Finding the Equation of a Straight Line: 30 Degrees with the Positive X-Axis and Passing Through (5, 1)

Understanding the Problem

We are given a scenario where we need to find the equation of a straight line that makes a 30-degree angle with the positive x-axis and passes through the point (5, 1). This problem involves several mathematical concepts including trigonometry, the point-slope form of a line, and basic algebra.

Calculating the Slope

The slope of a line can be determined using the angle it makes with the positive x-axis. The formula for the slope, m, given an angle θ is:

m tan(theta)

For θ 30 degrees, we use the tangent value of 30 degrees. The tangent of 30 degrees is given by:

tan(30^circ) frac{1}{sqrt{3}} approx 0.5774

Using the Point-Slope Form

The point-slope form of the equation of a line can be written as:

y - y_1 m(x - x_1)

Here, (x_1, y_1) is a point on the line. Given that our line passes through the point (5, 1), we substitute the values:

y - 1 frac{1}{sqrt{3}}(x - 5)

Simplifying the Equation

To simplify the equation, we distribute the slope:

y - 1 frac{1}{sqrt{3}}x - frac{5}{sqrt{3}}

Adding 1 to both sides of the equation, we get:

y frac{1}{sqrt{3}}x 1 - frac{5}{sqrt{3}}

Combining the terms on the right side:

y frac{1}{sqrt{3}}x (1 - frac{5}{sqrt{3}})

We simplify the constant term:

1 - frac{5}{sqrt{3}} frac{sqrt{3}}{sqrt{3}} - frac{5}{sqrt{3}} frac{sqrt{3} - 5}{sqrt{3}}

Thus, the equation of the line in slope-intercept form is:

y frac{1}{sqrt{3}}x frac{sqrt{3} - 5}{sqrt{3}}

Alternative Forms of the Equation

In addition to the slope-intercept form, the equation can be written in other forms. For instance, the standard form of the equation can also be derived:

sqrt{3}y - 1 x - 5

This can be rearranged to:

sqrt{3}y - x 1 - 5

sqrt{3}y - x -4

Each form provides a different perspective on the same line and can be useful in different contexts.