Problem 60 - Precalculus

Question

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

Step-by-Step Solution

Answer

The distance between the point and the line is \(\frac{3\sqrt{2}}{2}.\)

Step 1: Converting Line Equation

First, convert the equation of the line into standard form. The line equation provided is \(y = x + 2\). By rewriting, it can be changed to \(x - y + 2 = 0\). Here, A = 1, B = -1, and C = 2.

Step 2: Substitute Point Coordinates and Coefficients Into Formula

Now substitute the coordinates of the point (2,1) and coefficients into the distance formula. So, \(D = \frac{|1*2 + (-1)*1 + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|2 - 1 + 2|}{\sqrt{1 + 1}} = \frac{|3|}{\sqrt{2}} = \frac{3}{\sqrt{2}}.\)

Step 3: Rationalizing Denominator

Finally, to rationalize the denominator, multiply both numerator and denominator by \(\sqrt{2}\), which yields the distance \(D = \frac{3\sqrt{2}}{2}.\)

Key Concepts

Distance Formula

The distance formula is used to calculate the shortest distance from a point to a line. This concept is crucial in geometry and analytical mathematics, which helps to find how far apart two entities are. When working with a line in a two-dimensional space, you'll employ the specific formula: \\[\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\] \where: \

\
• \(A, B, C\) are the coefficients from the line equation in standard form \(Ax + By + C = 0\).
\
• \((x_1, y_1)\) represent the coordinates of the point.
\

\By plugging the values directly into this formula, you can obtain the perpendicular distance efficiently. The absolute value in the numerator ensures that the distance remains non-negative, irrespective of the relative positions of the line and the point. This formula simplifies the process by letting you avoid dealing with trigonometry, requiring only basic algebraic operations. Additionally, the square root in the denominator normalizes the direction vector’s length, ensuring consistency and accuracy in measurements regardless of the line's slope.

Standard Form of Line Equation

The standard form of a line equation provides a consistent way of expressing a line using three coefficients. For a line given by \(y = mx + b\), the standard form is structured as \(Ax + By + C = 0\). \

\
• 'A' denotes the coefficient of \(x\) and should preferably be a non-negative integer.
\
• 'B' represents the coefficient of \(y\), ensuring that 'By' contrasts with 'Ax' to maintain balance between the two variables.
\
• 'C' is the constant term which acts as an offset from the origin.
\

\The transformation from slope-intercept form, \(y = mx + b\), to standard form involves shifting terms from one side of the equation to the other. In this process, every term is adjusted to ensure that it fits perfectly into \(Ax + By + C = 0\). This methodology grants clarity in calculations, especially when coupled with formulas that require these coefficients for further computations. Moreover, the standard form is ideal for analyzing relationships between lines and checking for parallelism or perpendicularity due to its symmetric nature.

Rationalizing Denominators

Rationalizing denominators involves transforming an expression to remove any radicals (like square roots) from the denominator. This process makes expressions simpler and more coherent, especially useful when precise values are needed. In mathematics, expressions are more easily manipulated when free of radicals in the denominator due to the rational nature of whole numbers and integers. \For example, given a fraction like \(\frac{3}{\sqrt{2}}\), you would rationalize it by multiplying both the numerator and the denominator by \(\sqrt{2}\). This gives: \\[\frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\] \The rationale behind multiplying by \(\sqrt{2}\) is because \(\sqrt{2} \times \sqrt{2} = 2\), a whole number, clear of radicals. The end result, \(\frac{3\sqrt{2}}{2}\), is a cleaner version mathematically as it does not involve irrational numbers in the denominator. Historically, this method is a standard practice in mathematics to promote clarity and precision in results, ensuring expressions are in a more refined form suitable for further analysis and interpretation.