Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-4\), plot the given points in the coordinate plane and then find the distance between them. \((-1,5),(6,3)\)

Step-by-Step Solution

Answer

The distance between the points \((-1,5)\) and \((6,3)\) is \(\sqrt{53}\), approximately 7.28.

Step 1: Understand the Problem

We are given two points, \((-1,5)\) and \((6,3)\), and we need to find the distance between them in the coordinate plane.

Step 2: Recall the Distance Formula

The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) in the coordinate plane is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].

Step 3: Substitute the Coordinates

Substitute the given points \((-1,5)\) and \((6,3)\) into the distance formula. Here, \(x_1 = -1\), \(y_1 = 5\), \(x_2 = 6\), and \(y_2 = 3\): \[ d = \sqrt{(6 - (-1))^2 + (3 - 5)^2} \].

Step 4: Simplify the Expressions Inside the Square Root

Calculate the differences and square them: \(6 - (-1) = 7\) and \((3 - 5) = -2\). Square these results: \(7^2 = 49\) and \((-2)^2 = 4\).

Step 5: Complete the Distance Calculation

Add the squared differences and take the square root: \[ d = \sqrt{49 + 4} = \sqrt{53} \].

Step 6: Approximate if Necessary

The distance \(d\) simplifies to \(\sqrt{53}\), which is approximately 7.28 when rounded to two decimal places.

Key Concepts

Coordinate Plane

The coordinate plane is a two-dimensional surface where points can be plotted. It consists of a horizontal line called the x-axis and a vertical line called the y-axis. These axes intersect at a point called the origin, marked as \(0, 0\). The plane is divided into four quadrants:

• Quadrant I: Positive x and y values.
• Quadrant II: Negative x values, positive y values.
• Quadrant III: Negative x and y values.
• Quadrant IV: Positive x values, negative y values.

Each location on the plane is identified by a pair of coordinates \((x, y)\). For example, the point \((-1, 5)\) means starting at the origin, move left 1 unit along the x-axis (since -1 is negative) and 5 units up along the y-axis. Similarly, the point \(6, 3)\) means you move 6 units right and 3 units up. Knowing how to locate points helps us determine their positions in relation to each other, which is essential for distance calculation.

Distance Calculation

Distance calculation between two points in a coordinate plane relies on the Distance Formula. This formula calculates the length of the straight line segment connecting the two points. To apply this formula, you need the coordinates of both points:

• If you have points \((x_1, y_1)\) and \(x_2, y_2)\).
• The formula for their distance \(d\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This formula finds the difference between the x-coordinates ( you could say, their horizontal distance) and the y-coordinates (their vertical distance).
Then, squares both differences to eliminate negative results and to calculate the area of squares formed by these differences.
Lastly, it adds these squares to get the diagonal square length, and finally takes the square root to find the actual distance.
Remember, a calculator can help approximate the square root for practical use.

Distance Between Points

Finding the distance between two points involves substituting their coordinates into the Distance Formula. Using the points \((-1,5)\) and \(6,3)\), apply each value properly:

• Assign coordinates: \(x_1 = -1, y_1 = 5, x_2 = 6, y_2 = 3\).
• Substitute into the formula: \[ d = \sqrt{(6 - (-1))^2 + (3 - 5)^2} \]
• Calculate: \(6 - (-1) = 7\) and \(3 - 5 = -2\).
• Square the results: \(7^2 = 49\) and \((-2)^2 = 4\).
• Find the final distance: \[ d = \sqrt{49 + 4} = \sqrt{53} \] which is roughly 7.28.

Using this systematic approach ensures accurate results. Understanding each calculation step helps demystify the process and boosts your confidence in tackling similar math problems. Always double-check your calculations, as small arithmetic errors can lead to incorrect results.