Problem 49 - Algebra and Trigonometry

Question

Find a point on the \(y\) -axis that is equidistant from the points \((5,-5)\) and \((1,1) .\)

Step-by-Step Solution

Answer

The point on the y-axis is (0, -4).

Step 1: Define the Point on the y-axis

A point on the y-axis can be defined as \((0, y)\).

Step 2: Use the Distance Formula

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

Step 3: Calculate Distance from (5, -5) to (0, y)

Plugging the coordinates into the distance formula, we have: \[d_1 = \sqrt{(0-5)^2 + (y+5)^2} = \sqrt{25 + (y+5)^2}.\]

Step 4: Calculate Distance from (1, 1) to (0, y)

Plugging the coordinates into the distance formula, we have: \[d_2 = \sqrt{(0-1)^2 + (y-1)^2} = \sqrt{1 + (y-1)^2}.\]

Step 5: Set the Distances Equal

Since the point is equidistant from both points, set \[d_1 = d_2\]. Thus, \[\sqrt{25 + (y+5)^2} = \sqrt{1 + (y-1)^2}.\]

Step 6: Solve the Equation

Square both sides to eliminate the square roots, leading to: \[25 + (y+5)^2 = 1 + (y-1)^2.\] Expand both sides:\[25 + y^2 + 10y + 25 = 1 + y^2 - 2y + 1.\]Simplify: \\[y^2 + 10y + 50 = y^2 - 2y + 2.\] Subtract \(y^2\) from both sides: \[10y + 50 = -2y + 2.\]Add \(2y\) to both sides: \[12y + 50 = 2.\] Subtract \(50\) from both sides: \[12y = -48.\]Divide by \(12\): \[y = -4.\]

Step 7: Conclusion

The point on the y-axis that is equidistant from \((5, -5)\) and \((1, 1)\) is \((0, -4)\).

Key Concepts

Equidistant Points

When we talk about equidistant points, we mean a point that is the same distance from two or more different points. In this exercise, the goal is to find a point on the y-axis that is equidistant from the points \((5,-5)\) and \((1,1)\). To achieve this, we use the distance formula which allows us to calculate the distance between any two points in coordinate geometry. Once we calculate the distances from the y-axis point to each of the original points, we then equate these distances because "equidistant" means they should be equal.This concept is essential in geometry and algebra, where we often find a common ground or a balance between two positions or values. Think of it like a seesaw: for it to be perfectly balanced, both sides must be equidistant from the midpoint.

Algebra Problem

This exercise is an algebra problem that involves applying mathematical concepts and manipulations. Solving it began with setting an unknown variable \(y\) at the point \((0, y)\) on the y-axis. From there, the algebraic journey started with the application of the distance formula. By setting the distances equal, an equation is formed. The next steps involve expanding and simplifying this equation to isolate \(y\). The process requires familiar algebraic techniques such as expanding brackets, rearranging terms, and simplifying expressions. To solve for \(y\), key steps include:

• Expanding both sides of the equation
• Simplifying by canceling out common terms
• Isolating \(y\) by performing operations like addition and subtraction on both sides
• Finally, dividing by coefficients to solve for \(y\)

This methodical approach helps in solving many algebra problems by breaking them down into manageable steps.

Y-Axis

The y-axis is the vertical line in a coordinate plane where all x-values equal zero. In terms of algebraic equations, it is significant because it lets us focus purely on the y-coordinate. In this exercise, the point to find is on the y-axis, denoted as \((0, y)\). The work with the y-axis simplifies the problem because any point on this axis only has a y-coordinate to consider when calculating distances. This characteristic makes equations less complex as we're only addressing one variable.The y-axis helps narrow down possibilities to streamline the calculation. In many math problems, focusing on axes or specific lines can reduce the complexity of multi-variable problems by ensuring one variable remains constant.