Question

Find the equation of the line which is at a perpendicularl distance of 5 units form the origin and the angle made by the perpendicular with the positive x- axis is 30° . ​

Answer 1

Solution :- The required equation of the line is

$xcos \alpha + y \cos( \alpha ) = p$

Here, p = 5 units and

[tex] \alpha = 30°

Thus , the required equation of the given line is

xcos30 ° + y sin30° = 5

or, √3x + y = 10 Answer ✔

Answer 2

Answer:

Let the length of the perpendicular be p from the origin,

Hence, angle at origin will be 30 degree = alpha

X cos alpha + y sin alpha = p

X cos 30 + y sin 30 = p

x/2 + y square root 3/2 = p

x + y square root 3 = 2p    eq(1)

now in the triangle that will be made with x axis,

cos 30 = 2p/ square root 3

now in the triangle that will be with y axis,

cos 60 = 2p

Hence, area = 50/ square root 3

½ * 2p/square root 3 * 2p = 50/ square root 3

P^2 = 25

P = +_ 5

Hence, equation will be square root of (3x) + y = +_ 10