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Green University Of
Bangladesh
Name : Md. Toufiq Elahi
ID : 153002030
Department : Computer Science and
Engineering(CSE)
Topics
Parallel
&
Perpendicular
Parallel Lines :
 In geometry, parallel
lines are lines in a plane which do
not meet; that is, two lines in a
plane that do not intersect or
touch each other at any point are
said to be parallel.
Example :
Parallel lines in real
life
Perpendicular Lines :
 Perpendicular means "at
right angles". A line meeting
another at a right angle, or 90°
is said to be perpendicular to
it.
Example :
Perpendicular lines in
real life
Find the equation of line passing through(-5, 6)
and a) Parallel, b) Perpendicular, to 7x-8y=9 ?
 a ) Solution Parallel Line :
Given that,
7x-8y=9
=>7x-8y-9=0 -----------(1)
(1)Let the equation of the state line
parallel to (1) be,
7x-8y-k=0 ----------------(2)
Since (2) passes through (-5, 6) so
we get,
7(-5)-8(6)-k=0
=>-35 -48-k=0
:. K=-83
The required equation is,
7x-8y-(-83)=0
=>7x-8y+83=0
(Answer.)
b) Solution perpendicular Line :
Let the equation of straight line perpendicular to (1) be,
8x-7y-k=0 ------------------(3)
Since (3) passes through (-5, 6) so we get,
8(-5)+7(6)-k=0
=>-40+42-k=0
:. K=2
S0 the required equition is,
8x+7y-2=0
(Answer.)
Thank You

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Parallel And Perpendicular With Real Life Examples

  • 1. Green University Of Bangladesh Name : Md. Toufiq Elahi ID : 153002030 Department : Computer Science and Engineering(CSE)
  • 3. Parallel Lines :  In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. Example : Parallel lines in real life
  • 4. Perpendicular Lines :  Perpendicular means "at right angles". A line meeting another at a right angle, or 90° is said to be perpendicular to it. Example : Perpendicular lines in real life
  • 5. Find the equation of line passing through(-5, 6) and a) Parallel, b) Perpendicular, to 7x-8y=9 ?  a ) Solution Parallel Line : Given that, 7x-8y=9 =>7x-8y-9=0 -----------(1) (1)Let the equation of the state line parallel to (1) be, 7x-8y-k=0 ----------------(2) Since (2) passes through (-5, 6) so we get, 7(-5)-8(6)-k=0 =>-35 -48-k=0 :. K=-83 The required equation is, 7x-8y-(-83)=0 =>7x-8y+83=0 (Answer.)
  • 6. b) Solution perpendicular Line : Let the equation of straight line perpendicular to (1) be, 8x-7y-k=0 ------------------(3) Since (3) passes through (-5, 6) so we get, 8(-5)+7(6)-k=0 =>-40+42-k=0 :. K=2 S0 the required equition is, 8x+7y-2=0 (Answer.)