Problem 54 - Precalculus with Limits

Question

In Exercises 51-58, find the distance between the point and the line. \(\textit{Point}\) \((-2, 1)\) \(\textit{Line}\) \(x - y = 2\)

Step-by-Step Solution

Answer

The shortest distance from the point (-2, 1) to the line \(x - y - 2 = 0\) is \(2.5 \sqrt{2}\).

Step 1: Rewriting the line equation into standard form

The given line equation is \(x - y = 2\). We can rewrite it into standard form \(Ax + By + C = 0\) by subtracting \(x\) from both sides. This gives us \(x - y - 2 = 0\). Therefore, \(A = 1\), \(B = -1\) and \(C = -2\).

Step 2: Identify the coordinates of the point

The given point is \(-2, 1\), which means \(x = -2\) and \(y = 1\).

Step 3: Calculate the distance

The formula for the distance between a point and a line is:\[\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\]Inserting the known values yields:\[\frac{|(1)(-2) + (-1)(1) + (-2)|}{\sqrt{(1)^2 + (-1)^2}} \]which simplifies to:\[\frac{|-2 - 1 - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{5}{\sqrt{2}} = 2.5 \sqrt{2}\]

Step 4: Interpretation of the result

We find that the shortest distance from the point \(-2, 1\) to the line \(x - y - 2 = 0\) is \(2.5 \sqrt{2}\). This is a positive value, indicating that the point and line do not coincide and the point is not on the line.

Key Concepts

Distance formula

The distance formula is a mathematical equation used to determine the distance between two points in a plane. It relies on the Pythagorean theorem and is given by the formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Here,

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
• This formula calculates the straight-line distance or 'as-the-crow-flies' distance between them.

The distance formula is foundational in geometry as it helps measure the separation between points, which is essential in various applications, such as engineering and physics. When working with more complex geometric scenarios, such as the distance from a point to a line, this concept is extended and modified, as we will explore next.

Point-line distance

Finding the distance between a point and a line in coordinate geometry requires a specific formula that adapts the concept of perpendicularity. The shortest distance from a point to a line is always the perpendicular distance. The formula to find this distance is:\[\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\]where:

• \((x, y)\) are the coordinates of the point.
• \(Ax + By + C = 0\) is the standard form of the line equation.
• \(A\), \(B\), and \(C\) are coefficients from the line equation.

This formula calculates how far a point is from a line on the plane, effectively by measuring along a line perpendicular to the given line. Understanding this formula is key in problems where the "closeness" or "remoteness" of geometric entities is questioned.

Linear equations

Linear equations consist of terms that are either constants or the product of a constant and a single variable. They are foundational in mathematics as they describe straight lines in geometry. The general form of a linear equation is:\[Ax + By + C = 0\]where:

• \(A\), \(B\), and \(C\) are constants.
• \(x\) and \(y\) represent variables or coordinates on a plane.

One of the simplest forms of linear equations is the slope-intercept form, \(y = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. Converting equations between different forms is crucial when solving problems like finding point-line distances. Linear equations are integral to coordinate geometry, as they describe relationships between quantities and predict their behavior under certain conditions.

Coordinate geometry

Coordinate geometry, also known as analytic geometry, combines algebraic equations with geometric interpretations using a coordinate plane. It provides methods to describe geometric figures using algebra, which offers a way to solve geometric problems numerically.The coordinate plane is defined by two perpendicular axes:

• The horizontal axis (often called the x-axis)
• The vertical axis (often called the y-axis)

Points on this plane are defined by ordered pairs \((x, y)\) where \(x\) and \(y\) indicate the position along the axes. By using linear equations and the distance formula, coordinate geometry helps solve various spatial problems, including finding the distance between a point and a line, as illustrated in the original problem.Coordinate geometry opens up a world where numerical computation meets visual shape, bringing abstract concepts into tangible forms one can compute and graph.