradients for both lines are not equal, both lines are not parallel lines. Determine whether the following pairs of straight lines are parallel lines. (a) (b) 9 5 1 6      x y x y 3 4 6 4 3 2     y x x y 2 1 m m  5 6 5 , 9 5 6 , 1 6 ) ( 2 1 2 1            m m m x y m x y a 2 3 2 3 2 3 4 2 4 3 2 ) ( 1        m x y x y x y b 2 3 4 3 2 3 3 6 4 3 4 6 2        m x y x y y x 2 1 m m 
  • 8. Example 2 Find the equation of the straight line by using the coordinate (1, 4) with the gradient of -2. Find the gradient of the straight line. Find the equation of straight line which passes through (1, 4) and is parallel to line 5 4 2   x y 2 2 2 5 4 5 2 5 4 2 1         m x y x y x y     6 2 4 2 2 1 2 4 1 1               x y x y x y x x m y y   1 1 x x m y y   
  • 10. What is it about perpendicular lines?
  • 11. Gradient of Parallel & Perpendicular Lines by graphing
  • 12. Proof of Gradient of Perpendicular Lines
  • 13. Example 3 Since the product of the gradients for both lines is -1, thus, both lines are perpendicular to each other. Determine whether the following pairs of straight lines are perpendicular lines. 0 4 3 0 2 3       x y x y 1 2 1    m m 3 1 3 2 3 1 2 3 0 2 3 1         m x y x y x y 3 4 3 0 4 3 2         m x y x y   1 3 3 1     1 2 1    m m
  • 14. Example 4 Find the equation of the normal line by using the coordinate (1, 4) with the gradient of 0.5. Find the gradient of the straight line. Find the equation of straight line which passes through (1, 4) and is perpendicular to line 6 2    x y 2 6 2 1      m x y     2 7 2 1 4 2 1 2 1 1 2 1 4 1 2 1            x y x y x y x x m y y   1 1 x x m y y    Find the gradient normal of the straight line. 2 1 2 1 1 1 2 1 2 2 1          m m m m m
  • 15. Example 5 3 1 5 3 1 15 3 15 3 ) (        mCD x y x y x y a       3 4 3 1 2 3 2 3 1 2 3 1 2 3 1 , 2 , 1 1 1             x y x y x y x x m y y mAB A       15 3 6 9 3 3 3 6 6 , 3 , 3 3 1 1 1 1 2 1                     x y x y x y x x m y y D mAB mDE
  • 16. Example 5 4 40 10 5 45 9 45 9 5 15 3 3 5 3 1 15 3 : , 3 5 3 1 : ) (                   x x x x x x x x x y DE x y AB b     3 , 4 3 15 4 3 ? , 4 15 3 :           E y y y x when x y AB     4 , 7 4 7 6 2 8 1 3 2 2 , 4 2 1 2 2 , 2 1 3 , 4 2 , 2 2 2 2 2 2 2 2 2 2 1 2 1                              B y x y x y x y x y y x x M
  • 20. Example 7 2 3 24 2 3 48 3 2 48 2 3 1           m x y x y y x 3 2 2 3 1 1 2 2 1 2 2        m m m m mAB m       3 20 3 2 12 3 16 3 2 8 3 2 12 12 , 8 , 3 2 1 2 1             x y x y x y x x m y y B mAB
  • 21. Example 8   5 , 5 2 10 , 2 10 2 2 8 , 2 7 3 2 , 2 ) ( 2 1 2 1                           M M M y y x x M a             606 . 3 13 4 9 5 7 5 8 7 , 8 , 5 , 5 , ? ) ( 2 2 2 1 2 2 1 2              MP MP MP y y x x MP P M MP b
  • 23. TRY DISCUSS THE ANSWERS WITH YOUR TEACHER
  • 25. Self Practice 7.4, page 187, Q1-Q4
  • 26. Intensive Practice 7.2, page 190, Q1–Q6 & Q10–Q11
  • 27. Intensive Practice 7.2, page 190, Q1–Q7 & Q10-Q11
  • 28. A problem can be real, but it is your choice whether you want to stay in it or you get out of it. GOOD LUCK & ALL THE BEST