Question

Find all points on the \(x\) -axis that are 6 units from the point (4,-3)

Short answer

The points on the x-axis are \((4 + 3\sqrt{3}, 0)\) and \((4 - 3\sqrt{3}, 0)\).

Step-by-step solution

Understand the Problem

We need to find all points on the x-axis that are 6 units away from the point (4, -3). Points on the x-axis have the form (x, 0).

Use the Distance Formula

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

Set Up the Equation

Let the point on the x-axis be \((x, 0)\). Using the distance formula, the distance between (4, -3) and \((x, 0)\) is: \[ 6 = \sqrt{(x - 4)^2 + (0 + 3)^2} \].

Simplify the Equation

Square both sides to remove the square root: \[ 6^2 = (x - 4)^2 + 3^2 \]\[ 36 = (x - 4)^2 + 9 \].

Solve for x

Subtract 9 from both sides: \[ 36 - 9 = (x - 4)^2 \]\[ 27 = (x - 4)^2 \].Taking the square root of both sides, \[ \sqrt{27} = |x - 4| \]\[ \sqrt{27} = \pm (x - 4) \].

Solve the Positive and Negative Scenarios

For the positive case: \[ \sqrt{27} = x - 4 \]\[ x = 4 + \sqrt{27} \].For the negative case: \[ -\sqrt{27} = x - 4 \]\[ x = 4 - \sqrt{27} \].

Final Answers

\[ x = 4 + 3\sqrt{3} \] and \[ x = 4 - 3\sqrt{3} \].Thus, the points are \((4 + 3\sqrt{3}, 0)\) and \((4 - 3\sqrt{3}, 0)\).

Key concepts

Distance Between Points

In coordinate geometry, the distance formula is used to find the distance between two points in a plane. This formula is essential because it gives us a way to measure how far apart any two points are, regardless of their position.
The distance formula is derived from the Pythagorean Theorem, and it's written 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 you're measuring the distance between.
To visualize, imagine a right-angled triangle where the horizontal leg is the difference between the x-coordinates ( \((x_2 - x_1)\) ) and the vertical leg is the difference between the y-coordinates ( \((y_2 - y_1)\) ). The distance ( \(d\) ) is the hypotenuse of this triangle.

Solving Algebraic Equations

Solving algebraic equations is a fundamental skill in mathematics. This process involves finding the unknown variables that satisfy given equations. For example, in the exercise, to find the points on the x-axis that are a certain distance from another point, you set up and solve an equation.
In step 3 of the solution, we use the distance formula to create an equation: \[ 6 = \sqrt{(x - 4)^2 + (0 + 3)^2} \].
This equation helps us relate the unknown variable \((x)\) with the known distances.
Next, by manipulating the equation (squaring both sides, simplifying, etc.), we aim to isolate and solve for the variable. This strategy is consistent across solving various algebraic problems.

Square Root Simplification

Square root simplification involves reducing a square root to its simplest form. This is useful in many algebraic problems, including the one in our exercise.
In step 5, after squaring both sides of the equation: \[ 6^2 = (x - 4)^2 + 3^2 \], we simplify further to: \[ 27 = (x - 4)^2 \].
Taking the square root of both sides allows us to simplify: \[ \sqrt{27} = \pm (x - 4) \].
Recognize that \sqrt{27} can be further simplified to \3\sqrt{3}, as \27 = 3 \times 3 \times 3\. This simplification allows easier and more accurate final steps in solving the equation.

Coordinate Geometry Problems

Coordinate geometry combines algebra and geometry to solve problems involving points, lines, and shapes on the coordinate plane. For instance, finding a point that is a specific distance away from another involves both geometric understanding and algebraic manipulation.
Our specific coordinate geometry problem is finding points on the x-axis a distance of 6 units from (4, -3).
Using the distance formula and solving the algebraic equation gives us points \((4 + 3\sqrt{3}, 0)\) and \((4 - 3\sqrt{3}, 0)\).
This synthesis of concepts helps tackle a wide array of problems in coordinate geometry, making it a versatile and valuable mathematical tool.