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.
Well the first thing that came to my mind was to do rise/run and doing that I got 6/5 which I added together to get 11 but since that's obviously not correct I assumed it was a little under so the answer I got was 7.8
