Line Reflection After Translation: A Math Problem

Let's dive into a cool math problem, guys! We're going to figure out what happens when we move a line using something called translation. Specifically, we'll be looking at the line described by the equation $5x - 64 + 30 = 0$ and what its new position is after we shift it using a translation vector. Don't worry if some of these words sound a bit intimidating; we'll break it all down so it's super clear. Math can be fun, and we'll make sure this is!

Understanding the Basics

Before we jump into solving the problem, it's important to understand what we're dealing with. So, let's quickly recap the basic concepts. First, we need to get our heads around lines and their equations. Secondly, what exactly is a translation in mathematical terms? And thirdly, how does translation affect a line's equation?

Lines and Their Equations

Lines in a two-dimensional plane can be represented by linear equations. The most common form is the slope-intercept form, which looks like $y = mx + c$, where $m$ is the slope (the steepness of the line) and $c$ is the y-intercept (where the line crosses the y-axis). Another form is the standard form, which is $Ax + By + C = 0$. Our line in the problem, $5x - 64 + 30 = 0$, is in a slightly modified version of the standard form. Let's simplify it first: $5x - 34 = 0$. This equation represents a vertical line because it can be further simplified to $x = \frac{34}{5}$, meaning that for every point on this line, the x-coordinate is always $\frac{34}{5}$, regardless of the y-coordinate. Visualizing this is key – imagine a straight vertical line on a graph cutting through the x-axis at $\frac{34}{5}$.

What is Translation?

Translation, in mathematical terms, is simply moving a geometric shape from one place to another without rotating or resizing it. Think of it as sliding the shape. This movement is defined by a vector, which tells us the direction and distance to move the shape. In our problem, the translation vector is given as $T = \begin{pmatrix} 10 \\ -5 \end{pmatrix}$. This vector tells us to move the line 10 units to the right (positive x-direction) and 5 units down (negative y-direction). It’s like giving the line a little nudge in a specific direction.

How Translation Affects a Line's Equation

Now, the tricky part is figuring out how this movement affects the line's equation. For a vertical line, like the one we have, translation in the y-direction doesn't actually change the equation. Why? Because the equation only cares about the x-coordinate. Shifting it up or down doesn't change the fact that all points on the line have the same x-coordinate. However, translation in the x-direction does affect the equation. If we move the line to the right, we're changing the x-coordinate where the line intersects the x-axis. This change needs to be reflected in the equation. Essentially, we're looking for a new equation that represents the line after it has been slid 10 units to the right.

Solving the Problem Step-by-Step

Okay, now that we have all the background info, let's actually solve this problem! We'll take it one step at a time to make sure we don't miss anything.

1. Simplify the Original Equation

The original equation is $5x - 64 + 30 = 0$. Let's simplify this by combining the constant terms:

$5x - 34 = 0$

Now, let's isolate $x$ to make it even clearer what our line looks like:

$5x = 34$

$x = \frac{34}{5}$

So, our original line is a vertical line at $x = \frac{34}{5}$. This means every point on the line has an x-coordinate of $\frac{34}{5}$, no matter what the y-coordinate is.

2. Understand the Translation Vector

The translation vector is $T = \begin{pmatrix} 10 \\ -5 \end{pmatrix}$. This means we're shifting the line 10 units to the right and 5 units down. Remember, for a vertical line, the vertical shift doesn't change the equation, so we only need to worry about the horizontal shift.

3. Apply the Translation

Since we're moving the line 10 units to the right, we need to subtract 10 from the original x-coordinate. This is because if the original line was at $x = a$, moving it 10 units right means the new line will be at $x = a + 10$. To express this in the equation, we replace $x$ with $(x - 10)$ in the original equation.

So, let's think about this logically. Our original line is at $x = \frac{34}{5}$. After the translation, the new line will be 10 units to the right. That means the new x-coordinate will be:

$\frac{34}{5} + 10 = \frac{34}{5} + \frac{50}{5} = \frac{84}{5}$

So, the translated line will be at $x = \frac{84}{5}$.

4. Write the Equation of the Translated Line

Now we know the new line is a vertical line at $x = \frac{84}{5}$. To write the equation in the standard form (like the original equation), we can do the following:

$x = \frac{84}{5}$

Multiply both sides by 5 to get rid of the fraction:

$5x = 84$

Subtract 84 from both sides to get the equation in standard form:

$5x - 84 = 0$

5. The Answer

Therefore, the equation of the line after the translation is $5x - 84 = 0$. That's it! We've successfully found the new position of the line after it was translated.

Breaking Down the Logic Further

To make sure we really nail this concept, let’s explore the logic behind why we replace $x$ with $(x - 10)$ when translating to the right.

Thinking About Points

Imagine a point on the original line. Let's say a point on the line $x = \frac{34}{5}$ is $(\frac{34}{5}, 0)$. After the translation $T = \begin{pmatrix} 10 \\ -5 \end{pmatrix}$, this point will move to:

$(\frac{34}{5} + 10, 0 - 5) = (\frac{84}{5}, -5)$

Notice that the x-coordinate of the new point is $\frac{84}{5}$, which confirms our calculation earlier. The y-coordinate changed due to the vertical translation, but as we discussed, this doesn't affect the equation of a vertical line.

The General Rule

In general, when translating a graph by a vector $\begin{pmatrix} h \\ k \end{pmatrix}$, we replace $x$ with $(x - h)$ and $y$ with $(y - k)$ in the original equation. This rule works for any type of function, not just lines. It's a fundamental concept in transformations of graphs.

Why Does This Work?

The reason this works is that we're essentially asking: