Problem 17 - College Algebra and Calculus: An Applied Approach

Question

Find \(x\) such that the distance between the points is 15 . \((3,-4),(x, 5)\)

Step-by-Step Solution

Answer

The \(x\) value that makes the distance between the points (3,-4) and (\(x\), 5) to be 15 is \(x = 15\)

Step 1: Plug the Knowns into the Formula

We know that the known points are (3,-4) and (x,5). We also know that the distance between these points is given to be 15. Let's apply these details to the formula of distance. We have: \[15 = \sqrt{(x - 3)^2 + ((5 - -4)^2)}\].

Step 2: Simplify the Equation

First handle the easy subtraction inside parenthesis on the right side of the equation to get: \[15 = \sqrt{(x - 3)^2 + (9)^2}\], which further simplifies to \[15 = \sqrt{(x - 3)^2 + 81}\].

Step 3: Solve for x

Next, square both sides to get rid of the square root on the right side of the equation. This gives us: \[225 = (x - 3)^2 + 81\]. Move 81 to the left side of the equation to end up with \[144 = (x - 3)^2\]. Now take square root of both sides to get: \[12 = x - 3\]. Adding 3 to both sides gives us the value of x. So, the final outcome is: \(x = 15 \)

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This is a way to precisely describe the location of points. In the Cartesian coordinate system, points are defined by an ordered pair of numbers \(x, y\) where \(x\) and \(y\) are the coordinates of the point.

Coordinate geometry is fundamental in solving many types of problems, such as finding the distance between two points or the slope of a line. It can be especially helpful in real-world applications, like engineering and physics, where understanding location and movement is crucial.

In the specific case of determining the distance between two points in a plane, like in our exercise, we use the **distance formula**. The formula is derived from the Pythagorean theorem and is given by:

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

This formula helps us compute linear distances using the differences in their x and y coordinates.

Algebra

Algebra is a part of mathematics that deals with symbols and the rules for manipulating these symbols. In coordinate geometry, we use algebra to develop formulas and solve equations, like those that give us the distance between two points.

In our exercise, once we inserted our points into the distance formula, we were faced with an equation involving a square root. To solve the equation for \(x\), we needed to employ some algebraic manipulations:

• First, we simplified expressions within parentheses to make the calculations more straightforward.
• We then squared both sides to eliminate the square root, which led us to a quadratic equation.
• Finally, solving the quadratic equation gave us the desired value of \(x\).

Algebra allows us to rearrange and simplify expressions to isolate variables, which is exactly what was done to find \(x\). In broader terms, algebra is a versatile tool for solving all kinds of mathematical problems by transforming complex problems into simpler ones that are easier to manage.

Problem Solving

Problem-solving skills are critical in mathematics and involve understanding what is required to solve a problem and applying a series of logical steps to arrive at a solution. Let's break down the problem-solving process using the given exercise as an example:

• **Understand the Problem:** We needed to find the value of \(x\) that satisfies the distance condition between the two points.
• **Make a Plan:** Use the distance formula that applies to points in a coordinate plane.
• **Execute the Plan:** Plug in the known values into the formula, perform algebraic manipulations, and solve for \(x\).
  
• **Check the Solution:** Verify that the calculated value of \(x\) maintains the correct distance between the two points.

Successful problem solving requires the ability to organize information, analyze situations, and think critically to come up with a solution. By following a structured approach, we can tackle even complex mathematical problems efficiently.