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
>>
Student Questions
0
votes
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $$1/a^2 + 1/b^2 + 1/c^2 = 1/p^2$$
cbse
class12
bookproblem
sec-c
difficult
math
Share
asked
Nov 2, 2012
by
balaji.thirumalai
retagged
Mar 19, 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.
Related questions
0
votes
1
answer
Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin, then $\large\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} = \frac{1}{p^2}$.
asked
Nov 29, 2012
by
balaji.thirumalai
cbse
class12
bookproblem
ch11
misc
p499
q21
difficult
sec-b
math
0
votes
1
answer
If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$, then show that $ \large\frac{1}{p^2}$$= \large\frac{1}{a^2}$$ +\large\frac{1}{b^2}$
asked
May 18, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
exercise10-3
sec-c
q18
difficult
+1
vote
0
answers
A variable plane, which remains at a constant distance '3p' from the origin cuts the coordinate axes at A, B and C. Show that locus of the centroid of the triangle is \(\large \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2} .\)
asked
Jan 23, 2013
by
thanvigandhi_1
cbse
class12
modelpaper
2012
sec-c
q27
math
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
If $a$ and $b$ are roots of $x^2-3x+p=0$ and $c$ and $d$ are roots of $x^2-12x+q=0$, where $a,b,c,d$ form a G.P. then prove that $(q+p):(q-p)=17:15$
asked
Mar 31, 2014
by
rvidyagovindarajan_1
cbse
class11
ch9
bookproblem
misc
difficult
sec-c
q18
math
sequences-and-series
0
votes
1
answer
Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line $ x + y = 4$ may be at a distance of 3 units from this point
asked
May 21, 2014
by
thanvigandhi_1
cbse
class11
ch10
straight-lines
bookproblem
misc
sec-c
q16
difficult
0
votes
1
answer
The relation \(R\) in the set \(A\) of points in a plane given by \(R = \{(P, Q)\) : distance of the point \(P\) from the origin is same as the distance of the point \(Q\) from the origin\(\}\), is a) Reflexive only b) Symmetry C) both reflexive and symmetry but not transitive d) Is an equivalence relation. Also show that the set of all points related to a point Pis not equal to (0,0) is the circle passing through P with origin as centre.
asked
Jun 12, 2014
by
balaji.thirumalai
cl121
cbse
class12
bookproblem
ch1
sec1
q11
p6
easy
sec-a
math
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
...