General Equation Of A Line Passing Through A Point And Parallel To A Given Line

Hey guys! Let's dive into a fascinating problem in coordinate geometry – finding the general equation of a line. This isn't just about crunching numbers; it's about understanding the fundamental relationships between lines and points in a plane. We're going to break down the problem step-by-step, so you'll not only get the answer but also grasp the underlying concepts. So, buckle up and let’s embark on this mathematical journey together!

The Problem at Hand

Okay, so here's the problem we're tackling: What is the general equation of a line that passes through the point $(2,3)$ and is parallel to the line given by the parametric equations:

$$\begin{cases} x = 1 + 2\lambda \\ y = -2 - \lambda \end{cases}, \lambda \in R$$

We have four options to choose from:

A. $x + y - 4 = 0$ B. $x + 2y - 8 = 0$ C. $-2x - y + 7 = 0$ D. $-2x - y + 4 = 0

This might seem a bit daunting at first, but don't worry! We'll dissect it piece by piece. To start, we need to understand what each part of the problem is telling us.

Decoding the Problem

Before we jump into calculations, let's make sure we understand the key elements of the problem. We're dealing with lines and points in a 2D plane, so visualizing this can be super helpful. Imagine a coordinate grid – that's our playground for this problem.

• The Goal: We need to find the general equation of a line. Remember, the general form of a linear equation is $Ax + By + C = 0$, where A, B, and C are constants.
• The Given Point: We know the line passes through the point $(2,3)$. This is a crucial piece of information because any point on the line must satisfy the line's equation.
• The Parallel Line: We're given another line in parametric form. Parametric equations describe a line using a parameter (in this case, $\lambda$). The fact that our line is parallel to this one is super important – it tells us about the slope of our line.

Understanding Parametric Equations

Let's quickly recap parametric equations. The equations:

$$\begin{cases} x = 1 + 2\lambda \\ y = -2 - \lambda \end{cases}$$

tell us how to find any point $(x, y)$ on the line by plugging in different values for $\lambda$. Think of $\lambda$ as a dial that moves us along the line. Each value of $\lambda$ gives us a unique point on the line.

Parallel Lines and Slope

Here's a key concept: Parallel lines have the same slope. This is a fundamental geometric principle. So, to find the slope of our target line, we first need to find the slope of the given parallel line. Once we have the slope and a point, we can determine the equation of the line.

Step-by-Step Solution

Okay, let's roll up our sleeves and solve this problem. We'll break it down into manageable steps.

Step 1: Finding the Slope of the Given Line

We have the parametric equations:

$$\begin{cases} x = 1 + 2\lambda \\ y = -2 - \lambda \end{cases}$$

To find the slope, we need to express this line in the slope-intercept form ($y = mx + b$), where $m$ is the slope. To do this, we'll eliminate the parameter $\lambda$.

From the equation $x = 1 + 2\lambda$, we can solve for $\lambda$:

$$2\lambda = x - 1 \implies \lambda = \frac{x - 1}{2}$$

Now, substitute this value of $\lambda$ into the equation $y = -2 - \lambda$:

$$y = -2 - \frac{x - 1}{2}$$

Simplify this equation:

$$y = -2 - \frac{1}{2}x + \frac{1}{2}$$

$$y = -\frac{1}{2}x - \frac{3}{2}$$

Now we have the equation in slope-intercept form. The slope of the given line is $m = -\frac{1}{2}$.

Step 2: Using the Slope and the Point to Find the Equation

Since our line is parallel to the given line, it has the same slope, $m = -\frac{1}{2}$. We also know that our line passes through the point $(2, 3)$.

We can use the point-slope form of a linear equation:

$$y - y_1 = m(x - x_1)$$

where $(x_1, y_1)$ is the given point and $m$ is the slope.

Plug in the values $x_1 = 2$, $y_1 = 3$, and $m = -\frac{1}{2}$:

$$y - 3 = -\frac{1}{2}(x - 2)$$

Step 3: Converting to General Form

Now, let's convert this equation to the general form ($Ax + By + C = 0$).

Multiply both sides by 2 to get rid of the fraction:

$$2(y - 3) = -1(x - 2)$$

$$2y - 6 = -x + 2$$

Rearrange the terms to get the general form:

$$x + 2y - 8 = 0$$

Step 4: Matching with the Options

We've found the general equation of the line: $x + 2y - 8 = 0$.

Now, let's compare this with the given options:

A. $x + y - 4 = 0$ B. $x + 2y - 8 = 0$ C. $-2x - y + 7 = 0$ D. $-2x - y + 4 = 0

Our equation matches option B!

The Answer

Therefore, the general equation of the line passing through the point $(2,3)$ and parallel to the given line is:

B. $x + 2y - 8 = 0$

Why This Works: A Deeper Dive

Let's take a moment to reflect on why this method works. It's not just about blindly following steps; it's about understanding the underlying mathematical principles.

• Parametric to Cartesian: Converting from parametric equations to the slope-intercept form (or directly finding the slope) allows us to extract the directional information (the slope) from the parametric representation. Parametric equations are great for describing movement along a line, while the Cartesian form (like $y = mx + b$ or $Ax + By + C = 0$) is better for describing the line itself as a static entity.
• Parallel Lines and Slopes: The fact that parallel lines have the same slope is a cornerstone of Euclidean geometry. It allows us to transfer information from one line to another. This concept is crucial in many geometric problems.
• Point-Slope Form: The point-slope form is a powerful tool because it directly incorporates the slope and a point on the line. It's a natural way to build the equation of a line when you have this information.
• General Form: The general form is useful because it's a standard way to represent linear equations, making it easy to compare different lines and perform algebraic manipulations.

Common Pitfalls to Avoid

While solving this type of problem, it's easy to make a few common mistakes. Let's highlight them so you can avoid them in the future:

• Incorrectly Eliminating the Parameter: When converting from parametric to Cartesian form, make sure you eliminate the parameter $\lambda$ correctly. Double-check your algebra to avoid errors.
• Forgetting the Sign: A small sign error can completely change the equation of the line. Pay close attention to positive and negative signs throughout your calculations.
• Confusing Slopes of Parallel and Perpendicular Lines: Remember that parallel lines have the same slope, while perpendicular lines have slopes that are negative reciprocals of each other. Mixing these up is a common mistake.
• Not Converting to General Form: The problem specifically asks for the general equation. Make sure you convert your equation to the form $Ax + By + C = 0$ before comparing with the options.

Real-World Applications

Okay, so we've solved a math problem. But why is this useful in the real world? Well, linear equations and the concepts of parallelism and perpendicularity pop up in many different fields:

• Navigation: GPS systems and other navigation tools rely on coordinate systems and linear equations to determine positions and routes. Understanding parallel lines is crucial for maintaining a constant course.
• Computer Graphics: In computer graphics, lines are fundamental building blocks for creating images and animations. Understanding how to define and manipulate lines is essential.
• Engineering: Engineers use linear equations to model various systems, from the forces in a bridge to the flow of electricity in a circuit. Parallel and perpendicular lines play a key role in structural design.
• Economics: Linear equations are used to model supply and demand curves, cost functions, and other economic relationships. Understanding the slopes and intercepts of these lines can provide valuable insights.

Practice Makes Perfect

The best way to master these concepts is to practice! Try solving similar problems with different points and parallel lines. You can also explore problems involving perpendicular lines to further solidify your understanding.

Practice Problems

1. Find the general equation of a line passing through the point $(-1, 2)$ and parallel to the line given by $x = 3 - \lambda$, $y = 1 + 2\lambda$.
2. What is the equation of a line passing through the origin $(0, 0)$ and parallel to the line $2x - 3y + 5 = 0$?
3. Determine the equation of a line that passes through the point $(4, -1)$ and is parallel to the line defined by the points $(1, 1)$ and $(3, 5)$.

By working through these problems, you'll build your confidence and develop a deeper understanding of linear equations and their applications.

Conclusion: Mastering the Line

We've journeyed through the process of finding the general equation of a line, starting from parametric equations and a given point. We've dissected the problem, solved it step-by-step, and explored the underlying concepts. We've also looked at common pitfalls and real-world applications.

Remember, mathematics is not just about memorizing formulas; it's about understanding the relationships between different concepts. By grasping these relationships, you can tackle a wide range of problems with confidence.

So, keep practicing, keep exploring, and keep unraveling the mysteries of mathematics! You've got this!