Parallel Lines: Equations And Student Solutions

by Andrew McMorgan 48 views

Hey Plastik Magazine readers! Let's dive into a geometry problem that's super common in math classes: finding the equation of a line parallel to another. This is the kind of stuff you'll encounter whether you're brushing up on your math skills or helping your kids with their homework. We'll break down the problem step-by-step, check out what Trish and Demetri did, and see if they nailed it. Ready to roll?

The Problem: Finding the Equation of a Parallel Line

So, here's the deal, guys: We've got a geometry class facing a classic problem. The task? To find the equation of a line that's parallel to the line yβˆ’3=βˆ’(x+1)y - 3 = -(x + 1) and also passes through the point (4, 2). This kind of problem really tests your understanding of lines, slopes, and how they relate to each other. Remember, parallel lines are lines that never intersect, which means they have the same slope. Understanding this concept is key to solving this problem.

First, let's look at the given line: yβˆ’3=βˆ’(x+1)y - 3 = -(x + 1). This equation is written in what's called point-slope form. The point-slope form of a linear equation is a way of writing the equation using a point on the line (x1,y1)(x_1, y_1) and the slope of the line, mm. The general formula is yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1). To make things easier, we should rewrite it in slope-intercept form, which is y=mx+by = mx + b, where mm is the slope and bb is the y-intercept. To do this, we simplify the equation. Distribute the negative sign on the right side: yβˆ’3=βˆ’xβˆ’1y - 3 = -x - 1. Add 3 to both sides to isolate y: y=βˆ’x+2y = -x + 2. Now, the equation is in slope-intercept form. It's easy to see that the slope of this line is -1. This means any line parallel to this one will also have a slope of -1. Now, our mission is to find the equation of the parallel line that passes through the point (4, 2). To solve this problem, we need to know what parallel lines are. These are lines in the same plane that never intersect.

Knowing this, we know that two parallel lines have the same slope. Now, with the slope in hand, we are ready to figure out the answer and check our students' answers! To solve this problem, we'll be using both the slope-intercept form and the point-slope form of linear equations. This is where Trish and Demetri's answers come in. This problem provides a great opportunity to explore different forms of linear equations and how they relate to each other. Working through this problem reinforces the concepts of slope, parallel lines, and the different ways to represent a linear equation. Also, it gives us a chance to test our knowledge of math!

Trish's Solution: yβˆ’2=βˆ’1(xβˆ’4)y - 2 = -1(x - 4)

Trish jumped in with the equation yβˆ’2=βˆ’1(xβˆ’4)y - 2 = -1(x - 4). This equation is in point-slope form. Let's break it down, shall we? Remember the point-slope form: yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1). Here, Trish has identified the point (4, 2) which is correct, and used -1 as the slope, also correct, since parallel lines have the same slope. Let's simplify Trish's equation to see if it makes sense. Distribute the -1 on the right side: yβˆ’2=βˆ’x+4y - 2 = -x + 4. Then, add 2 to both sides: y=βˆ’x+6y = -x + 6. So, the equation Trish provided is indeed a line with a slope of -1, parallel to the original line, and it passes through the point (4, 2).

To make sure this is crystal clear, let's plug in the point (4, 2) into the equation y=βˆ’x+6y = -x + 6. If we substitute x = 4, we get: y=βˆ’4+6=2y = -4 + 6 = 2. Bingo! The point (4, 2) satisfies the equation, so it's on the line. In mathematical terms, Trish is spot on! Her method is correct, and her final answer, when simplified, meets all the requirements of the problem.

Let's recap what makes Trish's solution legit. First, it uses the correct slope (-1) because the lines are parallel. Second, it correctly uses the given point (4, 2). Third, the resulting equation, when simplified, is a linear equation. This is the perfect example of how to solve this kind of math problem. Trish's solution is a great illustration of understanding and applying the fundamental concepts of parallel lines and linear equations. She not only identified the right slope but also correctly used the point to construct the correct equation. Her answer perfectly meets all the required criteria. We can say that Trish totally crushed this problem! Great job, Trish! This is a simple but important problem that can be understood using the basics of algebra and geometry.

Demetri's Solution: y=βˆ’x+6y = -x + 6

Now, let's see what Demetri did. He gave us the equation y=βˆ’x+6y = -x + 6. This equation is in slope-intercept form, where the slope is -1 and the y-intercept is 6. This is perfect! The slope matches the original line and Trish's solution. So, the line Demetri provided is parallel to the original line. Let's check if the point (4, 2) is on this line, too. Substitute x = 4 into the equation: y=βˆ’4+6=2y = -4 + 6 = 2. So the point (4, 2) is indeed on Demetri's line.

This confirms that Demetri also found the correct equation for a line parallel to the original line and passing through the given point. Since this is in slope-intercept form, the slope is readily apparent, making it easy to confirm that the line is parallel. Both Trish and Demetri's equations represent the same line, just in different forms, which is awesome! Now, let’s dig a little deeper to fully understand why their answers are correct. The solution also provides a great example of how to solve similar problems. If the problem had a different point or a different slope, we would have to adapt to the new conditions. But the core concepts will remain the same. The process of converting the equation to slope-intercept form makes it easier to verify that the slope is correct and to find the y-intercept.

To summarize, Demetri’s solution is perfect! He correctly determined the slope and found the y-intercept by applying the given information about the point. By using the slope-intercept form, Demetri's answer shows a complete understanding of the relationship between parallel lines, slopes, and the point-slope method. If you're a student, use their solutions as a guide in your future math exercises, guys! Always check your final equations to ensure they satisfy all conditions of the problem.

Are the Students Correct?

Absolutely, both Trish and Demetri are correct! They both found the equation of a line that is parallel to yβˆ’3=βˆ’(x+1)y - 3 = -(x + 1) and passes through the point (4, 2). Trish's answer, in point-slope form, yβˆ’2=βˆ’1(xβˆ’4)y - 2 = -1(x - 4), and Demetri's answer, in slope-intercept form, y=βˆ’x+6y = -x + 6, are both valid and represent the same line. The key takeaway is that understanding the properties of parallel lines (same slope) and being able to apply the point-slope form and slope-intercept form is crucial. These equations are simply different ways of expressing the same line. The fact that they can express the same line in two different ways demonstrates a thorough understanding of the concept.

Also, it is essential to be comfortable with both forms. Knowing how to convert between the point-slope form and the slope-intercept form allows you to check your work and ensure accuracy. This is especially true when dealing with parallel lines, where understanding the slope is critical. These problems highlight the importance of understanding the concepts behind the equations, not just memorizing formulas. Remember, mastering these concepts helps solve more complex math problems and improves overall math skills. This is the kind of stuff that will make the difference! And that's all, folks! Hope this helps you understand parallel lines a bit better. Keep up the great work in your math classes! And, see you in the next one!