### Video Transcript

Given that the straight line passing through the points one, eight and negative six, 𝑘 is parallel to the 𝑥-axis, find the value of 𝑘.

Let’s solve this equation in two different ways. First, we’ll solve algebraically and then we’ll solve by graphing. Let’s highlight some important information first. The line we’re talking about is a straight line that is parallel to the 𝑥-axis; this tells us something about our line. Primarily, it tells us that our slope is zero. Lines that are parallel to the 𝑥-axis have 𝑦-values that do not change. The 𝑦-values stay the same, so they’re not moving up or down and the slope is zero.

Let’s look at how we would solve this problem with algebra using the function for a slope of a line. We find the slope of the line by solving 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. We can use the two points we’re given and plug them into this formula: 𝑥 one, 𝑦 one would be one, eight; 𝑥 two, 𝑦 two would be negative six, 𝑘. 𝑦 two equals 𝑘 minus 𝑦 one, which is eight, over 𝑥 two, which is negative six, minus 𝑥 one, which is one.

Now, there’s one other piece of information that we can include here. We already know what our slope will be. Our slope is zero. Now, we need to solve for 𝑘 to figure out what is the value of this 𝑘 point. Our first step was subtract one from negative six, and that’s negative seven. Now we need to work on isolating 𝑘. To do that, I’ll multiply both sides of the equation by negative seven. Negative seven times one over negative seven equals one. That cancels out. Negative seven times zero equals zero, and we’re left with zero equals 𝑘 minus eight. We will add eight to both sides to isolate 𝑘. Zero plus eight equals eight. 𝑘 equals eight.

Now, let’s try to solve by graphing. Okay, here’s a sketch of our graph. Let’s start by graphing the point that we were given: one, eight. Here’s where point one, eight would fall. Now, what do we know about this line? We know that it’s parallel to the 𝑥-axis, which means we know what it will look like. It’s a horizontal line.

Now, let’s try to graph our second point: negative six, 𝑘. We know that the 𝑥-value will be negative six. We also know that this point must fall on our line. That means the only 𝑦-value — the only value 𝑘 can be — is equal to eight. Negative six, eight is the second point that we were looking for. Both of these processes produce a solution of 𝑘 equal to eight; both of these processes help us to see that our 𝑘 value is equal to eight.