Translation Of Lines: Solving Image Equations

Hey guys! Today, we're diving into a super important topic in math: translation of lines. We'll be tackling some common questions about how lines shift when we move them around on a graph. This is a fundamental concept in geometry, and understanding it will help you ace your exams and build a solid foundation for more advanced topics. Let's break it down step by step!

1. Finding the Image of a Line After Translation

Okay, so the first question we're tackling is: If the line Y+X=0 is translated by t: $\begin{pmatrix} -1 \\ 2 \end{pmatrix}$, what is the equation of the image of the line? This might sound a bit intimidating at first, but trust me, it's totally doable! The key here is understanding what translation actually means.

Understanding Translation

Translation, in simple terms, is just moving a shape (in this case, a line) without rotating or resizing it. Think of it like sliding a piece of paper across a table. The translation vector $\begin{pmatrix} -1 \\ 2 \end{pmatrix}$ tells us exactly how to move the line. The top number (-1) tells us how much to move it horizontally (left if negative, right if positive), and the bottom number (2) tells us how much to move it vertically (up if positive, down if negative). So, in this case, we're moving the line 1 unit to the left and 2 units up.

The Math Behind It

Now, let's get to the math. To find the equation of the translated line, we need to consider how the coordinates of points on the line change after the translation. Let's say a point (x, y) lies on the original line Y + X = 0. After the translation, this point will move to a new location (x', y'). The relationship between the original coordinates (x, y) and the new coordinates (x', y') can be expressed as:

x' = x + (-1) => x = x' + 1 y' = y + 2 => y = y' - 2

Essentially, we're subtracting the translation vector components from the new coordinates to get back to the original coordinates. This is a crucial step in solving these types of problems.

Finding the Equation of the Image

Now, we'll substitute these expressions for x and y into the original equation of the line, Y + X = 0. This will give us the equation of the translated line in terms of x' and y':

(y' - 2) + (x' + 1) = 0

Simplify this equation:

y' + x' - 1 = 0

Finally, we can rewrite this equation in the standard form, replacing x' and y' with x and y to get the equation of the image:

Y + X - 1 = 0

So, the equation of the image of the line Y + X = 0 after the translation is Y + X - 1 = 0. See? Not so scary after all!

2. Translation with a Different Vector

Let's try another one! This time, the question is: What is the image of the line 5X-64+30=0 under the translation t : $\begin{pmatrix} 10 \\ -5 \end{pmatrix}$? Notice that the core concept is the same, but we have a different line and a different translation vector. Don't let the numbers intimidate you; we'll follow the same steps as before.

Simplifying the Original Equation

First, let's simplify the original equation of the line: 5X - 64 + 30 = 0. Combining the constants, we get:

5X - 34 = 0

This makes the equation a bit easier to work with.

Applying the Translation

Now, we apply the translation t : $\begin{pmatrix} 10 \\ -5 \end{pmatrix}$. This means we're moving the line 10 units to the right and 5 units down. Let (x, y) be a point on the original line, and let (x', y') be its image after the translation. The relationship between the coordinates is:

x' = x + 10 => x = x' - 10 y' = y - 5 => y = y' + 5

Notice how we're doing the opposite operation of the translation to get the original coordinates in terms of the new coordinates. This is key to substituting correctly.

Finding the Image Equation

Next, substitute these expressions for x and y into the simplified equation of the original line, 5X - 34 = 0:

5(x' - 10) - 34 = 0

Now, expand and simplify:

5x' - 50 - 34 = 0

5x' - 84 = 0

Finally, replace x' with x to get the equation of the image:

5X - 84 = 0

So, the image of the line 5X - 64 + 30 = 0 after the translation is 5X - 84 = 0. Great job!

3. General Equation of the Image Line

Let's tackle a slightly more general question: What is the equation of the image of the line after a transformation? This question is a bit broader, but the same principles apply. To answer this, we need a specific line equation and a specific transformation (like a translation). Let's consider a general case first and then look at specific examples.

General Approach

When dealing with the equation of the image of a line after a transformation, the core idea remains the same: find the relationship between the original coordinates (x, y) and the transformed coordinates (x', y'). This relationship will depend on the type of transformation (translation, rotation, reflection, etc.).

For a translation, as we've seen, this relationship is straightforward:

x' = x + a y' = y + b

where (a, b) is the translation vector. We then solve for x and y in terms of x' and y':

x = x' - a y = y' - b

And substitute these into the original equation of the line.

Example with a General Line Equation

Let's say we have a line with the general equation Ax + By + C = 0, and we translate it by the vector $\begin{pmatrix} a \\ b \end{pmatrix}$. Following the same procedure:

x = x' - a y = y' - b

Substitute into the original equation:

A(x' - a) + B(y' - b) + C = 0

Expand and simplify:

Ax' - Aa + By' - Bb + C = 0

Replacing x' and y' with x and y, the equation of the image is:

Ax + By - Aa - Bb + C = 0

This general form shows how the constants in the equation change after translation. The coefficients A and B, which determine the slope of the line, remain the same, but the constant term C changes to -Aa - Bb + C.

Specific Example

Let's make this more concrete with an example. Suppose our original line is 2x - 3y + 5 = 0, and we translate it by the vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Here, A = 2, B = -3, C = 5, a = 2, and b = -1. Plugging these values into our general equation for the image:

2x - 3y - (2 * 2) - (-3 * -1) + 5 = 0

Simplify:

2x - 3y - 4 - 3 + 5 = 0

2x - 3y - 2 = 0

So, the image of the line 2x - 3y + 5 = 0 after the translation is 2x - 3y - 2 = 0.

Key Takeaways

• Translation involves shifting a line without changing its orientation.
• The translation vector tells you how much to shift horizontally and vertically.
• To find the image equation, find the relationship between original and transformed coordinates.
• Substitute the expressions for x and y in terms of x' and y' into the original equation.
• Simplify the resulting equation to get the equation of the image.

Wrapping Up

So, there you have it! We've explored how to find the equations of lines after translation, covering specific examples and a general approach. Remember, the key is to understand the relationship between the original coordinates and the transformed coordinates. Once you've got that down, the rest is just algebra!

Keep practicing these types of problems, and you'll become a pro at translating lines in no time. If you have any questions, don't hesitate to ask. Happy problem-solving, guys! And remember, math is awesome! 🤓