Distance Between Two Points: Pythagorean Theorem

Hey guys! Ever wondered how to find the distance between two points on a graph? Well, buckle up, because we're about to dive into using the Pythagorean Theorem to do just that! We'll take the points $(-1,6)$ and $(5,-2)$ as an example. Let's break it down step-by-step so it's super easy to follow. Finding the distance between two points might seem daunting at first, but trust me, with a little Pythagorean magic, it becomes a piece of cake. The Pythagorean Theorem, usually written as $a^2 + b^2 = c^2$, is not just for triangles; it's a versatile tool that helps us in various mathematical scenarios, including coordinate geometry. The key is to visualize or construct a right triangle using our two points. Think of the distance between the x-coordinates and the distance between the y-coordinates as the legs of this triangle. The actual distance between the two points then becomes the hypotenuse, which we can easily find using our theorem. Trust me, understanding this connection opens up a whole new perspective on how geometry and algebra intertwine. To kick things off, let’s understand why this method works and what each component represents in our coordinate plane. First off, we need to plot these points, or at least visualize them, on a coordinate plane. Once you've got your points, imagine drawing a horizontal line from $(-1,6)$ to a point directly below $(5,-2)$. Then, draw a vertical line from $(5,-2)$ up to that horizontal line. Boom! You've got yourself a right triangle. The horizontal leg represents the difference in the x-coordinates, and the vertical leg represents the difference in the y-coordinates. Now, finding these lengths is just a matter of simple subtraction. This is where the beauty of coordinate geometry really shines—transforming geometric problems into algebraic calculations. And don’t worry if this sounds complicated; we’ll go through a detailed example to make sure you’ve got it down pat. So, stick around, and let's demystify this concept together!

Step-by-Step Calculation

Okay, let's get our hands dirty with the actual numbers. We need to find the lengths of the legs of our right triangle. Remember, the points are $(-1,6)$ and $(5,-2)$. First, let's find the horizontal distance (the difference in x-coordinates). We'll subtract the x-coordinates: $5 - (-1) = 5 + 1 = 6$. So, the length of our horizontal leg is 6. Next, we find the vertical distance (the difference in y-coordinates). Subtract the y-coordinates: $6 - (-2) = 6 + 2 = 8$. So, the length of our vertical leg is 8. Now we have a right triangle with legs of length 6 and 8. Time to bring in the Pythagorean Theorem: $a^2 + b^2 = c^2$. Here, $a = 6$ and $b = 8$, so we have $6^2 + 8^2 = c^2$. Calculate the squares: $36 + 64 = c^2$. Add them up: $100 = c^2$. Now, to find $c$, we take the square root of both sides: $c = \sqrt{100} = 10$. So, the distance between the points $(-1,6)$ and $(5,-2)$ is exactly 10! Isn't that neat? To reinforce this, imagine these steps visually on a coordinate plane. As you perform each subtraction, you're essentially measuring the span along each axis. These measurements perfectly translate into the lengths of the right triangle's legs, connecting the algebraic manipulation to the geometric reality. It’s this connection that makes the Pythagorean Theorem so powerful in coordinate geometry, allowing us to solve problems that might otherwise seem complex with relative ease. Furthermore, remember that the order in which you subtract the coordinates doesn't matter, as long as you're consistent. If you subtracted $(-1)$ from 5, make sure you subtract $(-2)$ from 6. This ensures you're finding the true difference and not a negative value, which would still give you the correct squared result thanks to squaring. The important thing is to keep the x and y coordinates paired correctly to avoid confusion. In summary, we found the horizontal and vertical distances, squared them, added them together, and took the square root to find the hypotenuse, which is the distance between the two points. Great job, guys! You're getting the hang of it. Keep practicing, and you'll be a pro in no time!

Rounding to the Nearest Hundredth

In our previous example, the distance came out to be a nice, clean 10. But what if it wasn't so perfect? What if we ended up with a square root that doesn't give us a whole number? That's where rounding comes in handy. Let's say, hypothetically, that after applying the Pythagorean Theorem, we found that $c^2 = 50$. Then, $c = \sqrt{50}$. Now, $\sqrt{50}$ isn't a whole number; it's approximately 7.07106781... If we need to round this to the nearest hundredth, we look at the digit in the thousandths place (the third digit after the decimal), which is 1 in this case. Since 1 is less than 5, we round down. That means we keep the hundredths digit as it is. So, $\sqrt{50}$ rounded to the nearest hundredth is 7.07. See how simple that is? When dealing with more complex numbers, rounding becomes particularly crucial to present results in a manageable and understandable format. For instance, in fields like engineering or physics, presenting a precise but overly long number might not be as useful as a rounded value that conveys the necessary accuracy without the extra clutter. The key here is to understand the level of precision that's required for the task at hand. If you're building a bridge, you'll likely need very precise measurements, whereas for a simple estimation, rounding to the nearest whole number might suffice. Moreover, the concept of significant figures comes into play when rounding. Significant figures are the digits that carry meaning contributing to its precision. When rounding, you should maintain the appropriate number of significant figures to reflect the accuracy of your original measurements. For example, if you start with values measured to three significant figures, your final answer should also be rounded to three significant figures. This ensures that you're not implying a level of precision that doesn't exist. Also, keep in mind that calculators and computers often display numbers with many decimal places, but it's up to you to decide how many of those digits are relevant and should be included in your final answer. Always consider the context of your problem and what level of precision is necessary. Remember, rounding is a tool to help us simplify and present data in a meaningful way. Now, let’s look at some more examples to help solidify your understanding of rounding to the nearest hundredth. These practical exercises will give you the confidence to tackle any rounding challenge that comes your way.

Practice Problems

Ready to put your skills to the test? Let's try a couple of practice problems to make sure you've got this down.

Problem 1: Find the distance between the points $(2, -3)$ and $(-4, 1)$. Round your answer to the nearest hundredth, if necessary.

Problem 2: Determine the distance between the points $(0, 5)$ and $(3, -1)$. Round your answer to the nearest hundredth, if necessary.

Solutions:

Problem 1:

1. Find the difference in x-coordinates: $-4 - 2 = -6$. The absolute value is $|-6| = 6$.
2. Find the difference in y-coordinates: $1 - (-3) = 1 + 3 = 4$.
3. Apply the Pythagorean Theorem: $6^2 + 4^2 = c^2$, so $36 + 16 = c^2$, which means $52 = c^2$.
4. Take the square root: $c = \sqrt{52} \approx 7.21110255...$
5. Round to the nearest hundredth: $c \approx 7.21$.

Problem 2:

1. Find the difference in x-coordinates: $3 - 0 = 3$.
2. Find the difference in y-coordinates: $-1 - 5 = -6$. The absolute value is $|-6| = 6$.
3. Apply the Pythagorean Theorem: $3^2 + 6^2 = c^2$, so $9 + 36 = c^2$, which means $45 = c^2$.
4. Take the square root: $c = \sqrt{45} \approx 6.70820393...$
5. Round to the nearest hundredth: $c \approx 6.71$.

How did you do? Did you get them right? Great job if you did! If not, don't worry; just go back and review the steps. The Pythagorean Theorem is your friend, and with a little practice, you'll be solving these problems like a pro. Remember, the key is to break it down into manageable steps and stay organized. And now that you've nailed these practice problems, you're well on your way to mastering the distance formula and the Pythagorean Theorem in coordinate geometry. Keep up the awesome work!

Conclusion

Alright, we've covered a lot today! We learned how to use the Pythagorean Theorem to find the distance between two points on a graph, and we practiced rounding our answers to the nearest hundredth. Remember, the Pythagorean Theorem ($a^2 + b^2 = c^2$) is a powerful tool that can be used in many different situations. By understanding how to apply it in coordinate geometry, you can solve a wide variety of problems. The ability to accurately calculate distances and round appropriately is fundamental in many areas of math and science, so the time you invest in mastering these concepts will pay off in the long run. Don't be afraid to revisit these concepts and practice regularly. Mathematics is like a muscle—the more you use it, the stronger it gets. So, keep challenging yourself with new problems and seeking out opportunities to apply what you've learned. Whether you're calculating the distance between cities on a map or determining the shortest path in a video game, the principles we've discussed today are universally applicable. Moreover, remember that learning is a process. It's okay to make mistakes along the way. The important thing is to learn from those mistakes and keep moving forward. Embrace the challenges, celebrate your successes, and never stop exploring the fascinating world of mathematics. As you continue your mathematical journey, remember that every problem is an opportunity to grow and deepen your understanding. And who knows, maybe one day you'll discover a new theorem or solve a problem that no one else has ever been able to crack. The possibilities are endless! So, keep that spark of curiosity alive and never lose your passion for learning. You've got this!