What is the standard form of the equation of a line for which the length of the normal segment to the origin is 8 and the normal makes an angle of 120° with the positive x-axis?

Slope of line = tan(120) = -tan(60) = - √3
Distance from origin = 8

Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0

Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0)  =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3

Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0

Check:
slope = -1/√3  ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8  ok.

Similarly C=-16 will satisfy the given conditions.

Answer  The required equations are
√3 x + y = ± 16 
in standard form.

You can conveniently convert to point-slope form if you wish.

The percent increase is 25%