Question

Find all points on the \(y\) -axis that are 6 units from the point (4,-3)

Short answer

Points: (0, -3 + 2Sqrt(5)) and (0, -3 - 2Sqrt(5)).

Step-by-step solution

Understand the problem

We need to find all points on the y-axis that are 6 units away from the point (4, -3). Points on the y-axis have coordinates of the form (0, y).

Set up the distance formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\text{Distance} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)\.

Apply the distance formula

Using point (4, -3) and a point on the y-axis (0, y), the distance formula becomes: \(\text{Distance} = \sqrt{(0-4)^2+(y+3)^2} = 6\).

Solve for y

Simplify the equation: \(\text{Distance} = \sqrt{16+(y+3)^2} = 6\).\ Square both sides to eliminate the square root: \(\text{36} = \text{16}+(y+3)^2\).\ So, \(36-16 = (y+3)^2\).\ \(20 = (y+3)^2\).\ Take the square root of both sides: \(y+3 = \pm\text{Sqrt(20)}\).\ Therefore: \(y = -3 \pm 2\text{Sqrt(5)}\).

Write the final answers

The y-coordinates are \(y = -3 + 2\text{Sqrt(5)}\) and \(y = -3 - 2\text{Sqrt(5)}\). The points are: (0, -3 + 2\text{Sqrt(5)}) and (0, -3 - 2\text{Sqrt(5)}).

Key concepts

Coordinates

In geometry, we use coordinates to locate points on a plane. Coordinates are written as pairs (x, y), where x represents the horizontal position and y represents the vertical position.
Points on the y-axis always have the x-coordinate equal to 0. For example, (0, 2), (0, -5), and (0, y) are all points on the y-axis.
This exercise requires finding y-values such that the distance between these y-axis points and another point is a specific value. Understanding this about coordinates helps in solving problems related to locations and distances on a plane.

Distance Between Points

The distance formula is a mathematical tool used to calculate the distance between two points in a plane. If you have two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance between these points can be calculated using:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula comes from the Pythagorean theorem applied to the horizontal and vertical distances between the points.
By substituting the coordinates of the points into this formula, you can find how far apart they are. In this exercise, we used the distance formula to determine points on the y-axis that are a fixed distance from another point.

Square Root

The square root is a fundamental concept in mathematics. It is the value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because 3 \times 3 = 9.
When we have an equation that includes a square, i.e., \((y + 3)^2 = 20\), solving for y involves taking the square root of both sides to simplify. This means \((y + 3) = \pm\sqrt{20}\).
In the distance formula, squaring the distances and summing them allows us to combine the horizontal and vertical distances efficiently.
Then, we use the square root to find the actual linear distance, making square roots essential for solving these types of problems.