Ask Questions, Get Answers
Menu
X
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
CBSE XII
Math
JEEMAIN
Math
Physics
Chemistry
Practice Test Series
CBSE XI
Math
NEET
Physics
Chemistry
Biology - XII
Biology - XI
Olympiad class V
Math - 5 Test Series
Olympiad class VI
Math - 5 Test Series
studyplans
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
CBSE XII
Math
JEEMAIN
Math
Physics
Chemistry
Practice Test Series
CBSE XI
Math
NEET
Physics
Chemistry
Biology - XII
Biology - XI
Olympiad class V
Math - 5 Test Series
Olympiad class VI
Math - 5 Test Series
mobile
exams
ask
sample papers
tutors
pricing
sign-in
Download our FREE mobile app with 1000+ tests for CBSE, JEE MAIN, NEET
X
Search
Topics
Want to ask us a question?
Click here
Browse Questions
Student Questions
Ad
Home
>>
CBSE XI
>>
Math
>>
Straight Lines
0
votes
Find the distance between parallel lines $ l(x+y)+p=0$ and $l(x+y)-r=0$.
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-a
q6-(ii)
easy
Share
asked
May 9, 2014
by
thanvigandhi_1
retagged
Jul 30, 2014
Please
log in
or
register
to add a comment.
Can you answer this question?
Do not ask me again to answer questions
Please
log in
or
register
to answer this question.
1 Answer
0
votes
Toolbox:
Distance between parallel lines is $d=\large\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}$
The given parallel lines are $l(x+y)+p=0\: and \: l(x+y)-r=0$
Here $A=l, B = l, C_1=p \:and \: C_2=r$.
Hence substituting the values in $d = \large\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ we get,
$ d = \large\frac{|p-(-r)|}{\sqrt{l^2+l^2}}$$ = \large\frac{|p+r|}{l\sqrt 2}$
Hence the required distance is $ \large\frac{1}{\sqrt 2}$$ \bigg| \large\frac{p+r}{l} \bigg|$ units.
answered
May 9, 2014
by
thanvigandhi_1
Please
log in
or
register
to add a comment.
Related questions
0
votes
1
answer
Find the distance between parallel lines \[\] $15x+8y-34=0$ and $ 15x+8y+31=0$
asked
May 9, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-a
q6-(i)
easy
0
votes
1
answer
Reduce the following equation into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive $x$ - axis. $ x-\sqrt 3 y +8=0$
asked
May 7, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-a
q3-(i)
medium
0
votes
1
answer
Reduce the following equation into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive $x$ - axis. $ y-2=0$
asked
May 7, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-a
q3-(ii)
medium
0
votes
1
answer
Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax+By+C=0$ is \[\] $\qquad \qquad A(x-x_1)+B(y-y_1)=0$.
asked
May 13, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-c
q11
difficult
0
votes
1
answer
If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $ x \cos \theta - y \sin \theta = k \cos 2\theta$ and $ x \sec \theta + y \: cosec \theta -k$, respectively, prove that $ p^2+4q^2=k^2$
asked
May 13, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-c
q16
difficult
0
votes
1
answer
Find angles between the lines $ \sqrt 3x+y=1$ and $x+\sqrt 3y=1$.
asked
May 10, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-b
q9
medium
0
votes
1
answer
Find the equation of the line parallel to the line $3x-4y+2=0$ and passing through the point $(-2, 3)$
asked
May 10, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-b
q7
medium
Ask Question
Tag:
Math
Phy
Chem
Bio
Other
SUBMIT QUESTION
►
Please Wait
Take Test
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
JEEMAIN
350+ TESTS
NEET
320+ TESTS
CBSE XI MATH
50+ TESTS
CBSE XII MATH
80+ TESTS
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...