Problem 35 - Algebra and Trigonometry

Question

Which of the points \(P(3,1)\) or \(Q(-1,3)\) is closer to the point \(R(-1,-1) ?\)

Step-by-Step Solution

Answer

Point Q(-1,3) is closer to point R(-1,-1).

Step 1: Understand the Problem

We need to determine which of the points \( P(3,1) \) or \( Q(-1,3) \) is closer to the point \( R(-1,-1) \). This involves calculating the distance from \( R \) to each of the two points and comparing the results.

Step 2: Recall the Distance Formula

The distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). We will use this to calculate the distance between \( R \) and each of the points \( P \) and \( Q \).

Step 3: Calculate Distance from Point P to R

Substitute \( x_1 = 3, y_1 = 1, x_2 = -1, y_2 = -1 \) into the formula. \[d = \sqrt{(-1 - 3)^2 + (-1 - 1)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]This is the distance from \( P \) to \( R \).

Step 4: Calculate Distance from Point Q to R

Substitute \( x_1 = -1, y_1 = 3, x_2 = -1, y_2 = -1 \) into the formula.\[d = \sqrt{(-1 + 1)^2 + (-1 - 3)^2} = \sqrt{(0)^2 + (-4)^2} = \sqrt{0 + 16} = \sqrt{16} = 4\]This is the distance from \( Q \) to \( R \).

Step 5: Compare the Distances

The distance from \( R \) to \( P \) is \( 2\sqrt{5} \) which is approximately \( 4.47 \). The distance from \( R \) to \( Q \) is \( 4 \). Since \( 4 \) is less than \( 4.47 \), point \( Q \) is closer to point \( R \).

Key Concepts

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, is a mathematical tool that links algebra and geometry. It allows us to use algebraic formulas to derive geometric properties. With coordinate geometry, we can:

• Locate points on a plane using coordinates, typically written as \( (x, y) \).
• Measure distances and angles between these points.
• Desribe shapes and size, among other applications.

To solve problems like determining the closest point to another, we utilize coordinate points that represent specific locations on the Cartesian plane. Each point has two values, an x-coordinate (horizontal) and a y-coordinate (vertical). This coordinate system simplifies understanding locations and various properties of shapes in a two-dimensional plane.

Calculating Distance Between Points

Understanding how to calculate the distance between two points is a fundamental concept in coordinate geometry. This is where the distance formula comes into play. The formula:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]is derived from the Pythagorean theorem and helps determine the straight-line distance between any two points, \( (x_1, y_1) \) and \( (x_2, y_2) \).
To apply this:

• Subtract the x-coordinates \( (x_2 - x_1) \) and square the result.
• Similarly, subtract the y-coordinates \( (y_2 - y_1) \) and square this result as well.
• Add these squared differences, and then take the square root of the sum.

This formula gives a precise measure of the distance between two points, which is crucial in various applications, such as navigation, engineering, and computer graphics.

Comparing Distances

Once you have calculated the distances between various points, comparing these distances allows you to draw conclusions about the positioning of the points relative to each other. When you wish to determine which point is closer to another, calculate each distance and then compare the resulting values.
For example, in the exercise, by using the distance formula, we calculated that the distance from \( R(-1,-1) \) to \( P(3,1) \) is \( 2\sqrt{5} \) while the distance to \( Q(-1,3) \) is 4.

• A smaller distance indicates proximity, showing one point is nearer to the reference point than the other.
• Thus, point Q is closer to point R since 4 is less than approximately 4.47 \( (2\sqrt{5} ≈ 4.47) \).

Analyzing and comparing such distances can be essential in practical applications, like finding the shortest path or determining proximity in navigation purposes.