Problem 7 - 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. $$(-2,-6) \text { and }(3,-4)$$

Step-by-Step Solution

Answer

The distance between the points (-2,-6) and (3,-4) is approximately 5.39

Step 1: Identify the coordinates

Point 1 has coordinates (-2,-6), so label \(x_1 = -2\) and \(y_1 = -6\). Point 2 has coordinates (3,-4), so label \(x_2 = 3\) and \(y_2 = -4\).

Step 2: Apply the distance formula

Now, you can substitute these values into the distance formula which is, \[ D = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 } \]. Substituting you get, \[ D = \sqrt{ (3 - (-2))^2 + ((-4) - (-6))^2} \].

Step 3: Simplify the equation

Simplify the equation to get the answer. After simplifying, \[ D = \sqrt{ 25 + 4 }\], \[ D = \sqrt{ 29 }\].

Step 4: Convert to decimal form

The final solution is in the simplified radical form. Converting the square root of 29 to decimal form and rounding off to 2 decimal places, gives approximately 5.39.

Key Concepts

Coordinate Geometry

Coordinate geometry is all about understanding points, lines, and shapes using a coordinate plane. This plane is like a big grid that helps us pinpoint exact locations with pairs of numbers called coordinates.

Each point on the plane is identified by an \(x\) and a \(y\) value, also known as an ordered pair. For example, in the exercise given, the points given are \( (-2, -6) \) and \( (3, -4) \). Here, \(x_1\) is \(-2\) and \(y_1\) is \(-6\); \(x_2\) is \(+3\) and \(y_2\) is \(-4\).

Using these coordinates, you can calculate distances, slopes, and even find out how different shapes behave. It's a vital part of understanding how geometry works in the real world.

Pythagorean Theorem

The Pythagorean theorem helps explain why the distance formula makes sense. Imagine you have two points plotted on a flat surface. You can draw a right triangle by using a horizontal line, a vertical line, and the direct line connecting the two points.

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides:

• Formula: \((a^2 + b^2 = c^2)\)
• \(a\) and \(b\) are the legs
• \(c\) is the hypotenuse

The distance formula is derived from this principle, where the length of the hypotenuse is the distance between the two points.

In our example, the horizontal distance is \((x_2 - x_1)\), and the vertical distance is \((y_2 - y_1)\). Plugging these differences into the Pythagorean theorem gives us the distance formula \(D = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 }\), which calculates the hypotenuse directly.

Radical Simplification

Radical simplification is the process of making a square root as simple as possible. This means breaking it down to its most basic form, without changing its value.

In the solution provided, we notice the expression \(\sqrt{29}\), which can't be simplified further as 29 is a prime number. So it stays as \(\sqrt{29}\) in the radical form. However, if the number inside the radical could be broken into smaller perfect squares, you could simplify it further.

• Example: \((\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2})\)

Simplifying radicals helps in working with roots more easily and usually shows the exact value that a calculator might approximate. For exercises like this, you would also express the radical in decimal form for practical use, which involves getting a calculator involved to find approximate values like 5.39 for \(\sqrt{29}\).

Always remember to check if the number inside the square root can be factored into squares for easier simplification!