Finding K Values For A Given Distance Between Two Points

In coordinate geometry, determining the distance between two points is a fundamental concept. The distance formula, derived from the Pythagorean theorem, provides a straightforward method to calculate this distance. In this article, we delve into a specific problem where we are given two points, $(2, k)$ and $(0, -6)$, and a desired distance of $\sqrt{5}$ between them. Our task is to find the values of $k$ that satisfy this condition. This exploration will not only reinforce the application of the distance formula but also highlight the algebraic techniques involved in solving such problems. Let’s embark on this mathematical journey to unravel the values of $k$ that meet the specified criteria.

Understanding the Distance Formula

The distance formula is a cornerstone of coordinate geometry, enabling us to calculate the distance between two points in a coordinate plane. It is derived directly from the Pythagorean theorem, which relates the sides of a right triangle. If we have two points, $(x_1, y_1)$ 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 essentially calculates the length of the hypotenuse of a right triangle whose legs are the horizontal and vertical distances between the two points. The difference in the x-coordinates, $(x_2 - x_1)$, represents the horizontal leg, and the difference in the y-coordinates, $(y_2 - y_1)$, represents the vertical leg. Squaring these differences, summing them, and then taking the square root gives us the distance between the points.

In the context of our problem, we are given the points $(2, k)$ and $(0, -6)$, and we want to find the values of $k$ such that the distance between these points is $\sqrt{5}$. This means we will set up the distance formula with these values and solve for $k$. The process involves substituting the coordinates into the formula, squaring both sides to eliminate the square root, and then solving the resulting quadratic equation. This application of the distance formula is crucial in various mathematical and real-world scenarios, from navigation to computer graphics.

Setting Up the Equation

To find the values of $k$ for which the distance between the points $(2, k)$ and $(0, -6)$ is $\sqrt{5}$, we begin by applying the distance formula. Let $(x_1, y_1) = (2, k)$ and $(x_2, y_2) = (0, -6)$. Plugging these values into the distance formula, we get:

$\sqrt{5} = \sqrt{(0 - 2)^2 + (-6 - k)^2}$

This equation represents the core of our problem. It states that the distance between the two points, calculated using the coordinates and the unknown $k$, must equal $\sqrt{5}$. The next step is to simplify this equation and solve for $k$. To do this, we first square both sides of the equation to eliminate the square root:

$(\sqrt{5})^2 = (\sqrt{(0 - 2)^2 + (-6 - k)^2})^2$

This simplifies to:

$5 = (0 - 2)^2 + (-6 - k)^2$

Now we have an equation without square roots, which is easier to manipulate algebraically. The next step involves expanding the terms and rearranging the equation into a standard form that we can solve. This setup is crucial for accurately finding the values of $k$ that satisfy the given distance condition. The equation setup is a critical phase in solving mathematical problems, ensuring that the subsequent algebraic manipulations lead to the correct solution.

Solving for k

After setting up the equation, the next step is to solve for $k$. We have the equation:

$5 = (0 - 2)^2 + (-6 - k)^2$

First, we simplify the terms inside the parentheses and the squares:

$5 = (-2)^2 + (-6 - k)^2$

$5 = 4 + (-6 - k)^2$

Now, we subtract 4 from both sides of the equation:

$1 = (-6 - k)^2$

To eliminate the square, we take the square root of both sides:

$\pm\sqrt{1} = -6 - k$

This gives us two possibilities:

$1 = -6 - k$ or $-1 = -6 - k$

Solving the first equation, $1 = -6 - k$, we add 6 to both sides:

$7 = -k$

$k = -7$

Solving the second equation, $-1 = -6 - k$, we add 6 to both sides:

$5 = -k$

$k = -5$

Thus, the values of $k$ that satisfy the equation are $-7$ and $-5$. This means that if the y-coordinate of the first point is either $-7$ or $-5$, the distance between the points $(2, k)$ and $(0, -6)$ will be $\sqrt{5}$. The solution for k involves careful algebraic manipulation and consideration of both positive and negative roots, ensuring we capture all possible values that satisfy the given condition.

Verifying the Solutions

To ensure the accuracy of our solutions, it is crucial to verify the values of $k$ we found. We determined that $k$ could be either $-7$ or $-5$. We will now substitute each of these values back into the distance formula to confirm that the distance between the points $(2, k)$ and $(0, -6)$ is indeed $\sqrt{5}$.

First, let's verify for $k = -7$. The points become $(2, -7)$ and $(0, -6)$. Applying the distance formula:

$d = \sqrt{(0 - 2)^2 + (-6 - (-7))^2}$

$d = \sqrt{(-2)^2 + (1)^2}$

$d = \sqrt{4 + 1}$

$d = \sqrt{5}$

This confirms that when $k = -7$, the distance between the points is $\sqrt{5}$.

Next, let's verify for $k = -5$. The points become $(2, -5)$ and $(0, -6)$. Applying the distance formula:

$d = \sqrt{(0 - 2)^2 + (-6 - (-5))^2}$

$d = \sqrt{(-2)^2 + (-1)^2}$

$d = \sqrt{4 + 1}$

$d = \sqrt{5}$

This also confirms that when $k = -5$, the distance between the points is $\sqrt{5}$.

The verification of solutions is a vital step in problem-solving, as it ensures that the calculated values satisfy the original conditions of the problem. By substituting the values back into the distance formula, we have confidently confirmed that our solutions, $k = -7$ and $k = -5$, are correct.

Conclusion

In this article, we successfully determined the values of $k$ for which the distance between the points $(2, k)$ and $(0, -6)$ is $\sqrt{5}$. We began by understanding the distance formula, a fundamental concept in coordinate geometry derived from the Pythagorean theorem. We then set up the equation by substituting the given points and distance into the formula. The next step involved solving for $k$, which required algebraic manipulation, including squaring both sides of the equation, simplifying terms, and solving the resulting quadratic equation. We found two possible values for $k$: $-7$ and $-5$.

To ensure the accuracy of our solutions, we performed a crucial step of verification. By substituting each value of $k$ back into the distance formula, we confirmed that both $k = -7$ and $k = -5$ indeed result in a distance of $\sqrt{5}$ between the given points. This process not only validated our calculations but also reinforced the importance of checking solutions in mathematical problem-solving.

The problem-solving approach we employed, from understanding the formula to verifying the solutions, is a valuable strategy applicable to a wide range of mathematical problems. This exercise highlights the interplay between geometric concepts and algebraic techniques, demonstrating how they work together to solve problems in coordinate geometry. The conclusion of our exploration reaffirms the correctness of our solutions and emphasizes the importance of a systematic approach in mathematics.