Finding the Equation of a Straight Line with Y-intercept and Angle of Inclination

The problem asks us to find the equation of a straight line given two key pieces of information:

• It cuts off an intercept of 2 from the negative direction of the y-axis.
• It is inclined at 30° with the positive direction of the x-axis.

Understanding the Given Information

Let's break down what the given information means in terms of the line's properties:

1. Y-intercept: An intercept of 2 from the negative direction of the y-axis means the line crosses the y-axis at the point (0, -2). So, the y-intercept (denoted by 'c') is -2.
2. Angle of Inclination: The angle of inclination (denoted by '\(\theta\)') is the angle formed by the line with the positive direction of the x-axis, measured counterclockwise. We are given that \(\theta = 30^\circ\).

Calculating the Slope of the Straight Line

The slope of a line (denoted by 'm') is related to its angle of inclination by the formula:

\(\qquad m = \tan(\theta)\)

In this case, the angle of inclination is \(30^\circ\). So, the slope is:

\(\qquad m = \tan(30^\circ)\)

We know the value of \(\tan(30^\circ)\) from trigonometry:

\(\qquad m = \frac{1}{\sqrt{3}}\)

Using the Slope-Intercept Form of a Line Equation

A common way to represent the equation of a straight line is the slope-intercept form, which is:

\(\qquad y = mx + c\)

where 'm' is the slope and 'c' is the y-intercept.

We have calculated the slope \(m = \frac{1}{\sqrt{3}}\) and the y-intercept is given as $c = -2$. Now, we substitute these values into the slope-intercept form:

\(\qquad y = \left(\frac{1}{\sqrt{3}}\right)x + (-2)\)

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

Converting to the General Form

The options provided are in the general form of a linear equation, which is $Ax + By + C = 0$. Let's rearrange our equation to match this form.

Start with the equation:

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

To eliminate the fraction, multiply the entire equation by \(\sqrt{3}\):

\(\qquad \sqrt{3}y = \sqrt{3}\left(\frac{x}{\sqrt{3}}\right) - 2\sqrt{3}\)

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

Now, move all terms to one side to get the general form $Ax + By + C = 0$:

\(\qquad 0 = x - \sqrt{3}y - 2\sqrt{3}\)

or

\(\qquad x - \sqrt{3}y - 2\sqrt{3} = 0\)

Comparing with Options

Let's compare our derived equation \(x - \sqrt{3}y - 2\sqrt{3} = 0\) with the given options:

• Option 1: \(x - 2\sqrt{3} y - 3\sqrt{2} = 0\) (Does not match)
• Option 2: \(x + 2\sqrt{3}y - 3\sqrt{2} = 0\) (Does not match)
• Option 3: \(x + \sqrt{3}y - 2\sqrt{3} = 0\) (Does not match)
• Option 4: \(x - \sqrt{3}y - 2\sqrt{3} = 0\) (Matches our derived equation)

Therefore, the equation of the straight line is \(x - \sqrt{3}y - 2\sqrt{3} = 0\).

Revision Table: Straight Line Equations

Additional Information: Slope and Angle

The slope of a straight line measures its steepness. A positive slope indicates the line rises from left to right, while a negative slope indicates it falls. The angle of inclination is always measured counterclockwise from the positive x-axis.

• If the line is parallel to the x-axis, its angle of inclination is \(0^\circ\) and its slope is \(\tan(0^\circ) = 0\).
• If the line is parallel to the y-axis, its angle of inclination is \(90^\circ\), and its slope \(\tan(90^\circ)\) is undefined. The equation of such a line is $x = k$, where k is a constant.
• The angle of inclination \(\theta\) always satisfies \(0^\circ \le \theta < 180^\circ\) for non-vertical lines.