Question

find the equation of the line on which the perpendicular from origin makes an angle of 30 degree with the x-axis and which forms a triangle of area 50 by root 3 square units with the axis

Answer 1

"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

"

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