Coordinate Plane Problem

milanproda wrote:The angle made by the line \sqrt{3}x+y+6=0, with y-axis is

1. 60 degrees
2. 30 degrees
3. 45 degrees
4. 120 degrees
5. None of the above


I was wondering, if I wanted to find the X intrecept, and made y equal to zero, the equation would become √3x=-6. If I squared the √3 and the -6, would the x intrecept become 36? Could I do that?

You have done some miscalculation. Otherwise the value of x-intercept cannot be 36.

Anyway, No need to square.
As √3x = -6 => x = -(6/√3) = -2√3
Thus x-intercept is -2√3.

We have x-intercept = -2√3 and from the equation y-intercept = -6.

Hence we can draw a figure like the following,



Now, if you know trigonometry, then tan(required angle) = (2√3/6) = 1/√3
Hence, required angle = 30 degrees.

And if you don't know trigonometry, the in the right-angled triangle ABC

• AC = 2√3
  BC = 6
  AB = 4√3

Thus the sides are in the ratio 1 : √3 : 2.
Hence it is a 30-60-90 triangle and angle ABC = 30 degrees

The correct answer is B.

milanproda wrote:Thank you for the reply, it makes much more sense now.

As for my question, I thought that if you squared a square root (in this case √3) that you would have to square everything else in the equation (-6)? So i presumed (falsey =( ) that √3x= -6^2 =3x=36 x=12.

Yes, your thought (the blue one) is correct but not your execution(the red one).

You should square x too.
Thus, √3x = -6
=> 3x² = 36
=> x² = 12
=> x = ±2√3

Thus squaring unnecessarily introduces x = 2√3, for which y is not zero. x = -2√3 is the correct one.