Problem 43 - 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: Identify the Problem

We need to find a point on the y-axis that is equidistant from two given points: (5, -5) and (1, 1). A point on the y-axis has coordinates of the form (0, y).

Step 2: Use Distance Formula

Calculate the distances from the point (0, y) to each of the given points using the distance formula. For a point (x1, y1) and (x2, y2), the distance is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step 3: Set Up Equations

For point (5, -5), the distance is \(\sqrt{(5-0)^2 + ((-5)-y)^2} = \sqrt{25 + (y+5)^2}\). For point (1, 1), the distance is \(\sqrt{(1-0)^2 + (1-y)^2} = \sqrt{1 + (y-1)^2}\).

Step 4: Equidistant Condition

Set the distances equal to each other because the point (0, y) is equidistant from both points. So, \(\sqrt{25 + (y+5)^2} = \sqrt{1 + (y-1)^2}\).

Step 5: Solve the Equation

Square both sides to eliminate the square roots: \[25 + (y+5)^2 = 1 + (y-1)^2\] Expand both squares: \[(y^2 + 10y + 25)\] and \[(y^2 - 2y + 1).\] Simplify and solve: \[25 + y^2 + 10y + 25 = 1 + y^2 - 2y + 1\] Combine like terms: \[50 + y^2 + 10y = y^2 - 2y + 2\] Subtract \(y^2\) from both sides: \[50 + 10y = -2y + 2\] Add 2y to both sides: \[50 + 12y = 2\] Finally, solve for \(y\) by subtracting 50 and dividing by 12: \[12y = -48\] \[y = -4\].

Step 6: Verify the Solution

Plug \(y = -4\) back into the context of the problem to ensure both distances are equal. Distance from (0, -4) to (5, -5) is \(\sqrt{25 + 1} = \sqrt{26}\), and from (0, -4) to (1, 1) is also \(\sqrt{1 + 25} = \sqrt{26}\). Hence, it is equidistant.

Key Concepts

Distance Formula

The distance formula is a crucial tool in coordinate geometry that helps us find the distance between any two points in the plane. To compute this, we use the formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This formula applies the Pythagorean theorem in a coordinate plane, where the difference in \(x\)-coordinates and \(y\)-coordinates forms a right triangle.

• The horizontal change is \(x_2 - x_1\).
• The vertical change is \(y_2 - y_1\).

By squaring these differences, summing them, and taking the square root, the formula calculates the shortest path between two points, known as Euclidean distance. In the given exercise, this formula helps us measure how far a point on the \(y\)-axis is from the points \((5, -5)\) and \((1, 1)\). By setting the distances equal, we can find where the third point lies.

Coordinate Geometry

Coordinate geometry, sometimes called analytical geometry, allows us to analyze geometric shapes and figures using a coordinate system. By using an \(xy\)-coordinate plane, we can represent points with coordinates \((x, y)\) and explore the spatial relationships between them. In this exercise, all points are represented in the 2-dimensional plane.

• A point on the \(y\)-axis has the form \((0, y)\). This is because for any point on the \(y\)-axis, the \(x\)-coordinate is always 0.
• Systems of equations can emerge from equidistant conditions between points.

The use of coordinate geometry allows us to construct equations based on geometric conditions, such as being equidistant, which can then be solved algebraically. In the context of the problem, coordinate geometry helps transform a visual problem into an algebraic one, which can be tackled with formulae like the distance formula.

Mathematical Problem Solving

Mathematical problem solving involves breaking down a problem into manageable steps. For this problem, we apply a systematic approach:

• Identify the specific requirement: find a point equidistant from two others.
• Use known methods, such as the distance formula, to set up equations.
• Transform these equations into a form that can be solved algebraically.
• Review solutions for consistency with initial conditions.

By setting up an equation where the distances from \((0, y)\) to each of the given points are equal, we reduce the original problem to an algebraic exercise. Simplifying this equation involves expanding, combining like terms, and solving linear equations, demonstrating comprehensive problem-solving techniques. Each transformation leads to a simpler expression, ultimately revealing the solution: the point \((0, -4)\), verified to be equidistant.