Coordinate Geometry Solutions: M1, M2, And Line D

Hey guys! Today, we're diving deep into a cool coordinate geometry problem. We've got two points, M1 and M2, and a line d with the equation x + 4y + 6 = 0. We need to tackle two parts: first, find the distance from M1 to line d, and second, figure out the equation of a line that passes through M1 and is parallel to line d. Let's break this down step-by-step, without any AI help, so we can really understand the nitty-gritty of these calculations. Get your notebooks ready!

Part A: Distance from Point M1 to Line d

Alright, let's get started with the first mission: calculating the distance from point M1(0, 3) to the line d: x + 4y + 6 = 0. This is a classic problem in coordinate geometry, and there's a handy formula for it. The distance from a point (x₀, y₀) to a line Ax + By + C = 0 is given by the formula:

$$d = \frac{|Ax₀ + By₀ + C|}{\sqrt{A² + B²}}$$

In our case, the point M1 is (x₀, y₀) = (0, 3), and the line d is x + 4y + 6 = 0. So, we can identify our coefficients: A = 1, B = 4, and C = 6. Now, let's plug these values into the formula. We substitute x₀ = 0 and y₀ = 3 into the numerator:

$|Ax₀ + By₀ + C| = |(1)(0) + (4)(3) + 6|$

Calculating this out, we get:

$|0 + 12 + 6| = |18| = 18$

So, the absolute value of the numerator is 18. Now, let's look at the denominator. We need to find the square root of A² + B²:

$\sqrt{A² + B²} = \sqrt{1² + 4²}$

This gives us:

$\sqrt{1 + 16} = \sqrt{17}$

Now, we combine the numerator and the denominator to find the distance:

$$d = \frac{18}{\sqrt{17}}$$

To make it look a bit cleaner, we can rationalize the denominator by multiplying both the numerator and the denominator by $\sqrt{17}$:

$$d = \frac{18\sqrt{17}}{17}$$

And there you have it! The distance from point M1 to the line d is $\frac{18\sqrt{17}}{17}$ units. Pretty straightforward when you know the formula, right? This formula is super useful and shows up a lot, so make sure you've got it down pat. It's all about identifying your point coordinates and the coefficients of your line equation and then carefully plugging them into the formula. Don't forget the absolute value in the numerator – distances are always positive!

Part B: Equation of a Line Through M1 Parallel to Line d

Now for the second part, guys: we need to find the equation of a line that passes through M1(0, 3) and is parallel to line d (x + 4y + 6 = 0). The key word here is parallel. Parallel lines have the same slope. So, our first step is to find the slope of line d. We can rewrite the equation of line d in slope-intercept form, which is y = mx + c, where 'm' is the slope.

Starting with x + 4y + 6 = 0, let's isolate y:

$4y = -x - 6$

Now, divide everything by 4:

$$y = \frac{-x - 6}{4}$$

$$y = -\frac{1}{4}x - \frac{6}{4}$$

$$y = -\frac{1}{4}x - \frac{3}{2}$$

From this, we can clearly see that the slope of line d (let's call it m_d) is -1/4. Since our new line needs to be parallel to line d, it must have the same slope. So, the slope of our new line (let's call it m_new) is also -1/4.

Now we have the slope (m_new = -1/4) and a point that the new line passes through (M1(0, 3)). We can use the point-slope form of a linear equation, which is $y - y₁ = m(x - x₁)$. Here, (x₁, y₁) is our point M1(0, 3) and m is our slope -1/4.

Let's plug in the values:

$y - 3 = -\frac{1}{4}(x - 0)$

$y - 3 = -\frac{1}{4}x$

To get the equation in a standard form (like Ax + By + C = 0 or y = mx + c), we can rearrange this. Let's add 3 to both sides to get it into slope-intercept form:

$$y = -\frac{1}{4}x + 3$$

And there we go! This is the equation of the line passing through M1 and parallel to line d. It's important to remember that parallel lines have identical slopes. So, if you're ever asked to find a line parallel to another, your first move should always be to find the slope of the given line. Then, use that same slope with the given point to construct the new line's equation. This process involves understanding the relationship between parallel lines and knowing how to use the point-slope or slope-intercept forms of linear equations.

Double-Checking Our Work

Let's do a quick sanity check, guys. For Part A, we found the distance from M1(0, 3) to x + 4y + 6 = 0 to be $\frac{18\sqrt{17}}{17}$. The formula used is standard, and the calculations seem correct. The numerator $|(1)(0) + (4)(3) + 6| = |18| = 18$, and the denominator $\sqrt{1^2 + 4^2} = \sqrt{17}$. So, $\frac{18}{\sqrt{17}}$ is right. Rationalizing gives $\frac{18\sqrt{17}}{17}$. Looks solid.

For Part B, we found the equation of the line through M1(0, 3) parallel to x + 4y + 6 = 0 to be $y = -\frac{1}{4}x + 3$. We correctly identified the slope of line d as -1/4 by converting it to slope-intercept form. A parallel line must have the same slope. Using the point-slope form with M1(0, 3) and slope -1/4 gave us $y - 3 = -\frac{1}{4}(x - 0)$, which simplifies to $y = -\frac{1}{4}x + 3$. This looks correct. A quick check: does the point M1(0, 3) satisfy this equation? If we plug in x=0, we get $y = -\frac{1}{4}(0) + 3 = 3$. Yes, it does! So, our point is indeed on the line. The slope is -1/4, which is the same as line d, confirming it's parallel.

Why Understanding These Concepts Matters

Guys, mastering these coordinate geometry concepts is super important. It's not just about solving textbook problems; it's about building a strong foundation for more advanced math and science. Understanding distances between points and lines, and the properties of parallel and perpendicular lines, are fundamental skills. These ideas pop up in calculus, physics (think vectors and forces), engineering, computer graphics, and so much more. Being able to derive equations of lines based on given conditions allows you to model real-world scenarios mathematically. For instance, if you're tracking the path of an object, you might need to find lines representing its trajectory or boundaries. The ability to perform these calculations manually, without relying on automated tools, ensures you truly grasp the underlying principles. It enhances your problem-solving abilities and makes you a more capable and confident mathematician. So, keep practicing, and don't shy away from these essential geometry skills!

Keep practicing these types of problems, and you'll become a coordinate geometry whiz in no time! Let me know if you have any other questions, and we'll tackle them together.