Distance Between Point And Line: A Simple Guide

Hey guys! Ever wondered how to find the distance between a point and a line? It might sound intimidating, but it's actually pretty straightforward once you get the hang of it. Today, we're going to break down exactly how to calculate the distance between the point $(3,5)$ and the line given by the equation $5x - 3y + 10 = 0$. Trust me, by the end of this guide, you'll be a pro at this!

Understanding the Formula

Before we dive into the specifics, let's quickly introduce the formula we'll be using. The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Where:

• $(x_0, y_0)$ are the coordinates of the point.
• $A$, $B$, and $C$ are the coefficients from the line's equation.

This formula might look a bit scary at first, but don't worry! We'll break it down step by step and apply it to our specific problem. Think of it as a recipe – once you have all the ingredients (the values of $A$, $B$, $C$, $x_0$, and $y_0$), you just plug them into the formula and voilà, you have the distance. This formula is derived from vector projections and ensures that we're finding the shortest distance (i.e., the perpendicular distance) from the point to the line. This is why it's so important to use the absolute value in the numerator, ensuring that the distance is always a positive value. Understanding the origin of the formula can also help you remember it better and apply it in different contexts. For instance, if you're dealing with lines and planes in three-dimensional space, a similar formula applies, but with an additional dimension to account for.

Identifying the Values

Okay, let's get practical. First, we need to identify the values from our problem:

• Point: $(x_0, y_0) = (3, 5)$
• Line: $5x - 3y + 10 = 0$

From the line's equation, we can see that:

• $A = 5$
• $B = -3$
• $C = 10$

Now that we have all the values, we are ready to plug them into the formula. Seriously, the hardest part is just remembering the formula and keeping track of what each variable represents. Make sure you double-check your values before plugging them in; a small mistake here can throw off your entire calculation. Remember, $A$, $B$, and $C$ come directly from the coefficients of $x$, $y$, and the constant term in the equation of the line, respectively. And $(x_0, y_0)$ is simply the point from which you want to measure the distance to the line. Once you've identified these values correctly, you're halfway there!

Plugging into the Formula

Now comes the fun part – plugging the values into the formula:

$d = \frac{|(5)(3) + (-3)(5) + 10|}{\sqrt{(5)^2 + (-3)^2}}$

Let's simplify this step-by-step.

First, calculate the numerator:

$|(5)(3) + (-3)(5) + 10| = |15 - 15 + 10| = |10| = 10$

Next, calculate the denominator:

$\sqrt{(5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$

So, the distance is:

$d = \frac{10}{\sqrt{34}}$

Alright, we're almost there! Plugging these values carefully is crucial. Pay close attention to the signs and make sure you're performing the operations in the correct order (PEMDAS/BODMAS is your friend here!). The numerator involves multiplying the coefficients of the line equation by the coordinates of the point and then adding the constant term. The absolute value ensures that we're always dealing with a positive distance. In the denominator, we're calculating the magnitude of the vector formed by the coefficients $A$ and $B$. This involves squaring each coefficient, adding them together, and then taking the square root. The denominator essentially normalizes the distance, ensuring that it's measured in the correct units. So, double-check each calculation to avoid any silly mistakes!

Simplifying the Result

We have $d = \frac{10}{\sqrt{34}}$. To get rid of the square root in the denominator, we can rationalize it:

$d = \frac{10}{\sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt{34}} = \frac{10\sqrt{34}}{34} = \frac{5\sqrt{34}}{17}$

So, the distance between the point $(3,5)$ and the line $5x - 3y + 10 = 0$ is $\frac{5\sqrt{34}}{17}$ units.

And that's it! We've successfully found the distance. Simplifying the result often involves rationalizing the denominator to present the answer in a cleaner form. This is especially important in mathematical contexts where a simplified answer is preferred. In our case, we multiplied both the numerator and denominator by $\sqrt{34}$ to eliminate the square root from the denominator. This gives us $\frac{10\sqrt{34}}{34}$, which can be further simplified by dividing both the numerator and denominator by their greatest common divisor, which is 2. This leaves us with the final simplified answer of $\frac{5\sqrt{34}}{17}$. Always remember to simplify your results as much as possible; it not only makes the answer look nicer but can also make it easier to work with in further calculations.

Wrapping Up

So, there you have it! Finding the distance between a point and a line isn't so scary after all. Just remember the formula, identify your values, plug them in, and simplify. You'll be solving these problems like a math wizard in no time! Remember, practice makes perfect, so try solving a few more examples to really solidify your understanding. You can even create your own problems by randomly choosing points and lines, and then calculating the distances. This will help you become more comfortable with the formula and the process. And don't be afraid to ask for help if you get stuck. There are plenty of resources available online, including videos and step-by-step guides. With a little bit of effort, you'll be a distance-finding pro in no time. Keep practicing, keep learning, and most importantly, have fun with math!