Square ABCD: Finding Vertex D And Lattice Points On Diagonal AC

Let's dive into this geometry problem where we're given three vertices of a square and tasked with finding the fourth vertex and identifying lattice points on a diagonal. Geometry can be tricky, but with a step-by-step approach, we can totally nail this! So, let's break down how to find the missing vertex and those special points on the diagonal.

a) Determining the Coordinates of Vertex D

Alright, so we have points A(-4, -2), B(2, -2), and C(2, 4), which are consecutive vertices of our square ABCD. To find the coordinates of vertex D, we need to use the properties of a square. Remember, in a square, all sides are equal in length, and all angles are 90 degrees. This means that sides AB and BC are perpendicular, and their lengths can help us locate point D. Let's break this down step by step to make sure we've got it.

First, let's visualize what we have. Points A and B have the same y-coordinate (-2), which means side AB is a horizontal line. Points B and C have the same x-coordinate (2), so side BC is a vertical line. This confirms that angle ABC is a right angle, which is exactly what we expect in a square. Knowing this right off the bat makes things a little easier, right? It's like finding the first piece of a puzzle!

Now, let’s figure out the length of sides AB and BC. The length of AB can be found by looking at the difference in x-coordinates of points A and B because they're on a horizontal line. So, the length of AB is |2 - (-4)| = 6 units. Similarly, the length of BC is the difference in y-coordinates of B and C since they're on a vertical line, which gives us |4 - (-2)| = 6 units. This is great news! It confirms that we're dealing with a square because the sides AB and BC are equal in length – each is 6 units long. Remember, in a square, all sides have to be the same length, so this is a crucial piece of information.

To find point D, we need to understand how it relates to the other points. Since ABCD is a square, side AD must be parallel to BC and equal in length, and side CD must be parallel to BA and also equal in length. Think of it like completing the square – we're just filling in the last corner. Because AD is parallel to BC (which is vertical), AD will also be a vertical line. Since the length of AD must be 6 units, we move 6 units up from point A. So, if A is at (-4, -2), then D’s x-coordinate will be the same as A’s, which is -4. To find the y-coordinate, we add 6 (the side length) to A’s y-coordinate: -2 + 6 = 4. Therefore, the coordinates of D are (-4, 4).

Alternatively, we could have looked at it from point C’s perspective. CD must be parallel to AB, which is a horizontal line. So, CD will also be a horizontal line. Since the length of CD must be 6 units, we move 6 units to the left from point C. If C is at (2, 4), then D’s y-coordinate will be the same as C’s, which is 4. To find the x-coordinate, we subtract 6 (the side length) from C’s x-coordinate: 2 - 6 = -4. This gives us the same coordinates for D: (-4, 4).

So, there we have it, by carefully considering the properties of a square and using the coordinates of the known vertices, we’ve successfully found the coordinates of vertex D. It’s like we’ve completed the puzzle – D sits snugly at (-4, 4), making ABCD a perfect square. This part of the problem really showcases how understanding the fundamental properties of geometric shapes can help us solve complex problems. It's not just about memorizing formulas, it’s about understanding how shapes behave and relate to each other. Now, let’s move on to the next part and tackle those lattice points!

b) Listing the Coordinates of All Lattice Points Lying on the Diagonal AC

Okay, guys, now we need to find the coordinates of all the lattice points that lie on the diagonal AC. A lattice point, just to refresh, is a point where both the x and y coordinates are integers. These are the nice, neat points that sit right on the gridlines of our coordinate plane. Finding these points on the diagonal AC involves a bit of coordinate geometry, but it's totally manageable. We're essentially looking for integer solutions along the line segment that connects A and C. Let’s break this down into smaller steps so it's super clear.

First things first, we need to figure out the equation of the line that passes through points A(-4, -2) and C(2, 4). Remember, to define a line, we often use the slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept. Finding the equation of the line is like creating a map for the diagonal AC; it tells us exactly where every point on the line is located. This will be our key to unlocking the lattice points.

Let's calculate the slope (m) first. The slope is the “steepness” of the line and is defined as the change in y divided by the change in x. Using points A and C, the slope m is calculated as (4 - (-2)) / (2 - (-4)) = 6 / 6 = 1. So, our line has a slope of 1, which means for every unit we move to the right, we move one unit up. This makes sense visually, as we can imagine the line rising steadily from A to C.

Now that we have the slope, we can use the point-slope form of a line equation, which is y - y1 = m(x - x1), where (x1, y1) is a known point on the line. Let’s use point A(-4, -2) as our known point. Plugging in the values, we get y - (-2) = 1(x - (-4)). Simplifying this, we have y + 2 = x + 4. Subtracting 2 from both sides gives us the equation of the line: y = x + 2. Boom! We've got our line equation. This equation is super important because it tells us the relationship between the x and y coordinates of every point on the line AC.

Now, we need to find the lattice points on the segment AC. Remember, lattice points are those with integer coordinates. So, we need to find integer pairs (x, y) that satisfy both the equation y = x + 2 and lie between points A and C. Think of it as filtering through the infinite points on the line to find the special ones that fit our criteria – integers only and within the segment AC.

Since x ranges from -4 (at point A) to 2 (at point C), we can test integer values of x within this range and see if the corresponding y values are also integers. This is a straightforward way to ensure we're only picking lattice points.

Let's start with x = -4. Using the equation y = x + 2, we get y = -4 + 2 = -2. This gives us the point (-4, -2), which is point A itself. So, that's one lattice point down!

Next, try x = -3. We get y = -3 + 2 = -1. The point is (-3, -1). This is another lattice point since both coordinates are integers. We're on a roll!

Now, let’s try x = -2. We get y = -2 + 2 = 0. The point is (-2, 0). Again, both coordinates are integers, so this is a lattice point.

If we continue this process for x = -1, we find y = -1 + 2 = 1, giving us the point (-1, 1). This is also a lattice point. For x = 0, we get y = 0 + 2 = 2, so the point is (0, 2), which is a lattice point as well. And for x = 1, we have y = 1 + 2 = 3, giving us the lattice point (1, 3).

Finally, for x = 2, we get y = 2 + 2 = 4, which gives us the point (2, 4), which is point C. So, we’ve reached the end of the segment! We've meticulously gone through each possible integer x-coordinate and found its corresponding integer y-coordinate.

Thus, the lattice points on the diagonal AC are: A(-4, -2), (-3, -1), (-2, 0), (-1, 1), (0, 2), (1, 3), and C(2, 4). Woohoo! We've successfully identified all the lattice points. This part of the problem really highlights the beauty of linking algebra (the line equation) with geometry (the coordinate plane). By understanding this connection, we can solve problems in a really elegant and efficient way. Plus, it's super satisfying to see how different parts of math fit together!

In summary, we’ve not only found the missing vertex of the square but also pinpointed all the lattice points along its diagonal. It's like we've solved a mini-mystery, and the solutions are the jewels we've uncovered along the way. These kinds of problems are more than just math exercises; they’re opportunities to sharpen our problem-solving skills and appreciate the interconnectedness of mathematical concepts. So, pat yourselves on the back, guys, for tackling this geometry challenge head-on! We’ve shown how a systematic approach and a solid understanding of basic principles can help us conquer even the trickiest geometric terrain. Keep practicing, keep exploring, and most importantly, keep enjoying the journey of learning mathematics!