Which Point Is Closer To E: C Or D?

Hey guys, welcome back to Plastik Magazine! Today, we're diving into a classic geometry problem that's super useful for understanding distances in a coordinate plane. We've got three points: $C(-6,3)$, $D(3,2)$, and $E(-2,1)$. The big question is, which point, $C$ or $D$, is closer to point $E$? Let's break it down and figure this out together!

Understanding the Distance Formula

Before we get our hands dirty with the numbers, let's quickly chat about how we actually measure the distance between two points on a 2D plane. You guys probably remember this from math class, but it's the distance formula, which is basically derived from the Pythagorean theorem. If you have two points, let's call them $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ between them is given by the formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Think of it like this: the difference in the x-coordinates gives you one leg of a right triangle, and the difference in the y-coordinates gives you the other leg. The distance between the two points is the hypotenuse of that triangle. So, we're just applying Pythagoras' theorem, $a^2 + b^2 = c^2$, where $a$ is the difference in x's, $b$ is the difference in y's, and $c$ is our distance $d$. Pretty neat, right? Knowing this formula is key to solving our problem today. We'll be using it twice: once to find the distance between $C$ and $E$, and again to find the distance between $D$ and $E$. Then, we just compare those two distances. The smaller one tells us which point is closer. It's a straightforward process, and once you get the hang of it, you'll be calculating distances like a pro!

Calculating the Distance Between C and E

Alright, let's get down to business and calculate the distance between our first pair of points: $C(-6,3)$ and $E(-2,1)$. Remember our trusty distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

For point $C$, our $(x_1, y_1)$ is $(-6, 3)$. For point $E$, our $(x_2, y_2)$ is $(-2, 1)$.

Let's plug these values into the formula:

• Difference in x-coordinates: $x_2 - x_1 = -2 - (-6) = -2 + 6 = 4$.
• Difference in y-coordinates: $y_2 - y_1 = 1 - 3 = -2$.

Now, we square these differences:

• $(x_2 - x_1)^2 = 4^2 = 16$.
• $(y_2 - y_1)^2 = (-2)^2 = 4$.

Add the squared differences together:

• $16 + 4 = 20$.

Finally, take the square root of the sum to find the distance, let's call it $d_{CE}$:

• $d_{CE} = \sqrt{20}$.

Now, we can simplify $\sqrt{20}$ if we want. We know that $20 = 4 \times 5$, so $\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$.

So, the distance between point $C$ and point $E$ is $2\sqrt{5}$. Keep this number handy, because we're going to do the same thing for point $D$ and point $E$ to see how they stack up!

Calculating the Distance Between D and E

Now it's time to calculate the distance between our second pair of points: $D(3,2)$ and $E(-2,1)$. We'll use the same distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

For point $D$, our $(x_1, y_1)$ is $(3, 2)$. For point $E$, our $(x_2, y_2)$ is $(-2, 1)$.

Let's plug these values into the formula:

• Difference in x-coordinates: $x_2 - x_1 = -2 - 3 = -5$.
• Difference in y-coordinates: $y_2 - y_1 = 1 - 2 = -1$.

Next, we square these differences:

• $(x_2 - x_1)^2 = (-5)^2 = 25$.
• $(y_2 - y_1)^2 = (-1)^2 = 1$.

Add the squared differences together:

• $25 + 1 = 26$.

Finally, take the square root of the sum to find the distance, let's call it $d_{DE}$:

• $d_{DE} = \sqrt{26}$.

So, the distance between point $D$ and point $E$ is $\sqrt{26}$. Now we have both distances: $d_{CE} = 2\sqrt{5}$ and $d_{DE} = \sqrt{26}$. The final step is to compare them.

Comparing the Distances and Drawing a Conclusion

We've calculated the distance from $C$ to $E$ as $d_{CE} = 2\sqrt{5}$ and the distance from $D$ to $E$ as $d_{DE} = \sqrt{26}$. To figure out which point is closer to $E$, we just need to compare these two values. It might not be immediately obvious which is smaller just by looking at them, so let's make it easier by squaring both distances. This is a neat trick because if $a > b > 0$, then $a^2 > b^2$. So, comparing $d_{CE}$ and $d_{DE}$ is the same as comparing their squares!

• Let's square $d_{CE}$: $(2\sqrt{5})^2 = 2^2 \times (\sqrt{5})^2 = 4 \times 5 = 20$.
• Let's square $d_{DE}$: $(\sqrt{26})^2 = 26$.

Now, we can clearly see that $20 < 26$. This means that $d_{CE}^2 < d_{DE}^2$, which implies that $d_{CE} < d_{DE}$.

Therefore, point $C$ is closer to point $E$ than point $D$ is. So, the answer to our question is that Point $C$ is closer to point $E$. This confirms option A.

It's pretty cool how we can use a simple formula to solve problems like this. Whether you're dealing with coordinates in a math test or thinking about distances in a game or a real-world scenario, understanding the distance formula is super valuable. Keep practicing, guys, and you'll become distance masters in no time!

Visualizing the Points

To really cement this in our minds, let's take a moment to visualize these points on a coordinate plane. Imagine a graph with an x-axis and a y-axis. Point $E$ is at $(-2,1)$. If you start at the origin (0,0), you move 2 units to the left and 1 unit up. That's where $E$ is.

Now, let's look at point $C(-6,3)$. From the origin, you move 6 units to the left and 3 units up. To get from $E$ to $C$, you'd move 4 units left (from -2 to -6) and 2 units up (from 1 to 3). That's our horizontal leg of 4 and vertical leg of 2 we calculated earlier.

Next, point $D(3,2)$. From the origin, you move 3 units to the right and 2 units up. To get from $E$ to $D$, you'd move 5 units to the right (from -2 to 3) and 1 unit up (from 1 to 2). That's our horizontal leg of 5 and vertical leg of 1 we calculated.

When you sketch this out, you can visually see that $E$ is located to the