/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Find all points on the \(y\) -ax... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all points on the \(y\) -axis that are a distance 6 from \(P(5,3)\)

Short Answer

Expert verified
The points are \((0, 3 + \sqrt{11})\) and \((0, 3 - \sqrt{11})\).

Step by step solution

01

Understanding the Problem

We need to find points on the y-axis, which means they are in the form of \((0, y)\). These points should be a distance of 6 units from the point \((5, 3)\).
02

Using the Distance Formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). We know \((x_1, y_1) = (5, 3)\), \((x_2, y_2) = (0, y)\), and the distance \(d = 6\).
03

Setting Up the Equation

Substitute the known values into the distance formula: \[ 6 = \sqrt{(0 - 5)^2 + (y - 3)^2} \]. Simplify it to: \[ 6 = \sqrt{25 + (y - 3)^2} \].
04

Eliminating the Square Root

Square both sides of the equation to eliminate the square root: \(36 = 25 + (y - 3)^2\).
05

Solving for y

Subtract 25 from both sides: \((y - 3)^2 = 11\). Take the square root of both sides: \(y - 3 = \pm \sqrt{11}\). Solve for \(y\): \(y = 3 + \sqrt{11}\) and \(y = 3 - \sqrt{11}\).
06

Identifying Points on the y-axis

The points on the y-axis that satisfy the condition are \((0, 3 + \sqrt{11})\) and \((0, 3 - \sqrt{11})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Points on the y-axis
When tackling problems related to points on the y-axis, remember that these points have a common characteristic: their x-coordinate is always zero. This simplifies our task significantly because when we say a point is on the y-axis, it will inherently be in the form of \((0, y)\). This trait makes it easy to focus only on the y-value when using the distance formula or any other calculations.

The y-axis is the vertical line on a graph that helps in determining the vertical position of a point. It's crucial, especially when combining it with other points on a coordinate plane, as it fixes the position along one of the axis lines.

Here's what you need to remember about points on the y-axis:
  • The x-coordinate is always zero.
  • They are vertically aligned.
  • The y-value changes but the x remains constant at zero.
Knowing this is very useful, especially in problems involving distances or equations, as you'll see precisely where changes need to occur.
Distance from a point
The distance from a specific point on a coordinate plane to another is calculated using the Distance Formula. This formula is an application of the Pythagorean theorem. When dealing with coordinates like \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

This formula is powerful. It allows us to measure exact displacement between two points, regardless of their position on the plane. In exercises like finding points from another point at a specific distance, it becomes crucial.

Important aspects to keep in mind:
  • The square root ensures that the distance is always non-negative.
  • Differences in coordinates \((x_2 - x_1)\) and \((y_2 - y_1)\) are squared to eliminate any direction inconsistency (as direction doesn't matter in distance).
  • This formula can be adjusted for any pair of points, including those lying directly on an axis like we have here in our problem.
Mastering this formula makes it much simpler to solve problems related to distances or finding specific points in coordinate geometry.
Solving equations with square roots
Equations involving square roots often come into play when applying the Distance Formula. In solving these, the key is to manipulate the equation to express the variable of interest. Here's how this works step-by-step:

1. **Setting Up Equation:** Start by setting up your equation based on what equals the expression under the square root.
2. **Eliminating the Square Root:** To eliminate the square root, square both sides of the equation. This will allow us to work with a simpler, linear (or quadratic) equation.
3. **Solving for the Variable:** Once the square root is gone, proceed to solve for your variable. This may involve algebraic operations such as addition, subtraction, multiplication, division, and dealing with any resulting quadratic expressions.

Like in our exercise, after substituting values, squaring both sides enabled us to solve for \(y\), leading to results like \(y = 3 + \sqrt{11}\) and \(y = 3 - \sqrt{11}\).

Here are some tips when dealing with equations involving square roots:
  • Squaring both sides of an equation can introduce extraneous solutions, so verify your answers.
  • Pay attention to signs especially when taking square roots since both positive and negative roots can be solutions.
  • Check your solutions in the original equation to ensure they are valid.
Understanding and solving such equations is fundamental to mastering coordinate geometry tasks.

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