Problem 12 - Precalculus Essentials

Question

Find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimals places. $$(2.6,1.3) \text { and }(1.6,-5.7)$$

Step-by-Step Solution

Answer

The distance between the two points (2.6, 1.3) and (1.6, -5.7) is approximately 7.07 units.

Step 1: Identify the coordinates

First, identify the x and y coordinates of each point. The coordinates for the first point are (x1, y1) = (2.6, 1.3) and for the second point are (x2, y2) = (1.6, -5.7).

Step 2: Apply the Distance Formula

Substitute these coordinates into the distance formula, \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). This becomes \(d=\sqrt{(1.6-2.6)^2+(-5.7-1.3)^2}\).

Step 3: Calculate the differences

Next, calculate the differences for each set of coordinates. This gives \(d=\sqrt{(-1)^2 + (-7)^2}\).

Step 4: Simplify and Solve for d

Simplify the equation and solve for 'd'. This gives \(d=\sqrt{1 + 49}\), which simplifies to \(\sqrt{50}\).

Step 5: Round Off

Round off the distance to two decimal places. This will depend on the simplified form of \(d\). If the precise form is used, \( \sqrt{50} \approx 7.07 \).

Key Concepts

Understanding Coordinates

In geometry, coordinates are used to locate points on a plane using a pair of numerical values. Each point on the plane has a set of coordinates expressed as \(x, y\). The first number is the x-coordinate, which tells you how far along the horizontal axis the point is. The second number is the y-coordinate, which indicates how far along the vertical axis the point is.

In our example:

• The first pair of coordinates is \(2.6, 1.3\), where 2.6 is the x-coordinate, and 1.3 is the y-coordinate.
• The second pair of coordinates is \(1.6, -5.7\), where 1.6 is the x-coordinate, and -5.7 is the y-coordinate.

Knowing how to read and interpret these values is crucial to using the distance formula effectively. The formula helps you calculate how far apart these two points are on the plane by considering these coordinates.

Radical Simplification

When you solve for distance using the distance formula, often you'll end up with a number under a square root sign, known as a radical expression. Simplifying this radical is essential for making calculations easier or more concise.

In our exercise, after applying the distance formula, we ended up with \(\sqrt{50}\\). Numbers like 50 can often be expressed in simpler terms by factoring out perfect squares.

• The number 50 can be broken down into 2 and 25, where 25 is a perfect square.
• You can simplify \(\sqrt{50}\\) to \(\5\sqrt{2}\). This means multiplying 5 by the square root of 2, since \(\\sqrt{25} = 5\).

Even if the result of your calculation will eventually be rounded, simplifying the radical first helps ensure that all estimations are as accurate as possible. This is a vital step in any mathematical solution involving radicals.

Rounding Decimals

After simplifying a radical in the distance calculation, sometimes the expression needs to be converted into a decimal form, especially when a more understandable number is required. Decimal rounding is used to trim lengthy decimal sequences down to a manageable number of decimal places, while still retaining an accurate approximation.

In rounding to two decimal places, you always observe the third decimal place to decide how to round the second. If it's 5 or greater, you round up; otherwise, you round down.
Using our example value of \(\sqrt{50}\\), the decimal equivalent, when calculated, is approximately 7.071. If we round this to two decimal places, it becomes 7.07.

• Note that 7.071 was rounded down to 7.07 because the third decimal place (1) was less than 5.

Rounding decimals is often the final step in expressing a distance calculation that can be easily communicated and understood.