Problem 62 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (1,4) && y=4 x+2\end{array}$$

Step-by-Step Solution

Answer

The distance between the point (1,4) and the line y = 4x + 2 is \(\frac{2\sqrt{17}}{17}\).

Step 1: Identify the necessary parameters

From the given problem, the line is represented as y = 4x+2, let's rearrange it into the generic line form ax + by + c = 0. In rearranging, we obtain -4x + y - 2 = 0. From this, a equals -4, b equals 1, and c equals -2. The point given is (1,4). Hence, x1 equals 1 and y1 equals 4.

Step 2: Substitute the parameters into the formula

Next, substitute the values of a, b, c, x1 and y1 into the formula \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}\). That results into \(d = \frac{|-4*1 + 1*4 - 2|}{\sqrt{(-4)^2+1^2}}\). Perform the necessary calculations.

Step 3: Find the absolute value and simplify the equation

Performing the absolute value operations, we get \(d = \frac{|-2|}{\sqrt{17}}\). Subsequently, calculating the absolute value yields \(d = \frac{2}{\sqrt{17}}\). Lastly, rationalizing the denominator leads to the final answer.

Step 4: Final answer

The distance between the point and the line is \(\frac{2\sqrt{17}}{17}\).

Key Concepts

Line Equation

Understanding the equation of a line is crucial when finding the distance from a point to that line. In the exercise, the equation given is of the form \(y = 4x + 2\). This is a linear equation, representing a line in slope-intercept form, where the slope is the coefficient of \(x\) and the intercept is the constant term. For distance calculations, we often need to rearrange it into the standard line equation \(ax + by + c = 0\).

In our case, rearranging \(y = 4x + 2\) gives us \( -4x + y - 2 = 0\), providing us with the values \(a = -4\), \(b = 1\), and \(c = -2\) necessary for further calculations. Understanding how to manipulate this equation is essential in many geometry and algebra problems.

Point Coordinates

A point in a Cartesian plane is defined by its coordinates, which are usually written as a pair \( (x, y) \). In our textbook exercise, the point provided is \( (1, 4) \) where \(1\) is the x-coordinate and \(4\) is the y-coordinate. These coordinates pinpoint the location of the point relative to the origin \( (0,0) \).

In context, these coordinates (\(x_1\) and \(y_1\)) are substituted into the distance formula. The x-coordinate helps in understanding the horizontal position, while the y-coordinate reveals the vertical position, both of which are fundamental when relating a point to a line.

Absolute Value

The absolute value of a number is its distance from zero on the number line, regardless of the direction. It's denoted by two vertical bars, like \( |x| \), and it always returns a non-negative value. For example, both \( |-2| \) and \( |2| \) are equal to \(2\).

In our exercise, absolute value helps us ensure that the distance, which cannot be negative, is calculated correctly. We applied the absolute value to the numerator of the distance formula, which resulted in \( | -4*1 + 1*4 - 2 | \), simplifying down to \( | -2 | \), and thereafter evaluated as \( 2 \).

Rationalizing the Denominator

Rationalizing the denominator in a fraction means removing any irrational numbers, like square roots, from the denominator. This process involves multiplying the numerator and the denominator by a number that will eliminate the square root in the denominator.

For instance, to rationalize \( \frac{2}{\sqrt{17}} \), we multiply the top and bottom by \( \sqrt{17} \) to get \( \frac{2\sqrt{17}}{17} \), which is the final step in our textbook solution. This process doesn't change the value of the fraction, but it does make it cleaner and often easier to understand or further manipulate in algebraic expressions.