Problem 40 - Precalculus: Functions and Graphs

Question

Find an equation for the set of points in an xy-plane that are equidistant from the point \(P\) and the line \(L\) $$P(5,-2) ; \quad k y=4$$

Step-by-Step Solution

Answer

The equation is \( x^2 - 10x + 12y = -13 \).

Step 1: Understanding the Components

We need to find an equation for the set of points that are equidistant from the point \( P(5, -2) \) and the line \( L: y = 4 \). This implies that any point \( (x, y) \) on this set is the same distance from both \( P \) and \( L \).

Step 2: Distance Formula from a Point

The distance between a point \( (x, y) \) and \( P(5, -2) \) is given by the formula \( d = \sqrt{(x-5)^2 + (y + 2)^2} \).

Step 3: Distance Formula from a Line

The distance from a point \( (x, y) \) to the line \( y = 4 \) can be calculated using the formula \( d = |y - 4| \). This is the vertical distance between the point and the horizontal line.

Step 4: Setting Up the Equation

Since points \( (x, y) \) should be equidistant from \( P \) and the line, set the distances equal: \[ \sqrt{(x-5)^2 + (y+2)^2} = |y - 4| \]

Step 5: Eliminating the Square Root and Absolute

Square both sides to eliminate the square root: \[ (x-5)^2 + (y+2)^2 = (y-4)^2 \] Expand and simplify both sides of the equation.

Step 6: Expand and Simplify

Expanding the equation we get: \[ (x-5)^2 + (y+2)^2 = y^2 - 8y + 16 \] Expanding further gives: \[ x^2 - 10x + 25 + y^2 + 4y + 4 = y^2 - 8y + 16 \] Cancel \( y^2 \) from both sides, and rearrange the terms: \[ x^2 - 10x + 29 + 12y = 16 \]

Step 7: Solving for y

Finally, simplify the equation for convenience which results in: \[ x^2 - 10x + 12y = -13 \]

Key Concepts

Distance Formula

The distance formula is a crucial tool to calculate how far apart two points are in a coordinate plane, often the xy-plane. It is derived from the Pythagorean theorem and is used when the coordinates of two points are known. For two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) between them is calculated as follows:

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

In this exercise, we specifically applied it to find the distance from any point on the plane \((x, y)\) to a specific point \((5, -2)\). The resulting formula was:

• \(d = \sqrt{(x-5)^2 + (y+2)^2}\)

This formula will help you determine how far any given point \( (x, y) \) on the xy-plane is from the fixed point \( P \). Breaking down problems into smaller parts and using the distance formula effectively makes it easier to solve complex geometric questions like locating equidistant points.

XY-Plane

The xy-plane is a two-dimensional space where each point is determined by a pair of numerical coordinates. These coordinates are often written as \( (x, y) \), with \( x \) representing the horizontal position, and \( y \) representing the vertical position.

• The horizontal axis is called the x-axis.
• The vertical axis is called the y-axis.
• The point where both axes meet is called the origin, given by \( (0,0) \).

In solving problems involving points on an xy-plane, like in our exercise, understanding the layout of the plane is essential. It helps in visualizing the problem, whether determining how distances compute or aligning them with geometric features.

Geometric Locus

In geometry, a geometric locus is a set of points that satisfy certain conditions or rules. These points could form lines, curves, or shapes based on the set criteria. Imagine a geometric locus like a path or a boundary where points have a specific relationship to other objects.

• In the given exercise, we want our geometric locus (call it line or curve) to have all points equidistant from a fixed point and a line.
• The established condition led us to equal the distance formulas, resulting in a curve that satisfies both conditions.

Understanding a geometric locus is vital, as it forms the basis of connecting complex points under uniform rules. By setting the distances equal between a point and a line, one achieves this consistency in spatial arrangement, aligning with the rules/orders the locus should follow.

Horizontal Line

A horizontal line in the xy-plane is one that runs parallel to the x-axis and maintains a constant y-coordinate for all its points. In simpler terms, it means all points on that line have the same height above or below the x-axis.

• The equation for a horizontal line is \( y = c \), where \(c\) is a constant, in this case, \y = 4\.
• In calculating distances, the use of the absolute difference between the y-values gives the vertical distance from a point to the line.

In our problem, the line \(y=4\) plays a critical role, acting as a boundary from which we measure equidistance to satisfy the conditions of our geometric locus. This straightforward approach simplifies complex geometrical analysis by breaking down the spatial separation vertically.