Test Index

KEAM 2010 Math Paper

© examsnet.com

Question : 116 of 120

Marks: +1, -0

The equation of one of the lines parallel to

4x - 3y = 5

and at a unit distance from the point (-1,-4) is

$3x + 4y - 3 = 0$
$3x + 4y + 3 = 0$
$4x - 3y + 3 = 0$
$4x - 3y - 3 = 0$
$4x - 3y - 4 = 0$

Solution:

The equation of line parallel to

4x - 3y = 5

is

4x - 3y + \lambda = 0

Now, the distance of point

(-1,-4)

from line

4x - 3y + \lambda = 0

is

\;\frac{4\cdot(-1)-3\cdot(-4)+\lambda}{\sqrt{4^{2}+(-3)^{2}}}

=\;\frac{8+\lambda}{5}

According to questions

\;\frac{8+\lambda}{5}=1

\Rightarrow\;\;\lambda=-3

\therefore

Equation of required line is

4x - 3y - 3 = 0

© examsnet.com

Go to Question:

Prev Question

Next Question