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Magazine Chapter 2: Problem 28
Find all points on the \(x\) -axis that are a distance 5 from \(P(-2,4)\)
Short Answer
Expert verified
The points are (1, 0) and (-5, 0).
Step by step solution
01
Understand the Concept of Distance
To find a point on the \(x\)-axis that is a specified distance from a given point, we use the distance formula. The distance from a point \((x_1, y_1)\) to \((x_2, y_2)\) can be calculated as \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Here, \(y_2\) will be 0 since the point is on the \(x\)-axis.
02
Set up the Equation
Plug in the known values: \(d = 5, (x_1, y_1) = (-2, 4)\), and \((x_2, y_2) = (x, 0)\) into the distance formula: \[ 5 = \sqrt{(x - (-2))^2 + (0 - 4)^2} \] Simplified, this becomes: \[ 5 = \sqrt{(x + 2)^2 + 16} \]
03
Solve the Equation for \(x\)
Square both sides to eliminate the square root: \[ 25 = (x + 2)^2 + 16 \] Subtract 16 from both sides to isolate the square term: \[ 9 = (x + 2)^2 \] Take the square root of both sides: \[ x + 2 = \pm 3 \] This gives us two possible equations: \(x + 2 = 3\) and \(x + 2 = -3\).
04
Determine the Points
Solve each equation from Step 3 separately: For \(x + 2 = 3\): \[ x = 3 - 2 \] \[ x = 1 \] For \(x + 2 = -3\): \[ x = -3 - 2 \] \[ x = -5 \] The points are \((1, 0)\) and \((-5, 0)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Geometry
Coordinate geometry, often called analytic geometry, is a branch of mathematics that uses algebraic principles to study geometric relationships. It involves representing geometric shapes using a coordinate system, which allows us to analyze and solve problems involving distances and angles.
In problems involving distances and locations, coordinate geometry provides a valuable framework. It gives us the powerful distance formula and helps visualize points in space, like finding points on the \( x \)-axis.
- Coordinates are expressed as ordered pairs \( (x, y) \), where \( x \) represents the horizontal position and \( y \) represents the vertical position of a point.
- The use of coordinates allows us to apply algebraic techniques to geometric problems.
In problems involving distances and locations, coordinate geometry provides a valuable framework. It gives us the powerful distance formula and helps visualize points in space, like finding points on the \( x \)-axis.
Solving Equations
Solving equations is a key part of mathematics, where we find the value of unknown variables that make the equation true. In the context of coordinate geometry, solving equations helps us find specific points that meet certain geometric conditions.
Understanding how to isolate variables and apply algebraic rules is crucial for finding solutions and determining coordinates.
- To solve equations like those derived from the distance formula, you often need to isolate the variable of interest. This involves manipulating the equation algebraically to find the value of the unknown.
- In our example, setting-up the distance formula equation: \[ 5 = \sqrt{(x + 2)^2 + 16} \] led us to solve for \( x \) by squaring both sides of the equation and using algebraic techniques.
Understanding how to isolate variables and apply algebraic rules is crucial for finding solutions and determining coordinates.
Distance Between Points
The distance between points in a coordinate plane can be found using the distance formula. This formula is derived from the Pythagorean Theorem and measures the straight-line distance between two points.
The formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) calculates the distance by considering the changes in the x and y direction between two points.
In summary, the distance formula is essential for any task that involves measuring distances between points in coordinate geometry.
The formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) calculates the distance by considering the changes in the x and y direction between two points.
- In the original exercise, understanding this formula allowed us to determine how far away a point on the \( x \)-axis was from \( P(-2,4) \).
- By plugging in the known coordinates and solving, we found the specific x-values where another point satisfied the given distance.
In summary, the distance formula is essential for any task that involves measuring distances between points in coordinate geometry.
X-axis
The \( x \)-axis is the horizontal axis in a coordinate plane, along which the value of \( y \) is 0. All points on this axis are represented as \( (x, 0) \).
In our solution, using the property of the \( x \)-axis, we successfully found two points \( (1, 0) \) and \((-5, 0) \) that are equidistant from \( P(-2,4) \).
- In problems involving distances from the \( x \)-axis, knowing the basic property that \( y = 0 \) is crucial. It simplifies calculations as it eliminates any vertical distance.
- Finding points a specific distance from another point, like \( P(-2,4) \), involves setting the distance formula with \( y_2 \) as zero.
In our solution, using the property of the \( x \)-axis, we successfully found two points \( (1, 0) \) and \((-5, 0) \) that are equidistant from \( P(-2,4) \).
