Reduce the equation `sqrt3y+y+2=0` to
(i) slope-intercept form and final the slope and y-intercept.
(ii) intercepts form and find the intercepts on the axes.

Explanation

(a) `sqrt3x+y+2=0`
`y=-sqrt3x-2`
This is the slope interxcept from of the line.
`therefore` Slope `=-sqrt3` and y intercept `=-2`
(b)` sqrt2x+y+2=0`
`((x)/(-2))/(sqrt3)+(y)/(-2)=1`
This is the intercept from of the given line
`therefore` x-intercept `=(-2)/(sqrt3)` and y-intercept `=-2`
(c ) `sqrt3x+y+2=0`
`sqrt3x+y=-2`
`implies(sqrt3)/(sqrt((sqrt3)^(2)+1^(2)))+(y)/(sqrt((sqrt3)^(2)+1^(2)))=(-2)/(sqrt((sqrt3)^(2)+1^(2)))`
`implies(sqrt3)/(2)xx+y/2=(-2)/(2)`
`implies(sqrt3)/(2)xx+1/2y=-1`
`implies(-sqrt3)/(2)-1/2y=1`
This is the normal form of the given line.
`therefore cos omega=(-sqrt3)/(2),sin omega=(-1)/(2),,p=+1.`
Since `cos omega and sin omegta` both are-negative, `omega` lies in `3^(rd)` quadrant.
`cos omega=(-sqrt3)/(2)=cos 210^(@)`
`omega=210^(@)`