logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the equation of line parallel to $3x-4y+2=0$ and at a distance 2 from it?

$\begin{array}{1 1}(a)\;3x-4y+12=0,3x-4y-8=0\\(b)\;3x-4y-8=0,3x-4y-8=0\\(c)\;3x-4y+10=0,3x-4y+8=0\\(d)\;3x+4y+12=0,3x+4y-8=0\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
For a line to be parallel to line $3x-4y+2=0$ it should have the same slope
$\therefore $ The required line will be $3x-4y+c=0$
Now,
Distance between two parallel lines is $\large\frac{\mid c_1-c\mid}{\sqrt{a^2+b^2}}$
Hence $2=\large\frac{\mid 2-c\mid}{\sqrt{3^2+4^2}}$
$\therefore \pm =\large\frac{2-c}{5}$
Taking +ve sign
$10-2=-c$
$c=-8$
Taking -ve sign
$-10=2-c$
$c=12$
$\therefore$ Required equations are $3x-4y+12=0,3x-4y-8=0$
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...