Calculate Distance Between Two Parallel Lines

Hey guys! Today, we're diving deep into a classic math problem: finding the distance between two lines. Specifically, we'll be tackling the equations $-2x + 6y - 9 = 0$ and y = rac{1}{3}x - 2. This might sound a bit daunting at first, but trust me, once we break it down, it's totally manageable. We'll explore the concepts, go through the steps, and make sure you feel confident about solving this type of problem. So, grab your notebooks, and let's get started on this journey to master the distance between lines!

Understanding the Problem: Distance Between Lines

So, what exactly does it mean to find the distance between two lines? When we talk about the distance between two geometric objects, we're generally looking for the shortest possible distance between any point on the first object and any point on the second object. For two lines, this shortest distance is measured along a line segment that is perpendicular to both lines. This concept is super important, so keep it in mind. Now, not all pairs of lines have a constant distance between them. If the lines intersect, the distance at the intersection point is zero, and it increases as you move away. However, if the lines are parallel, the distance between them is constant everywhere. This is the scenario we're dealing with in our problem. The lines we're given are $-2x + 6y - 9 = 0$ and y = rac{1}{3}x - 2. The first step to solving this is to determine if these lines are indeed parallel. How do we do that? We look at their slopes! In the equation $y = mx + c$, $m$ represents the slope. For the second line, y = rac{1}{3}x - 2, the slope is clearly rac{1}{3}. Now, let's look at the first line, $-2x + 6y - 9 = 0$. To find its slope, we need to rearrange it into the $y = mx + c$ form. So, we get $6y = 2x + 9$, which simplifies to y = rac{2}{6}x + rac{9}{6}, or y = rac{1}{3}x + rac{3}{2}. Aha! The slope of the first line is also rac{1}{3}. Since both lines have the same slope, they are indeed parallel. This is a crucial confirmation, guys, because the method for finding the distance is specific to parallel lines. If they weren't parallel, the problem would be about finding the distance between intersecting lines, which is zero at the point of intersection.

Preparing the Line Equations

Alright, so we've confirmed our lines are parallel. Awesome! Now, to find the distance between them, it's super helpful if both equations are in the same standard form. We already did some work on the first line, $-2x + 6y - 9 = 0$, and rearranged it to y = rac{1}{3}x + rac{3}{2}. The second line is given as y = rac{1}{3}x - 2. Both are in the slope-intercept form ($y = mx + c$), which is great. However, for the formula we'll be using, it's often easier if they are in the general form $Ax + By + C = 0$. Let's convert our slope-intercept forms into this general form. For the first line, we have $-2x + 6y - 9 = 0$. This is already in the desired format, with $A = -2$, $B = 6$, and $C_1 = -9$. For the second line, y = rac{1}{3}x - 2, we can multiply the entire equation by 3 to get rid of the fraction: $3y = x - 6$. Now, rearrange this to get $x - 3y - 6 = 0$. Oh, wait a minute! When we converted the first line, we got $A = -2$ and $B = 6$. In the second line, we got $A = 1$ and $B = -3$. These coefficients for $x$ and $y$ don't quite match up, and that's important for the distance formula. We need the coefficients of $x$ and $y$ to be the same (or proportional and then made identical) in both equations. Let's go back. The first line is $-2x + 6y - 9 = 0$. The second line is y = rac{1}{3}x - 2. We found the slope rac{1}{3} for both. Let's stick with the general form $Ax + By + C = 0$ for both. For the first line, we have $-2x + 6y - 9 = 0$. We can divide this whole equation by $-2$ to simplify it and make the coefficient of $x$ equal to 1. This gives us x - 3y + rac{9}{2} = 0. Now, let's take the second line, y = rac{1}{3}x - 2. Rearranging it gives $x - 3y - 6 = 0$. Great! Now both lines are in the form $Ax + By + C = 0$ with $A=1$, $B=-3$. The first line is x - 3y + rac{9}{2} = 0, so C_1 = rac{9}{2}. The second line is $x - 3y - 6 = 0$, so $C_2 = -6$. Having the equations in this form, $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, is key for using the distance formula between parallel lines.

The Formula for Distance Between Parallel Lines

Now that we've got our parallel lines in a nice, clean form, x - 3y + rac{9}{2} = 0 and $x - 3y - 6 = 0$, we can talk about the magic formula for finding the distance between them. This formula is specifically designed for parallel lines that are expressed in the general form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$. The distance $d$ between these two lines is given by: d = rac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}. See? It's pretty straightforward once you have the equations in the right shape. Here, $A$ and $B$ are the coefficients of $x$ and $y$ in either equation (since they are the same for parallel lines), and $C_1$ and $C_2$ are the constant terms from each equation. It's important that the $A$ and $B$ coefficients are identical in both equations for this formula to work directly. If they were just proportional (e.g., $x+2y+3=0$ and $2x+4y+5=0$), you'd first need to multiply one of the equations by a constant to make the $A$ and $B$ values match. In our case, we have $A=1$, $B=-3$, C_1 = rac{9}{2}, and $C_2 = -6$. So, we just plug these values into the formula. The numerator is the absolute difference between the constants: |C_1 - C_2| = |rac{9}{2} - (-6)|. The denominator involves the square root of the sum of the squares of the coefficients $A$ and $B$: $\sqrt{A^2 + B^2} = \sqrt{1^2 + (-3)^2}$. Let's calculate these parts step-by-step to avoid any slip-ups. This formula essentially represents the length of the perpendicular segment connecting the two lines. It's derived using concepts of geometry and vector calculus, but for us, knowing and applying it is the main goal. It elegantly captures the shortest distance, which, as we discussed, is constant for parallel lines.

Calculating the Distance

Let's put our formula into action! We have $A=1$, $B=-3$, C_1 = rac{9}{2}, and $C_2 = -6$. Plugging these into the distance formula d = rac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}:

First, calculate the numerator: |C_1 - C_2| = |rac{9}{2} - (-6)| = |rac{9}{2} + 6|. To add these, we need a common denominator. $6$ is the same as rac{12}{2}. So, |rac{9}{2} + rac{12}{2}| = |rac{21}{2}| = rac{21}{2}.

Next, calculate the denominator: $\sqrt{A^2 + B^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.

Now, divide the numerator by the denominator: d = rac{rac{21}{2}}{\sqrt{10}}.

To simplify this, we can multiply the numerator by the reciprocal of the denominator: d = rac{21}{2\sqrt{10}}.

This is the exact distance. However, math problems often ask for a decimal approximation, and our answer choices are in decimals. So, let's convert this to a decimal. First, let's rationalize the denominator by multiplying the numerator and denominator by $\sqrt{10}$: d = rac{21\sqrt{10}}{2\sqrt{10}\sqrt{10}} = rac{21\sqrt{10}}{2 imes 10} = rac{21\sqrt{10}}{20}.

Now, we need the approximate value of $\sqrt{10}$. It's about $3.162$. So, $d \approx \frac{21 imes 3.162}{20} = \frac{66.402}{20} = 3.3201$.

Let's double-check the calculations. Numerator: |rac{9}{2} - (-6)| = |rac{9}{2} + rac{12}{2}| = |rac{21}{2}| = 10.5. Denominator: $\sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10} \approx 3.16227766$. Distance d = rac{10.5}{3.16227766} \approx 3.320127.

Looking at our answer choices: A. 3.34 B. 3.32 C. 3.33 D. 3.35

Our calculated value, approximately 3.3201, is closest to 3.32. So, option B seems to be the correct answer, guys!

Conclusion

And there you have it! We successfully found the distance between the two lines $-2x + 6y - 9 = 0$ and y = rac{1}{3}x - 2. The key steps were:

1. Confirming Parallelism: We checked if the slopes of the two lines were equal. If they are, the lines are parallel, and we can proceed.
2. Standardizing Equations: We converted both line equations into the general form $Ax + By + C = 0$, ensuring that the coefficients $A$ and $B$ were identical in both.
3. Applying the Formula: We used the formula for the distance between parallel lines: d = rac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.
4. Calculation: We plugged in the values and performed the arithmetic, including converting the exact answer to a decimal approximation.

This process might seem like a lot at first, but with practice, it becomes second nature. Understanding why we do each step, like making sure the lines are parallel or getting them into the same $Ax+By+C=0$ format, really helps solidify the concept. Remember, the distance between parallel lines is the shortest, perpendicular distance, and the formula gives us that exact value. Keep practicing these types of problems, and you'll be a math whiz in no time. If you ever encounter lines that aren't parallel, remember the distance is zero at their intersection point. Great job working through this problem, everyone!