Problem 9 - 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. $$(0,-3) \text { and }(4,1)$$

Step-by-Step Solution

Answer

The distance between the points (0,-3) and (4,1) is approximately 5.66 or the precise value in radical form is \(4\sqrt{2}\).

Step 1. Identify the coordinates

First, identify the x-coordinates and y-coordinates of each point. Let (0,-3) be \( (x_1, y_1) \) and (4,1) be \( (x_2, y_2)\). So, \(x_1 = 0\), \(y_1 = -3\), \(x_2 = 4\) and \(y_2 = 1\).

Step 2. Apply the Distance Formula

Next, substitute the coordinates into the distance formula \[d = \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\]. By plugging the values in, we get \[d = \sqrt{{(4 - 0)^2 + (1 -(-3))^2}}\].

Step 3. Simplify the expression

Simplify the expression. That gives, \[d =\sqrt{{(4)^2 + (4)^2}} = \sqrt{{16 + 16}} = \sqrt{{32}}\]. The simplified radical form of \(\sqrt{32}\) is \(4\sqrt{2}\).

Step 4. Round to two decimal places

Lastly, round off the decimal value of the distance up to two decimal places. The decimal form of \(\sqrt{32}\) or \(4\sqrt{2}\) is approximately 5.66.

Key Concepts

Coordinate Geometry

Coordinate geometry is a branch of geometry where we use a coordinate system to describe and analyze geometric shapes and their properties. In simple terms, it involves using numbers to represent points on a plane. Think of it as using a map, where streets and avenues create a grid, and each intersection point is identified using coordinates, like a pair of numbers.

In our exercise, we are given two points: (0, -3) and (4, 1). These points are represented by their coordinates, where each number represents a position on the x-axis or y-axis.

• The first number in the pair, often called x, tells you how far along the horizontal axis the point is.
• The second number in the pair, named y, shows the point's position on the vertical axis.

By understanding coordinate geometry, we can easily compute distances, find midpoints, and explore the geometric relationships between multiple points.

Simplified Radical Form

When calculating distances, sometimes the result is a radical or root, like a square root, which doesn't resolve to a simple whole number. That's where the simplified radical form becomes important. This is the process of breaking down a radical expression to its simplest form, which makes calculations easier to handle.

Let's look at our distance calculation, which gives us \[ \sqrt{32} \]. To simplify this, we need to factor it into perfect squares. We know that 32 can be broken into \[16 \times 2 \]. Since 16 is a perfect square (the square of 4), we can rewrite \[\sqrt{32}\] as \[\sqrt{16} \times \sqrt{2} = 4\sqrt{2} \].

Why is simplified radical form important? Besides making numbers more manageable, it helps in estimating sizes and values. This form is particularly useful in geometry, physics, and many areas of science where exact precision is needed, but working with decimals is not as clean or straightforward.

Distance Calculation

To determine the straight-line distance between two points in a coordinate plane, we use the distance formula. It's a mathematical representation derived from the Pythagorean theorem, which applies to right triangles.

The formula is:\[d = \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\]Here's a breakdown:

• The expression \((x_2 - x_1)\) measures how much the points differ horizontally, while \((y_2 - y_1)\) measures the vertical difference.
• By squaring these differences, we ensure that any negative values become positive (distances are always positive), and we prepare to use the Pythagorean theorem, as the squared sides of a triangle add up in this method.
• Finally, taking the square root finds the actual distance, making it 'real world' like the tape measure we might use to find a length.

In our example involving (0,-3) and (4,1), substituting these coordinates into the formula gave us a straightforward calculation. Resulting first as \[\sqrt{32}\], and simplifying to \[4\sqrt{2}\], it was straightforward to then use decimal equivalents for practical tasks, approximating it to about 5.66 when necessary for rounding.