Mastering the Equation of Perpendicular Lines

When it comes to college-level algebra, mastering the fundamentals can make all the difference—especially for exams like the College Algebra CLEP. One area that often trips students up is figuring out the equation of a line, particularly when dealing with perpendicular lines. But don't worry! We're about to break it down step-by-step in a way that hopefully makes it all feel a little less daunting. You know what I'm talking about—the kind of math that just seems to click once you get a grasp on the basics. So let’s dive right in!

First up, we’re tackling a specific prompt: What is the standard form of the equation of a line that passes through the point (-2, 5) and is perpendicular to the line given by 2x + 3y = 7? If that sounds a little complex, don't sweat it. We’ll break it down together.

To start, we need to identify the slope of our given line. Ah, slopes—a key ingredient in the recipe for equations! Rearranging the line equation 2x + 3y = 7 into slope-intercept form, we can isolate y to find the slope. By subtracting 2x from both sides and then dividing everything by 3, we see that the slope is -2/3. Here comes the fun part: the slope of the line we need to find must be the negative reciprocal of this slope. So what’s the negative reciprocal of -2/3? That would be 3/2! Who knew math could be so twisty?

Now that we’ve got our slope, it’s time to harness the power of the point-slope form of a line, which is laid out as y - y1 = m(x - x1). In our case, (x1, y1) is the point (-2, 5) and m is our slope 3/2. Plugging these values into the formula, we end up with:

y - 5 = (3/2)(x + 2).

Next, let's simplify this baby down. If we distribute that 3/2, we find:

y - 5 = (3/2)x + 3.

The next move is to isolate y. So, adding 5 to both sides yields:

y = (3/2)x + 8.

We’re almost there—but hold on! This isn’t yet in the standard form we need. The standard form of a line is structured as Ax + By = C. Now, here’s a trick to eliminate that lingering fraction. We can multiply through by 2 to smooth things out heaps. So let’s do that:

2y = 3x + 16.

But wait, we still have to rearrange this so it fits into standard form. We want to get everything on one side, so how about we move 3x to the left?

-3x + 2y = 16.

If we want A to be positive, we can multiply through by -1, leading us to our final answer:

3x - 2y = -16.

Okay, I know we just went on a bit of a journey, but stick with me. This is just one example of how to find equations of lines—perpendicular or otherwise. Understanding the relationships between different components in algebra can be immensely rewarding. As you continue to prep for your College Algebra CLEP exam, keeping your skills sharp with practice problems and little tricks like this can definitely help you feel more confident come test day. So grab a pencil, tackle some practice questions, and get those math skills shining!