Problem 21 - College Algebra

Question

For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth. \((19,12)\) and \((41,71)\)

Step-by-Step Solution

Answer

The distance is \(\sqrt{3965}\) or approximately 62.96.

Step 1: Write the Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In this case, the points are \((19, 12)\) and \((41, 71)\) so, \(x_1 = 19\), \(y_1 = 12\), \(x_2 = 41\), and \(y_2 = 71\).

Step 2: Substitute Data Points into Formula

Substitute each coordinate value into the distance formula: \[d = \sqrt{(41 - 19)^2 + (71 - 12)^2}\]Calculate the differences inside the parentheses: \[d = \sqrt{22^2 + 59^2}\]

Step 3: Calculate Squares

Calculate the squares of the differences: \[22^2 = 484\]and\[59^2 = 3481\]Substitute them back into the radical: \[d = \sqrt{484 + 3481}\]

Step 4: Sum the Values Under the Radical

Now, add the squared differences: \[484 + 3481 = 3965\]Substitute the sum back into the distance formula: \[d = \sqrt{3965}\]

Step 5: Simplify the Radical Expression

Find the simplest radical form of \(\sqrt{3965}\), if applicable. Here,\(\sqrt{3965}\) is approximately equal to 63. This cannot be simplified further.

Step 6: Calculate the Approximate Distance

Use a calculator to find the approximate value of \(\sqrt{3965}\):\[\sqrt{3965} \approx 62.964\]Rounding this to the nearest hundredth, we obtain:\[62.96\]

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that studies geometric figures using a coordinate system. It allows us to describe shapes and their properties by using algebraic equations. In coordinate geometry, points are defined using pairs of numbers, often referred to as

• coordinates
• ordered pairs

These pairs represent the horizontal (x-axis) and vertical (y-axis) positions on a two-dimensional plane. For example, a point represented by \((19, 12)\) indicates it is 19 units along the x-axis and 12 units up on the y-axis.

Coordinate geometry is fundamental in finding various measurements like distance and slope between points. By applying basic algebraic manipulation, you can determine exact spaces between points without needing to draw or visualize the figures.

Pythagorean Theorem

The Pythagorean theorem is an essential principle in mathematics that helps us find distances in a coordinate system. It's based on a right triangle relationship in which the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

• Formula: \( a^2 + b^2 = c^2 \)
• Here, \(c\) is the hypotenuse, while \(a\) and \(b\) represent the other two sides.

This theorem is the backbone of the distance formula used in coordinate geometry. For example, when calculating the distance between two points, you essentially create a right triangle where the horizontal and vertical differences between points act as the two shorter sides of the triangle.

By following this theorem, you can easily find the diagonal distance, which is crucial for various applications in geometry and physics.

Distance Calculation

Distance calculation between two points in coordinate geometry is straightforward using the distance formula. This mathematical approach helps determine the shortest path between two locations on a coordinate plane. The formula is derived from the Pythagorean theorem and is written as:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

In this formula,

• \( (x_1, y_1) \) and \( (x_2, y_2) \)represent the coordinates of the two points, with the differences in \(x\)and \(y\)being calculated first.
• These differences are squared to eliminate any negative values, ensuring all distances remain positive credit to the principle of absolute values.

Combining these squared differences under a square root gives the precise distance between the two points. This formula is widely used in navigation, physics, and computer graphics, linking algebra with spatial concepts.

Radicals and Simplification

Radicals deal with root expressions, like square roots, that often arise in distance calculations. Simplifying radical expressions means reducing them to their most basic form without changing their value. This occurs when you calculate square roots, and the process can be divided into:

• Identifying perfect squares: This involves factoring out any perfect squares under the radical to facilitate simplification.
• Approximation: If exact measurement isn't possible, a rough decimal approximation can be sought using calculators.

For instance, when you calculated the distance \(\sqrt{3965}\), you checked if there were any perfect squares that divide it, which there weren't. The value was then approximated to around \(62.964\).
Simplification of radicals often makes results more usable in practical contexts. However, maintaining exactness (or expressing in terms of simplest radicals) is pivotal in mathematics to retain the accuracy needed for further computations.