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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
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
ParulG:
hey..thanks but can u explain after equation no. 1
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