Question - Finding the Equation of a Line with a Given Point and Angle

The equation of a line in point-slope form is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \((x_1, y_1)\) is a point on the line.

We are given the point \((1, 2)\) and that the line makes a \(30^\circ\) angle with the x-axis. The slope \(m\) of the line is the tangent of the angle it makes with the x-axis, thus \(m = \tan(30^\circ) = \frac{\sqrt{3}}{3}\).

Substituting the point and the slope into the point-slope form equation, we get:

\(y - 2 = \frac{\sqrt{3}}{3}(x - 1)\)

Multiplying both sides by 3 to clear the fraction, we get:

\(3(y - 2) = \sqrt{3}(x - 1)\)

Expanding, we have:

\(3y - 6 = \sqrt{3}x - \sqrt{3}\)

Adding 6 and \(\sqrt{3}\) to both sides gives us the final equation:

\(3y = \sqrt{3}x + 6 - \sqrt{3}\)

Or we can express it in standard form \(Ax + By = C\) by rearranging terms:

\(\sqrt{3}x - 3y = -6 + \sqrt{3}\)