Equation Of A Perpendicular Line Passing Through A Point

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Hey guys! Let's dive into a common problem in algebra: finding the equation of a line that's perpendicular to another line and passes through a specific point. This is a classic exercise that combines our understanding of linear equations, slopes, and point-slope form. We'll break it down step-by-step, making sure everyone can follow along. So, grab your pencils, and let's get started!

Understanding the Problem

In this scenario, the main objective is to find the equation of a line that satisfies two crucial conditions. First, it must be perpendicular to a given line, which in our case is 2xβˆ’3y=52x - 3y = 5. Second, this line needs to pass through a specific point, namely (2,βˆ’1)(2, -1). To tackle this, we'll need to utilize our knowledge of slopes and how they relate in perpendicular lines, as well as the various forms of linear equations.

The initial line, 2xβˆ’3y=52x - 3y = 5, provides us with a starting point. To work with it effectively, we need to determine its slope. This is where converting the equation into slope-intercept form (y=mx+by = mx + b) becomes handy. Once we have the slope of the given line, we can easily find the slope of any line perpendicular to it. Remember, perpendicular lines have slopes that are negative reciprocals of each other. This is a fundamental concept in coordinate geometry.

The given point, (2,βˆ’1)(2, -1), is our anchor. It’s the specific location through which our new line must pass. This is where point-slope form comes into play. Knowing a point and the slope, we can construct the equation of the line using this form. Finally, we can convert the equation into a standard form to match the answer choices provided. So, in essence, we're playing a game of connect-the-dots with slopes and points to arrive at the equation of our desired line. It might sound complex, but breaking it down into these steps makes the process quite manageable and, dare I say, fun!

Step 1: Find the Slope of the Given Line

The first step in solving this problem is to determine the slope of the given line, which is represented by the equation 2xβˆ’3y=52x - 3y = 5. To easily identify the slope, we need to convert this equation into the slope-intercept form, which is y=mx+by = mx + b. Here, mm represents the slope of the line, and bb is the y-intercept. This form is super useful because it explicitly tells us the slope and where the line crosses the y-axis. So, let's get this equation transformed!

To convert 2xβˆ’3y=52x - 3y = 5 into slope-intercept form, we need to isolate yy on one side of the equation. We'll start by subtracting 2x2x from both sides. This gives us βˆ’3y=βˆ’2x+5-3y = -2x + 5. Now, to completely isolate yy, we'll divide both sides of the equation by βˆ’3-3. This step is crucial because it ensures that the coefficient of yy becomes 1, which is what we need for the slope-intercept form. Remember to divide each term on the right side by βˆ’3-3 to maintain the equation's balance.

After dividing by βˆ’3-3, we get y=βˆ’2xβˆ’3+5βˆ’3y = \frac{-2x}{-3} + \frac{5}{-3}, which simplifies to y=23xβˆ’53y = \frac{2}{3}x - \frac{5}{3}. Aha! Now our equation is in slope-intercept form. By looking at the coefficient of xx, we can clearly see that the slope (mm) of the given line is 23\frac{2}{3}. This is a crucial piece of information. It's like finding the key to a lock – we need this slope to figure out the slope of the line that's perpendicular to it. So, with this slope in hand, we're ready to move on to the next step. It’s all about taking these little steps, guys, and the big picture becomes much clearer!

Step 2: Determine the Slope of the Perpendicular Line

Now that we've successfully found the slope of the given line, which is 23\frac{2}{3}, our next crucial step is to determine the slope of the line perpendicular to it. This involves a neat little trick that relies on the properties of perpendicular lines. Remember, two lines are perpendicular if their slopes are negative reciprocals of each other. This means we need to flip the fraction and change the sign.

The concept of negative reciprocals is fundamental in coordinate geometry. If one line has a slope of mm, a line perpendicular to it will have a slope of βˆ’1m-\frac{1}{m}. It’s like the slopes are perfectly balanced opposites, ensuring the lines meet at a right angle. Think of it as a mathematical dance where one line perfectly complements the other to create a 90-degree angle. So, with this in mind, let's find the slope of our perpendicular line.

Given that the slope of the original line is 23\frac{2}{3}, we need to find its negative reciprocal. To do this, we first flip the fraction, which gives us 32\frac{3}{2}. Then, we change the sign. Since 23\frac{2}{3} is positive, its negative reciprocal will be negative. Therefore, the slope of the line perpendicular to the given line is βˆ’32-\frac{3}{2}. See how easy that was? We just flipped and negated! This new slope is super important because it will guide us in creating the equation of our perpendicular line. We're one step closer to the solution, guys! Knowing this slope and the point it passes through, we're now equipped to build the equation of the line.

Step 3: Use the Point-Slope Form

Alright, we've got the slope of our perpendicular line, which is βˆ’32-\frac{3}{2}, and we know it passes through the point (2,βˆ’1)(2, -1). Now comes the fun part: using the point-slope form to construct the equation of the line. The point-slope form is a fantastic tool in these situations, and it’s expressed as yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is a point on the line and mm is the slope.

The point-slope form is essentially a customized equation builder. It takes a specific point and a specific slope and molds them into the equation of the line that fits those conditions. It's like a tailor who takes your measurements and creates a perfectly fitting suit. In our case, the point (2,βˆ’1)(2, -1) and the slope βˆ’32-\frac{3}{2} are our measurements, and we're about to tailor-make the equation of our perpendicular line. This form is particularly handy because it directly incorporates the information we have, making the process straightforward and efficient.

To apply the point-slope form, we'll substitute our known values. We have x1=2x_1 = 2, y1=βˆ’1y_1 = -1, and m=βˆ’32m = -\frac{3}{2}. Plugging these into the formula, we get yβˆ’(βˆ’1)=βˆ’32(xβˆ’2)y - (-1) = -\frac{3}{2}(x - 2). Notice how we carefully placed each value in its corresponding spot? This is crucial to getting the correct equation. Now, let's simplify this expression. The yβˆ’(βˆ’1)y - (-1) becomes y+1y + 1, and we have y+1=βˆ’32(xβˆ’2)y + 1 = -\frac{3}{2}(x - 2). We're on the verge of having our equation, but we need to simplify it further and convert it into a more standard form. It's like we've assembled the pieces, and now we're putting the finishing touches on our equation. So, let's move on to the next step and tidy things up!

Step 4: Simplify the Equation

Okay, we've arrived at the equation y+1=βˆ’32(xβˆ’2)y + 1 = -\frac{3}{2}(x - 2), and now it's time to simplify this equation and get it into a more recognizable form. Our goal here is to eliminate the parentheses and fractions, and then rearrange the equation into a standard form, which will help us match it with the answer choices given. This step is all about algebraic manipulation – a bit like untangling a knot to reveal a clean, neat line.

First, let’s get rid of those parentheses. We'll distribute the βˆ’32-\frac{3}{2} across the (xβˆ’2)(x - 2) term. This means multiplying βˆ’32-\frac{3}{2} by both xx and βˆ’2-2. When we multiply βˆ’32-\frac{3}{2} by xx, we get βˆ’32x-\frac{3}{2}x. And when we multiply βˆ’32-\frac{3}{2} by βˆ’2-2, the negatives cancel out, and we get +3Γ—22+\frac{3 \times 2}{2}, which simplifies to +3+3. So, our equation now looks like y+1=βˆ’32x+3y + 1 = -\frac{3}{2}x + 3.

Next, we want to eliminate the fraction. To do this, we can multiply every term in the equation by the denominator, which is 2. This will clear out the fraction and give us a cleaner equation to work with. Multiplying each term by 2, we get 2(y+1)=2(βˆ’32x)+2(3)2(y + 1) = 2(-\frac{3}{2}x) + 2(3), which simplifies to 2y+2=βˆ’3x+62y + 2 = -3x + 6. We're making great progress, guys! The equation is looking much simpler now.

Finally, let's rearrange the terms to get the equation into a standard form, typically either Ax+By=CAx + By = C. We'll add 3x3x to both sides to get 3x+2y+2=63x + 2y + 2 = 6. Then, we'll subtract 2 from both sides to isolate the constant term on one side. This gives us 3x+2y=43x + 2y = 4. Ta-da! We've simplified the equation and put it into a standard form. It’s like we’ve polished a rough stone and revealed a sparkling gem of an equation.

Step 5: Match the Equation with the Answer Choices

We've successfully simplified our equation to 3x+2y=43x + 2y = 4. Now, the final step is to match this equation with the answer choices provided in the problem. This is where we get to see if all our hard work has paid off! It’s like fitting the last piece of a puzzle – a satisfying moment when everything clicks into place.

Looking back at the problem, we have the following answer choices:

A. 3xβˆ’2y=73x - 2y = 7

B. 3x+2y=43x + 2y = 4

C. 2x+3y=12x + 3y = 1

D. 2xβˆ’3y=72x - 3y = 7

Comparing our equation, 3x+2y=43x + 2y = 4, with these options, we can see a clear match. Option B, 3x+2y=43x + 2y = 4, is exactly what we derived. Hooray! This means we've correctly navigated through the steps, from finding the slope of the given line to simplifying the equation of the perpendicular line. It’s a testament to the power of breaking down a problem into manageable steps. We started with a line and a point, and through careful calculations and transformations, we've found the equation of the line that fits perfectly. This is the essence of problem-solving in mathematics – a journey from the given to the solution, with each step building upon the previous one. So, guys, give yourselves a pat on the back; we've nailed it!

Final Answer

After carefully working through the steps, we've successfully found the equation of the line that is perpendicular to 2xβˆ’3y=52x - 3y = 5 and passes through the point (2,βˆ’1)(2, -1). The equation we derived is 3x+2y=43x + 2y = 4.

Therefore, the correct answer is:

B. 3x+2y=43x + 2y = 4

We did it! This problem showcased how understanding fundamental concepts like slopes, perpendicular lines, and the point-slope form can lead us to the solution. Remember, guys, math isn't just about memorizing formulas; it's about understanding the relationships and using them to solve problems. Keep practicing, and you'll become math whizzes in no time!