Problem 66 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (-2,8) && y=-3 x+2\end{array}$$

Step-by-Step Solution

Answer

The distance between the point (-2, 8) and the line \(y=-3x+2\) is \(\frac{6\sqrt{10}}{5}\).

Step 1: Identify the Coefficients and Point Coordinates

From the equation of the line (\(y = -3x + 2\)), we identify A=-3, B=1, and C=-2. The coordinates of the point are (-2, 8).

Step 2: Substitute the Values into the Distance Formula

Substitute the known values into the formula for distance between a point and a line. \[ d = \frac{|Ax1+By1+C|}{\sqrt{A^2 + B^2}} \]. Substitute A=-3, B=1, C=-2, x1=-2, y1=8 to get: \[ d = \frac{|(-3*-2)+(1*8)+(-2)|}{\sqrt{(-3)^2 + (1)^2}} \], which simplifies to \[ d = \frac{|6+8-2|}{\sqrt{9+1}} \].

Step 3: Calculate the Distance

Calculating the values in the formula results in \[d = \frac{12}{\sqrt{10}}\], which simplifies to \[d = \frac{6\sqrt{10}}{5}\].

Key Concepts

Point-Line Distance Formula

Understanding the Point-Line Distance Formula is crucial for solving geometry problems involving the shortest path from a point to a line. This formula provides us with a precise measurement of that shortest distance. The formula is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

In this formula, \((x_1, y_1)\) are the coordinates of the given point, and \(A\), \(B\), and \(C\) are the coefficients from the general form of the line's equation, \(Ax + By + C = 0\). To comprehend this better, consider the line's equation as a balance of distances, where the numerator of the formula calculates a scaled version of the distance by involving the coordinates of the point and the line's parameters. The denominator normalizes this value, taking into account the slope of the line given by \(A\) and \(B\).

Coordinate Geometry

Coordinate Geometry, also known as analytic geometry, is a bridge between algebra and geometry where we use a coordinate system to geometrically represent algebraic equations. It allows us to translate geometric problems into algebraic ones that we can solve using equations and formulae.

In coordinate geometry, the positioning of any point can be described with an ordered pair of numbers, such as \((-2, 8)\), where the first number represents the horizontal distance from the origin, and the second number represents the vertical distance. Lines are also described using equations like \(y = -3x + 2\), which narrate a relationship between all points on the line. The interplay between these points and lines through equations is what enables the solving of complex geometric problems, making coordinate geometry a foundational element in the field of mathematics, especially in pre-calculus.

Pre-calculus

Pre-calculus is the study of mathematical concepts and skills that prepare students for calculus, the mathematical study of change. It involves a wide variety of topics including functions, algebra, trigonometry, and geometry. Within the realm of Pre-calculus, using the point-line distance formula exemplifies how algebraic and geometric understandings are applied in unison to solve problems before approaching the limit-based calculations in calculus.

Problems like finding the distance between a point and a line provide a practical application of Pre-calculus concepts. The solution requires knowledge of simplifying expressions, manipulating square roots, and understanding absolute values. These skills are not only essential for solving such geometry problems but also serve as a stepping stone for the more intricate concepts encountered in calculus.