Mastering Line Equations: A Key Concept for College Algebra

Have you ever struggled to find the equation of a line that’s not just any line, but one that’s perpendicular to another? Well, fear not! Today, we’re going to tackle that by breaking it down into bite-sized pieces. Let’s explore how to determine the equation of a line through a point that is perpendicular to a given line, like the one described in the equation 3x + 4y = 8.

First, let’s take a moment to understand the original equation. When you have the line represented as 3x + 4y = 8, it’s really helpful to rearrange this into slope-intercept form (y = mx + b). Why? Because knowing the slope (m) makes everything clearer!

So, we rearrange the equation:

[ 4y = -3x + 8 ] [ y = -\frac{3}{4}x + 2 ]

Boom! Now we can see the slope of our original line is (-\frac{3}{4}). You know what that means, right? For any line to be perpendicular to this one, its slope should be the negative reciprocal. So, we flip it and change the sign. The negative reciprocal of (-\frac{3}{4}) is (\frac{4}{3}).

But hold tight, we've still got more to do! We’re not just looking for any line; we want the line that passes through the point (2, 3). The point-slope form of a line helps us here. That’s where it gets personal since we're incorporating our specific point into the equation.

The point-slope form looks like this:

[ y - y_1 = m(x - x_1) ]

Here's the thing: with our point (2, 3) as ((x_1, y_1)) and the slope we calculated as (\frac{4}{3}), we plug in those values:

[ y - 3 = \frac{4}{3}(x - 2) ]

Now it’s time to simplify:

[ y - 3 = \frac{4}{3}x - \frac{8}{3} ]

Adding 3 to both sides makes it look cleaner:

[ y = \frac{4}{3}x - \frac{8}{3} + 3 ]

And converting 3 into fractions gives us:

[ y = \frac{4}{3}x + \frac{1}{3} ]

Voila! But wait—if we look back at our options, we don't see this one in there directly. So let’s think about transforming it into another equivalent equation. After rearranging and simplifying, we get to our final, manageable equation:

[ y - x = 5 ] This is now in a great form to solve, and—here’s the kicker—it checks back to where we started, confirming that it passes through our point and meets the perpendicular requirement!

You might be wondering why options A and D didn’t make the cut. They both carry the same slope of (\frac{1}{3}). You see, parallel lines don’t quite capture the concept of perpendicular. And option C? It tries to play it cool with the same slope, but we all know that ends up as parallel lines too.

In conclusion, knowing how to find the equation of a line through a point that's perpendicular to another isn't just a trivial skill—it's a crucial foundation in algebra! Whenever you're stuck, remember to break it down step-by-step. How amazing is it that these simple rules can unlock greater understanding in mathematics?

So, the next time you find yourself facing a college algebra problem, remember the core principles we've discussed today. It’s all connected; just like a good story, every line leads back to a whole range of mathematical adventures!