Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point on the \(x\)-axis that is equidistant from \((3,1)\) and \((6,4)\).

Step-by-Step Solution

Answer

The point is (7,0).

Step 1: Define the Problem

We need to find a point on the \(x\)-axis that is equidistant from two given points \((3,1)\) and \((6,4)\). A point on the \(x\)-axis has coordinates \((x, 0)\).

Step 2: Use the Distance Formula

The distance from a point \((x_1, y_1)\) to another point \((x_2, y_2)\) is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). We will set the distances from \((x, 0)\) to the points \((3, 1)\) and \((6, 4)\) equal, since the point \((x, 0)\) is equidistant from these two points.

Step 3: Set Up the Equation

The distance from \((x, 0)\) to \((3, 1)\) is \(\sqrt{(x - 3)^2 + (0 - 1)^2}\) and the distance from \((x, 0)\) to \((6, 4)\) is \(\sqrt{(x - 6)^2 + (0 - 4)^2}\). Set these equal to each other:\[ \sqrt{(x - 3)^2 + 1^2} = \sqrt{(x - 6)^2 + 16} \]

Step 4: Simplify the Equation

Squaring both sides of the equation to eliminate the square roots, we get:\[ (x - 3)^2 + 1 = (x - 6)^2 + 16 \]

Step 5: Expand Both Sides

Expand the squares:\[ (x^2 - 6x + 9) + 1 = (x^2 - 12x + 36) + 16 \]

Step 6: Simplify

Simplify both sides:\[ x^2 - 6x + 10 = x^2 - 12x + 52 \]

Step 7: Cancel Terms

Subtract \(x^2\) from both sides to eliminate \(x^2\):\[ -6x + 10 = -12x + 52 \]

Step 8: Solve for x

Add \(12x\) to both sides:\[ 6x + 10 = 52 \]Subtract 10 from both sides:\[ 6x = 42 \]Divide by 6:\[ x = 7 \]

Step 9: Conclusion

The point on the \(x\)-axis that is equidistant from \((3,1)\) and \((6,4)\) is \((7,0)\).

Key Concepts

Distance Formula

The distance formula is crucial for finding the space between two points in a coordinate plane. It comes from the Pythagorean theorem and is represented as:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Here, \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the two points.
This formula allows us to calculate the direct line distance, or Euclidean distance, between them.
In our exercise, this method is used to ensure that a point on the \(x\)-axis is equidistant from two other points.
To find this point, you compute the distance from the \(x\)-axis point \(x, 0\) to each of the given points, ensuring their distances are equal.
Using the distance formula ensures accurate calculations and is widely applicable in various geometric and real-world problems.

Coordinate Geometry

Coordinate geometry, sometimes referred to as analytic geometry, is fascinating because it allows us to use algebra to understand geometric concepts.
In coordinate geometry, points are plotted on a plane using pairs of numbers, called coordinates.

• Every point is defined by an \(x\)\-(horizontal position) and a \(y\)\-(vertical position).
• Lines, curves, and shapes can be described using equations.
• Analyzes spatial relationships and can solve problems dealing with distance, midpoints, and lines.

In this exercise, we're using the concept of a coordinate plane to locate a point on the \(x\)-axis.
Points in this example are defined relative to this axis, showing how coordinate geometry helps in solving practical problems by placing abstract concepts into visual coordinates.

Solving Equations

Understanding how to solve equations is a vital skill in algebra. Equations represent mathematical statements where two expressions are equal, involving variables, constants, and operators like addition or multiplication.To solve an equation:

• Combine like terms and simplify both sides of the equation wherever possible.
• Use algebraic manipulation, such as adding, subtracting, multiplying, or dividing both sides by the same number.
• Isolate the variable to one side to solve for it.

In our specific problem, we ended up with an equation derived from setting equal the distances calculated using the distance formula:\[(x - 3)^2 + 1 = (x - 6)^2 + 16\]By squaring, expanding, and simplifying, we isolate \(x\). The key steps involve basic math principles, reinforcing the significance of maintaining balance in equations and operations.
Solving such equations gives precise answers and is an invaluable tool in numerous fields such as physics, engineering, and economics.