Finding the Sum of Intercepts of a Line

The problem asks us to find the sum of the x and y intercepts of a straight line. We are given information about the line's distance from the origin and the angle its normal makes with the positive x-axis. This information is perfect for using the normal form of the equation of a line.

Understanding the Normal Form of a Line

The equation of a line in normal form is given by:

\(x \cos \alpha + y \sin \alpha = p\)

where:

• \(p\) is the perpendicular distance of the line from the origin.
• \(\alpha\) is the angle that the perpendicular (normal) from the origin to the line makes with the positive direction of the x-axis.

In this question, we are given:

• Perpendicular distance from origin, \(p = 4\) units.
• Angle the normal makes with the positive x-axis, \(\alpha = 15^\circ\).

Calculating Cosine and Sine of 15°

To use the normal form, we need the values of \(\cos 15^\circ\) and \(\sin 15^\circ\). We can calculate these using trigonometric identities for the difference of angles:

\(\cos (A - B) = \cos A \cos B + \sin A \sin B\)

\(\sin (A - B) = \sin A \cos B - \cos A \sin B\)

Using \(15^\circ = 45^\circ - 30^\circ\):

\(\cos 15^\circ = \cos (45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ\)

\(\cos 15^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

\(\sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ\)

\(\sin 15^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}\)

Writing the Equation of the Line

Now, substitute \(p = 4\), \(\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}\), and \(\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}\) into the normal form equation:

\(x \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + y \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = 4\)

Multiply both sides by 4 to simplify:

\(x (\sqrt{6} + \sqrt{2}) + y (\sqrt{6} - \sqrt{2}) = 16\)

Finding the Intercepts

The x-intercept is the value of \(x\) when \(y=0\). Substitute \(y=0\) into the equation:

\(x (\sqrt{6} + \sqrt{2}) + 0 (\sqrt{6} - \sqrt{2}) = 16\)

\(x (\sqrt{6} + \sqrt{2}) = 16\)

\(x = \frac{16}{\sqrt{6} + \sqrt{2}}\)

To rationalize the denominator, multiply the numerator and denominator by the conjugate \(\sqrt{6} - \sqrt{2}\):

\(x = \frac{16}{\sqrt{6} + \sqrt{2}} \times \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{16(\sqrt{6} - \sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} = \frac{16(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{16(\sqrt{6} - \sqrt{2})}{4}\)

\(x\text{-intercept} = 4(\sqrt{6} - \sqrt{2})\)

The y-intercept is the value of \(y\) when \(x=0\). Substitute \(x=0\) into the equation:

\(0 (\sqrt{6} + \sqrt{2}) + y (\sqrt{6} - \sqrt{2}) = 16\)

\(y (\sqrt{6} - \sqrt{2}) = 16\)

\(y = \frac{16}{\sqrt{6} - \sqrt{2}}\)

To rationalize the denominator, multiply the numerator and denominator by the conjugate \(\sqrt{6} + \sqrt{2}\):

\(y = \frac{16}{\sqrt{6} - \sqrt{2}} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \frac{16(\sqrt{6} + \sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} = \frac{16(\sqrt{6} + \sqrt{2})}{6 - 2} = \frac{16(\sqrt{6} + \sqrt{2})}{4}\)

\(y\text{-intercept} = 4(\sqrt{6} + \sqrt{2})\)

Calculating the Sum of Intercepts

The sum of the intercepts is the x-intercept plus the y-intercept:

\(\text{Sum} = x\text{-intercept} + y\text{-intercept}\)

\(\text{Sum} = 4(\sqrt{6} - \sqrt{2}) + 4(\sqrt{6} + \sqrt{2})\)

\(\text{Sum} = 4\sqrt{6} - 4\sqrt{2} + 4\sqrt{6} + 4\sqrt{2}\)

\(\text{Sum} = (4\sqrt{6} + 4\sqrt{6}) + (-4\sqrt{2} + 4\sqrt{2})\)

\(\text{Sum} = 8\sqrt{6} + 0\)

\(\text{Sum} = 8\sqrt{6}\)

The sum of the intercepts of the line is \(8\sqrt{6}\).

Revision Table: Line Equation Concepts

Additional Information: Trigonometric Values and Line Properties

Understanding special angles and trigonometric identities is crucial for solving many geometry and coordinate geometry problems. The values of \(\cos 15^\circ\) and \(\sin 15^\circ\) were derived using angle subtraction formulas. Similarly, values for other angles like \(75^\circ\), \(105^\circ\), etc., can be found. Also, the relationship between different forms of the line equation allows converting from one form to another based on the given information.

For example, once we have the equation in the normal form \(x \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + y \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = 4\), we can convert it to the intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) to directly read the intercepts. Dividing the entire equation by 4:

\(\frac{x}{4 / \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)} + \frac{y}{4 / \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)} = 1\)

\(\frac{x}{16 / (\sqrt{6} + \sqrt{2})} + \frac{y}{16 / (\sqrt{6} - \sqrt{2})} = 1\)

This directly shows the x-intercept \(a = \frac{16}{\sqrt{6} + \sqrt{2}}\) and the y-intercept \(b = \frac{16}{\sqrt{6} - \sqrt{2}}\), which we calculated earlier by setting y=0 and x=0 respectively. The sum \(a+b\) is what was required.