Graphing A Line: Find Points Given Slope & Intercept

Hey guys! Let's tackle a common math problem: graphing a line when you know a point it passes through and its slope. In this case, we're helping Vera, who wants to graph a line that goes through the point (0, 2) and has a slope of 2/3. The question is: which points could Vera use to graph this line? We have a few options to choose from: (-3, 0), (-2, -3), (2, 5), (3, 4), and (6, 6). Let's break this down step by step so you can easily solve similar problems!

Understanding Slope and Intercept

Before we dive into the options, let's make sure we're all on the same page about slope and intercept. These are fundamental concepts in linear equations and graphing. The slope of a line tells us how steep the line is and in what direction it's going. Mathematically, it's defined as the "rise over run," which means the change in the y-coordinate divided by the change in the x-coordinate between any two points on the line. A slope of 2/3 means that for every 2 units we move up (rise), we move 3 units to the right (run). The y-intercept is the point where the line crosses the y-axis. In our problem, Vera's line passes through (0, 2), which means the y-intercept is 2. This is super helpful because it gives us a starting point for graphing the line.

The slope-intercept form of a linear equation is y = mx + b, where 'm' represents the slope and 'b' represents the y-intercept. In Vera's case, the slope (m) is 2/3, and the y-intercept (b) is 2. So, the equation of the line is y = (2/3)x + 2. This equation is our key to checking which points lie on the line. If we plug in the x and y coordinates of a point into this equation and it holds true, then that point is on the line. If it doesn't, then the point is not on the line. Understanding this relationship between the equation of a line and the points that lie on it is crucial for graphing lines accurately. Now that we've refreshed our understanding of slope and intercept, we can confidently move on to evaluating the given points.

When we are trying to find points on a line, it's not enough to just guess and check. We need a systematic way to determine whether a point satisfies the equation of the line. This is where the equation y = (2/3)x + 2 becomes our best friend. For each point, we'll substitute the x-coordinate into the equation and see if the resulting y-value matches the y-coordinate of the point. If it does, then the point is on the line; if it doesn't, then it's not. This process is the foundation for accurately graphing lines and solving linear equations. It ensures that we're not just relying on visual estimations, but on solid mathematical principles. So, let's roll up our sleeves and start plugging in those coordinates!

Checking the Options

Let's go through each option and see if it lies on the line. We'll use the equation y = (2/3)x + 2.

Option 1: (-3, 0)

For the point (-3, 0), we substitute x = -3 into the equation: y = (2/3)(-3) + 2. This simplifies to y = -2 + 2, which gives us y = 0. Since the y-coordinate we calculated (0) matches the y-coordinate of the point (-3, 0), this point does lie on the line. This confirms that when x is -3, y is indeed 0, so Vera could use this point to graph her line. Keep track of this one; it's a valid option!

The calculation we performed here highlights the importance of following the order of operations and handling fractions correctly. Multiplying (2/3) by -3 involves multiplying the numerator (2) by -3, resulting in -6, and then dividing by the denominator (3), giving us -2. Adding this to the y-intercept (2) leads to the final y-value of 0. This meticulous approach ensures that we arrive at the correct answer and avoid errors that can arise from hasty calculations. The fact that the calculated y-value matched the given y-coordinate of the point (-3, 0) solidifies our confidence in this point being on the line.

Option 2: (-2, -3)

Next, let's check the point (-2, -3). Substituting x = -2 into the equation, we get: y = (2/3)(-2) + 2. This simplifies to y = -4/3 + 2. To add these, we need a common denominator, so we rewrite 2 as 6/3. Now we have y = -4/3 + 6/3, which gives us y = 2/3. The y-coordinate we calculated (2/3) does not match the y-coordinate of the point (-2, -3). Therefore, this point does not lie on the line. This demonstrates that not every point will fit the line's equation, even if it seems close at first glance.

This calculation emphasizes the need for precision when dealing with fractions. The fraction -4/3 represents a value between -1 and -2, and adding it to 2 requires a careful conversion to a common denominator. The process of converting 2 to 6/3 allows us to directly add the numerators and arrive at the correct result of 2/3. This careful attention to detail is what sets apart accurate solutions from approximations. The significant difference between the calculated y-value (2/3) and the given y-coordinate (-3) clearly indicates that this point is not on the line, reinforcing the importance of the algebraic method in verifying points.

Option 3: (2, 5)

Let's move on to the point (2, 5). Substituting x = 2 into the equation, we get: y = (2/3)(2) + 2. This simplifies to y = 4/3 + 2. Again, we need a common denominator, so we rewrite 2 as 6/3. Now we have y = 4/3 + 6/3, which gives us y = 10/3. Converting this improper fraction to a mixed number, we get y = 3 1/3. The y-coordinate we calculated (3 1/3) does not match the y-coordinate of the point (2, 5). Thus, this point does not lie on the line.

This step highlights the importance of being comfortable with fractions and mixed numbers. Converting the improper fraction 10/3 to the mixed number 3 1/3 helps us quickly see that the calculated y-value is quite different from the given y-coordinate of 5. This difference immediately tells us that the point (2, 5) is not on the line. The ability to perform these conversions efficiently saves time and reduces the chances of making errors. Just like in the previous step, the algebraic method has provided a clear and definitive answer, further emphasizing its reliability in determining whether a point lies on a line.

Option 4: (3, 4)

Now, let's check the point (3, 4). Substituting x = 3 into the equation, we get: y = (2/3)(3) + 2. This simplifies to y = 2 + 2, which gives us y = 4. Since the y-coordinate we calculated (4) matches the y-coordinate of the point (3, 4), this point does lie on the line. This provides another valid point that Vera could use to graph her line. Remember to keep track of this one!

This calculation is a great example of how the slope and y-intercept work together to define the line. The slope of 2/3 dictates how the y-value changes with respect to the x-value, and the y-intercept provides the starting point. In this case, when x is 3, the slope contributes 2 to the y-value, and adding the y-intercept of 2 brings us to a total y-value of 4. The fact that the calculated y-value perfectly matches the given y-coordinate reinforces our understanding of the line's equation and its behavior. The simplicity of this calculation also demonstrates the elegance and efficiency of the algebraic method when dealing with linear equations.

Option 5: (6, 6)

Finally, let's check the point (6, 6). Substituting x = 6 into the equation, we get: y = (2/3)(6) + 2. This simplifies to y = 4 + 2, which gives us y = 6. Since the y-coordinate we calculated (6) matches the y-coordinate of the point (6, 6), this point does lie on the line. This gives us our third point that Vera could use. We've nailed it!

This final calculation ties everything together and confirms our understanding of the problem. When x is 6, the slope of 2/3 contributes 4 to the y-value, and adding the y-intercept of 2 gives us a total y-value of 6. The perfect match between the calculated and given y-coordinates provides a satisfying conclusion to our point-checking process. This step also underscores the consistency and reliability of the algebraic method in identifying points on a line. With this point confirmed, we have successfully identified all the points that Vera could use to graph her line.

Conclusion

So, after checking all the options, we found that the points (-3, 0), (3, 4), and (6, 6) lie on the line that passes through (0, 2) and has a slope of 2/3. Vera could confidently use these points to graph her line accurately. Remember, the key is to use the equation y = mx + b to verify whether a point lies on the line. Just plug in the x-coordinate, calculate the y-coordinate, and see if it matches the given y-coordinate. You've got this!

By systematically checking each point against the line's equation, we have demonstrated a robust method for solving these types of problems. This approach not only provides the correct answer but also reinforces our understanding of the underlying mathematical concepts. From understanding slope and y-intercept to performing accurate calculations with fractions, each step in the process contributes to a deeper grasp of linear equations. With these skills in your toolkit, you'll be well-prepared to tackle more complex graphing problems in the future. So keep practicing, and remember, the key to success is a solid understanding of the fundamentals.