Problem 11 - Precalculus

Question

Which point is farther from the origin? (a) (3,-2) or \(\left(4, \frac{1}{2}\right)\) (b) (-6,7) or (9,0)

Step-by-Step Solution

Answer

The point \( \left(4, \frac{1}{2}\right) \) is farther for (a), and (-6, 7) is farther for (b).

Step 1: Understand the problem context

To determine which point is farther from the origin, we need to calculate the distance of each point from the origin (0,0).

Step 2: Distance formula

The distance formula from a point \((x, y)\) to the origin \((0,0)\) is \(d = \sqrt{x^2 + y^2}\). Calculate the distance for each point.

Step 3: Calculate distance for point (3, -2)

Using the formula: \[d = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\approx 3.61.\]

Step 4: Calculate distance for point \((4, \frac{1}{2})\)

Using the formula: \[d = \sqrt{4^2 + \left(\frac{1}{2}\right)^2} = \sqrt{16 + \frac{1}{4}} = \sqrt{16.25}\approx 4.03.\]

Step 5: Compare distances for part (a)

For points (3, -2) and \(\left(4, \frac{1}{2}\right)\), since \(4.03 > 3.61\), point \(\left(4, \frac{1}{2}\right)\) is farther from the origin.

Step 6: Calculate distance for point (-6, 7)

Using the formula: \[d = \sqrt{(-6)^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85}\approx 9.22.\]

Step 7: Calculate distance for point (9, 0)

Using the formula: \[d = \sqrt{9^2 + 0^2} = \sqrt{81} = 9.\]

Step 8: Compare distances for part (b)

For points (-6, 7) and (9, 0), since \(9.22 > 9\), point (-6, 7) is farther from the origin.

Key Concepts

Distance Formula

The distance formula is a mathematical expression used to calculate the distance between two points in a coordinate plane. It comes in handy particularly when you need to find how far a point is from the origin. The origin in a coordinate plane is the point (0,0). When using the distance formula, you measure directly from one point to another, allowing you to see this distance as the crow flies, so to speak.

The formula is expressed as:

• \[d = \sqrt{x^2 + y^2}\]

Here, \(d\) stands for distance, while \(x\) and \(y\) represent the coordinates of the point.
To illustrate, let's use the point (3, -2). First, square each value: \(3^2 = 9\) and \((-2)^2 = 4\). Next, sum these values: \(9 + 4 = 13\). Finally, find the square root: \(\sqrt{13} \approx 3.61\). This means the point (3, -2) is approximately 3.61 units away from the origin. This systematic approach using the distance formula helps students understand point-to-origin calculations clearly.

Coordinate Geometry

Coordinate geometry, often termed analytical geometry, is a branch of geometry where algebra is used to resolve geometric problems. This discipline involves representing points, lines, and shapes in a plane using ordered pairs (x, y). Coordinate geometry combines tools from both algebra and geometry, allowing for improved understanding and problem-solving capabilities.

Key components of coordinate geometry include:

• Points - Defined by coordinates on a graph.
• Lines - Made up of points and usually graphed as linear equations.
• Distance - Measured using the distance formula.

Detecting the distance of a point from the origin (0,0) involves these principles, making this area of study essential for comprehending how geometric configurations work in algebraic terms. Using coordinates simplifies complex spatial problems into algorithms and formulas that are straightforward to handle. When comparing points like \( -6, 7 \) and \( 9, 0 \), coordinate geometry allows you to instantly calculate and compare which is farther using algebraic techniques. This organized method translates visual geometric information into numbers and symbols, easing the problem-solving process.

Point Comparison

When comparing points in coordinate geometry, especially in relation to the origin, the goal is to ascertain which point is farthest away. This is done by calculating the distance of each point from the origin and comparing these distances. This process is straightforward and relies heavily on the distance formula, allowing for unbiased and clear comparisons.

Steps to compare include:

• Calculate distance for each point using the distance formula \(d = \sqrt{x^2 + y^2}\).
• Compare the resulting values.
• Determine which value is greater to identify the farther point.

For example, with points like (3, -2) and \(4, \frac{1}{2}\), calculating their distance and then comparing will indicate which one is farther from \(0, 0 \). In real-world applications, this comparison can help in fields ranging from computer graphics to geography to determine placement or routing, highlighting the practical utility of such mathematical techniques.