IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

ishan.maheshwari

specify the method to solve the below given question.( as i know the answer but not the correct method to solve it )

find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes tan

= ( 5 / 12 ) WITH THE POSITIVE DIRECTION OF X AXIS.

ans : 12x + 5y =39

priyesh

use the normal form of equation of line which is

xcos

+ ysin

= p

where

is the angle which the perpendicular makes with with the positive axis of x & p is the perpendicular distance from the origin to the line.

therefore

x12/13 + y5/13 = 3

therefore 12x + 5y = 39

shivamk

HEY, This is a very simple question if u know how to write the normal form of equation of a line. Now if p is the perpendicular distance of the line from the origin & if

is the inclination (ie angle made with the positive direction of x-axis) of this

& not the line then equation of a line is given by xcos

+ysin

= p. This is a standard result.Using this we since tan

=5/12 . cos

=12/13 & sin

=5/13. substituting this in the normal form of eq of a line we get the ans to be 12x+5y=39.

PLZzzzzzzzzzz RATE ME .

him26.89

i have an alternative.

let the eqn be y=(5/12)x +c.

prepen. from origin:

c/(13)/12 =3.

hence the eqn is 12x+5y=39.

rajafox

use either the normal form or use general form.

normal form
xcos(o) + ysin(o) = p where p is distance from origin

Ask & Discuss Questions with Community & Experts