Graphing A Line: Find Points Using Slope And Coordinates

Hey guys, let's dive into the awesome world of graphing lines! Today, we've got a cool problem where Vera wants to draw a line, and she's given us a starting point and the slope. Our mission? To find three other points that totally belong on this line. This isn't just about getting the answer; it's about understanding why these points work, so you can tackle any line-graphing challenge that comes your way. We'll break down how the slope and the given point work together to define the entire line, and how you can use that to find any point you need. So, grab your virtual pencils, and let's get plotting!

Understanding the Building Blocks: Point and Slope

Alright, so Vera's line starts at the point (0, 2). This is super important because it's our anchor, the spot where the line definitely exists. Think of it as the starting line for our graph. Now, the other crucial piece of information is the slope, which is given as $\frac{2}{3}$. What does slope even mean, right? In simple terms, it's the 'rise over run' of the line. For every 'run' (horizontal change) you take on the graph, you get a certain 'rise' (vertical change). A slope of $\frac{2}{3}$ means that for every 3 units you move to the right (run = +3), you move 2 units up (rise = +2). Conversely, if you move 3 units to the left (run = -3), you move 2 units down (rise = -2). This ratio is constant for the entire line. It's the secret sauce that tells us the direction and steepness of the line. Without both a point and a slope, we wouldn't be able to uniquely define a line. But since we have both, we can figure out any other point on this specific line. It’s like having a map and a compass – you know exactly where you are and which way to go!

How to Find More Points Using the Slope

Now that we've got our starting point $(0, 2)$ and our slope $\frac{2}{3}$, we can actually find other points on Vera's line. The slope $\frac{2}{3}$ tells us that for every increase of 3 in the x-coordinate, the y-coordinate increases by 2. Conversely, for every decrease of 3 in the x-coordinate, the y-coordinate decreases by 2. Let's use our starting point $(0, 2)$ and apply this rule:

• Moving Forward: If we add 3 to the x-coordinate (0 + 3 = 3) and add 2 to the y-coordinate (2 + 2 = 4), we get a new point (3, 4). This point should be on the line.
• Moving Forward Again: Let's do it again from (3, 4). Add 3 to the x-coordinate (3 + 3 = 6) and add 2 to the y-coordinate (4 + 2 = 6). This gives us another point (6, 6).
• Moving Backward: What if we go in the opposite direction? From our starting point $(0, 2)$, let's subtract 3 from the x-coordinate (0 - 3 = -3) and subtract 2 from the y-coordinate (2 - 2 = 0). This gives us the point (-3, 0).

So, based on our starting point $(0, 2)$ and the slope $\frac{2}{3}$, we've found three points: (3, 4), (6, 6), and (-3, 0). These are the points that Vera could use to graph her line. It's all about consistently applying that rise-over-run rule from a known point. Pretty neat, huh?

Verifying the Points: Why They Work

We've identified three potential points: $(-3,0)$, $(2,5)$, and $(3,4)$, $(6,6)$, and $(-3,0)$. But why do these specific points work, and why don't others? Let's re-examine the relationship between our starting point $(0,2)$ and the slope $\frac{2}{3}$. The slope formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$. We know our starting point $(x_1, y_1) = (0, 2)$ and our slope $m = \frac{2}{3}$. Let's test each of the potential points provided in the options to see if they satisfy this relationship with our starting point.

• Option A: $(-3,0)$ Let's use $(x_2, y_2) = (-3, 0)$. The slope between $(0, 2)$ and $(-3, 0)$ would be: $m = \frac{0 - 2}{-3 - 0} = \frac{-2}{-3} = \frac{2}{3}$. Since this matches Vera's given slope, $(-3,0)$ is a valid point.

• Option B: $(-2,-3)$ Let's use $(x_2, y_2) = (-2, -3)$. The slope between $(0, 2)$ and $(-2, -3)$ would be: $m = \frac{-3 - 2}{-2 - 0} = \frac{-5}{-2} = \frac{5}{2}$. This slope ($\frac{5}{2}$) is not equal to Vera's given slope ($\frac{2}{3}$), so $(-2,-3)$ is not on the line.

• Option C: $(2,5)$ Let's use $(x_2, y_2) = (2, 5)$. The slope between $(0, 2)$ and $(2, 5)$ would be: $m = \frac{5 - 2}{2 - 0} = \frac{3}{2}$. This slope ($\frac{3}{2}$) is not equal to Vera's given slope ($\frac{2}{3}$), so $(2,5)$ is not on the line.

• Option D: $(3,4)$ Let's use $(x_2, y_2) = (3, 4)$. The slope between $(0, 2)$ and $(3, 4)$ would be: $m = \frac{4 - 2}{3 - 0} = \frac{2}{3}$. Since this matches Vera's given slope, $(3,4)$ is a valid point.

• Option E: $(6,6)$ Let's use $(x_2, y_2) = (6, 6)$. The slope between $(0, 2)$ and $(6, 6)$ would be: $m = \frac{6 - 2}{6 - 0} = \frac{4}{6} = \frac{2}{3}$. Since this matches Vera's given slope, $(6,6)$ is a valid point.

So, the three points that Vera could use to graph the line are $(-3,0)$, $(3,4)$, and $(6,6)$. These points all maintain the $\frac{2}{3}$ slope relative to the starting point $(0,2)$, confirming they lie on the same line. It's all about consistency with that slope ratio!

Visualizing the Line on a Graph

Now, let's picture this on a graph, guys. Imagine the standard x-y coordinate plane. Vera's starting point is $(0,2)$. This is where the y-axis and the line intersect (the y-intercept, how cool is that?!). From this point, we apply our slope, $\frac{2}{3}$.

If we move right 3 units (from x=0 to x=3) and up 2 units (from y=2 to y=4), we land on the point $(3,4)$. See how that fits? It’s like taking steps: 3 steps to the right, 2 steps up.

Let's take another set of steps from $(3,4)$. Move right another 3 units (from x=3 to x=6) and up another 2 units (from y=4 to y=6). We arrive at $(6,6)$. This point continues the pattern, showing it's definitely on the same line.

What about going the other way? From our starting point $(0,2)$, if we move left 3 units (from x=0 to x=-3) and down 2 units (from y=2 to y=0), we arrive at $(-3,0)$. This shows that the line extends in both directions, and $(-3,0)$ is just as valid as the points we found by moving forward.

When you plot these points – $(0,2)$, $(-3,0)$, $(3,4)$, and $(6,6)$ – on graph paper, you'll see they all line up perfectly, forming a straight line. The slope $\frac{2}{3}$ dictates the angle and direction, and the point $(0,2)$ anchors it. If you were to try and plot a point like $(-2,-3)$ or $(2,5)$, you'd notice it just doesn't fall on that same straight path. They'd be off on their own, not part of Vera's line. This visual check is a great way to confirm your answers and really solidify your understanding of how points, slope, and lines are interconnected. It’s like seeing the whole picture come together!

Conclusion: The Power of Slope and Coordinates

So, there you have it! We've figured out how to find points on a line when given a starting point and a slope. Vera wants to graph a line that passes through $(0,2)$ and has a slope of $\frac{2}{3}$. By understanding that the slope $\frac{2}{3}$ means 'rise 2 for every run 3', we can find other points on the line. We can move from the given point $(0,2)$ by adding 3 to the x-coordinate and 2 to the y-coordinate repeatedly. This gave us $(3,4)$ and $(6,6)$. We also found that moving in the opposite direction, by subtracting 3 from x and subtracting 2 from y, gives us $(-3,0)$.

We rigorously checked these points using the slope formula, confirming that the slope between $(0,2)$ and each of these points is indeed $\frac{2}{3}$. This ensures they lie on the same line. Options B and C did not satisfy this condition, proving they are not on Vera's line. Therefore, the three points Vera could use to graph her line are $(-3,0)$, $(3,4)$, and $(6,6)$. This method works for any line where you have a point and a slope. It's a fundamental concept in algebra and geometry, and mastering it will make graphing lines a breeze! Keep practicing, guys, and you'll be a graphing pro in no time!