Rotating A Line: Finding The New Equation

Oct 24, 2025 by SLV Team 42 views

Hey guys! Let's dive into a cool math problem today: figuring out what happens when you spin a line around a point. Specifically, we’re going to tackle the question: What’s the equation of the line you get when you rotate the line $5x - 2y + 5 = 0$ by $90^{\circ}$ around the point $(4,1)$ ? This might sound a bit intimidating at first, but don't worry, we'll break it down step by step. Think of it like giving a line a little spin on a dance floor – we just need to figure out where it lands!

Understanding Rotations and Lines

Before we jump into the nitty-gritty, let's quickly recap what it means to rotate a line. Imagine you have a line drawn on a piece of paper, and you stick a pin at a certain point (our center of rotation). If you spin the paper, the line will move, but it will still be a straight line. The key thing that changes is its orientation, or how it's slanted. The center of rotation acts like the pivot point, and the angle of rotation tells us how much we're spinning the line.

Now, let’s talk about lines themselves. We often describe lines using equations like $5x - 2y + 5 = 0$ . This equation basically tells us the relationship between the x and y coordinates of any point that lies on the line. So, when we rotate the line, we're essentially changing this relationship, and we need to find the new equation that describes the rotated line.

To get started, it's crucial to understand the relationship between the original line and its rotated image. A rotation preserves the fundamental properties of a line—it remains straight. However, its orientation in the coordinate plane changes. This change in orientation is what we need to capture mathematically. The original line's equation provides a starting point, and the rotation parameters (angle and center) guide us in determining the new equation. We'll leverage geometric principles and algebraic manipulations to achieve this, ensuring a clear and accurate solution. Remember, the core concept is to transform the coordinates of points on the line based on the rotation, and then derive the equation that fits these transformed points.

Step-by-Step Solution

Okay, let’s get our hands dirty and solve this problem! Here’s a breakdown of the steps we’ll take:

Find the slope of the original line: The equation $5x - 2y + 5 = 0$ is in the standard form of a linear equation. We need to rewrite it in the slope-intercept form ( $y = mx + b$ ) to easily identify the slope ( $m$ ).
Determine the slope of the rotated line: Rotating a line by $90^{\circ}$ changes its slope. The new slope will be the negative reciprocal of the original slope. Think of it as flipping the fraction and changing the sign.
Find a point on the original line: To find the equation of the rotated line, we need a point that lies on it. We can pick any value for $x$ and solve for $y$ in the original equation.
Rotate the point: Now, we need to rotate the point we found in the previous step around the center of rotation $(4,1)$ . This involves using rotation transformation formulas.
Write the equation of the rotated line: We now have the slope of the rotated line and a point on it. We can use the point-slope form of a linear equation to write the equation of the rotated line.

Step 1: Finding the Slope of the Original Line

Let’s rewrite the equation $5x - 2y + 5 = 0$ in slope-intercept form ( $y = mx + b$ ).

Subtract $5x$ and $5$ from both sides: $-2y = -5x - 5$
Divide both sides by $-2$ : $y = \frac{5}{2}x + \frac{5}{2}$

So, the slope of the original line ( $m_1$ ) is $\frac{5}{2}$ .

Step 2: Determining the Slope of the Rotated Line

Rotating a line by $90^{\circ}$ means the new slope ( $m_2$ ) is the negative reciprocal of the original slope.

$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{5}{2}} = -\frac{2}{5}$

Thus, the slope of the rotated line is $-\frac{2}{5}$ . This step highlights the fundamental geometric principle that a 90-degree rotation transforms the slope of a line into its negative reciprocal. This is because the rotation effectively swaps the rise and run of the line while also changing its direction (positive to negative or vice versa). Understanding this relationship is key to solving rotation problems in coordinate geometry. Furthermore, it reinforces the connection between visual transformations and their algebraic representations, a core theme in linear algebra and geometry.

Step 3: Finding a Point on the Original Line

Let’s pick a convenient value for $x$ , say $x = 1$ , and plug it into the original equation to solve for $y$ .

$5(1) - 2y + 5 = 0$
$10 - 2y = 0$
$2y = 10$
$y = 5$

So, the point $(1, 5)$ lies on the original line.

Step 4: Rotating the Point

Now, this is where things get a little more interesting. We need to rotate the point $(1, 5)$ around the center of rotation $(4, 1)$ by $90^{\circ}$ . The formulas for rotating a point $(x, y)$ around a center $(h, k)$ by an angle $\theta$ are:

$x' = (x - h)\cos(\theta) - (y - k)\sin(\theta) + h$
$y' = (x - h)\sin(\theta) + (y - k)\cos(\theta) + k$

In our case, $(x, y) = (1, 5)$ , $(h, k) = (4, 1)$ , and $\theta = 90^{\circ}$ . Remember that $\cos(90^{\circ}) = 0$ and $\sin(90^{\circ}) = 1$ .

Let's plug in the values:

$x' = (1 - 4)(0) - (5 - 1)(1) + 4 = -4 + 4 = 0$
$y' = (1 - 4)(1) + (5 - 1)(0) + 1 = -3 + 1 = -2$

So, the rotated point is $(0, -2)$ . This rotation transformation is a fundamental concept in linear algebra and has wide applications beyond geometry, such as in computer graphics and robotics. The formulas we used are derived from matrix transformations, offering a powerful tool for handling rotations and other linear transformations in higher dimensions. By understanding these formulas, we not only solve this specific problem but also gain insight into a more general framework for geometric transformations.

Step 5: Writing the Equation of the Rotated Line

We now have the slope of the rotated line ( $m_2 = -\frac{2}{5}$ ) and a point on it $(0, -2)$ . We can use the point-slope form of a linear equation:

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

Plugging in the values, we get:

$y - (-2) = -\frac{2}{5}(x - 0)$
$y + 2 = -\frac{2}{5}x$

To get rid of the fraction, multiply both sides by 5:

$5(y + 2) = -2x$
$5y + 10 = -2x$

Finally, rearrange the equation into the standard form:

$2x + 5y + 10 = 0$

The Answer

So, the equation of the line resulting from rotating the line $5x - 2y + 5 = 0$ by $90^{\circ}$ about the point $(4,1)$ is $2x + 5y + 10 = 0$ . Awesome job, guys! We successfully navigated through this problem by breaking it down into manageable steps and applying the principles of coordinate geometry and transformations. Remember, the key is to understand the underlying concepts and how they connect to each other.

Key Takeaways

Rotating a line changes its slope, and the new slope after a $90^{\circ}$ rotation is the negative reciprocal of the original slope.
Rotation transformations involve specific formulas that help us find the new coordinates of a point after rotation.
The point-slope form of a linear equation is a powerful tool for writing the equation of a line when we know its slope and a point on it.
Breaking down complex problems into smaller, manageable steps makes them easier to solve.

In summary, this problem showcases the interplay between geometry and algebra, highlighting how geometric transformations can be expressed and analyzed using algebraic tools. The solution not only provides a specific answer but also reinforces fundamental mathematical concepts that are applicable in various fields, from engineering to computer science. The ability to visualize and mathematically represent geometric transformations is a valuable skill, and this example serves as a solid foundation for tackling more complex problems in the future. Remember, practice makes perfect, so keep exploring and challenging yourself with similar problems!

I hope this explanation was helpful! Keep practicing, and you'll become a pro at these types of problems in no time! Let me know if you have any other questions. Happy problem-solving!