### Video Transcript

Find the value of 𝑦 such that the straight line passing through three, negative 12 and negative five, three 𝑦 is perpendicular to the 𝑦-axis.

This is a problem. It can be solved in two ways. It be solved algebraically or using a graphical method. We can start by solving it algebraically. In order to do this, we’re gonna have to consider this formula, which gives us that 𝑚 is equal to 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one where 𝑚 is actually the slope of our line.

What this formula actually means is that 𝑚, our slope, is equal to the change in 𝑦 divided by the change in 𝑥. And we’re gonna use this to solve our problem algebraically. But when we look back at the question, we have to think how we gonna use this to help us solve it algebraically.

Well first of all, we need to know that we can use this formula. And we know that we can because it’s a straight line. Then the key phrase to solving this problem is ’perpendicular’ or, more importantly, it says that it is perpendicular to the 𝑦-axis. And here my little sketch in pink, I have given an example of a line that is perpendicular to the 𝑦-axis. And what we mean by perpendicular is we mean that right angles too.

So we can see that the pink line is at right angles to the 𝑦-axis. Okay, so we’re gonna come back to what I said before. I said that 𝑚, our slope, is equal to the change in 𝑦 divided by the change in 𝑥. So using this, if we look at what we’ve got in the sketch, we can actually see that there is in fact, in the pink line, no change in 𝑦.

Because it’s a horizontal line, the values of 𝑦 never increase or decrease. It just stays the same. So therefore, with our example, we can rewrite that like this where 𝑚 is equal to zero because the change in 𝑦 is zero, there is no change in 𝑦, divided by our change in 𝑥. And from this, we can actually calculate our slope because we know that zero divided by anything is always going to be zero.

So therefore we can say the slope of our line is gonna be equal to zero. And it’s this key bit of information that’s gonna allow us to solve this problem algebraically. Excellent! So we can now get on with finding a value of 𝑦. So the first thing we have was we have our pair of values. And what we’re gonna do first is we’re gonna label each of our coordinates. And this is gonna enable us to work out where they go when we put them back into our formula.

So here they are, the coordinates that have been labelled. So we get 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two. So now we can substitute these values into our formula. And having substituted our values into our formula, we get our slope 𝑚 is equal to three 𝑦 minus negative 12 divided by negative five minus three. Now we can actually tidy it up, which is gonna give us 𝑚 is equal to three 𝑦 plus 12.

And it gonna be plus 12 cause, again, beware of the negative signs. We’ve got minus negative 12. So that’s gonna equal plus 12. And that’s all divided by negative eight. Okay! Great! We’re now at a stage where we can actually use the slope that we know. And if we substitute 𝑚 is equal to zero into our formula, okay so we now have the equation zero is equal to three 𝑦 plus 12 all divided by negative eight. Okay, so I’m just gonna bring over here on the right-hand side to solve it.

Our first step is to multiply both side by negative eight. So that’s gonna give us zero, cause zero multiplied by by negative eight still zero, is equal to three 𝑦 plus 12. The next stage is to subtract 12 from each side, which gives us negative 12 is equal to three 𝑦. And then we do the final step, which is divide both sides by three which gives us negative four is equal to 𝑦 or just how I’ve written it which is 𝑦 is equal to negative four.

I just like to keep the 𝑦 on the left-hand side. And therefore, we found the value of 𝑦 such that the straight line that passes through our two points is perpendicular to the 𝑦-axis. As I mentioned earlier, there is also a graphical way of solving this kinda problem which I’m just gonna show you just to kinda check the answer, to check if it’s correct, but also as a method that you could possibly use if you’re solving this kind of problem in the future.

I’ve left the final part of our algebraic method just on right-hand side just so that we have the answer we can compare them when we’ve completed the graphical method. So here I’ve drawn a sketch of our axes. So now I’m gonna consider where our points would be on our axes. Well our first point would be at three, negative 12, while our second point is negative five, three 𝑦.

Before we think about where this is, we can remember two key points from our question: firstly, that it is a straight line passing through our two points; and secondly, that straight line’s perpendicular to 𝑦-axis, which means that our second point is gonna have to lie on the same horizontal line as our first point.

So therefore it would lie here. So this enables us to have a line which is horizontal and perpendicular to the 𝑦-axis. So therefore, we can now say that 𝑦 one must be equal to 𝑦 two. And therefore, we use this to find our missing 𝑦 value as negative 12 must be equal to three 𝑦.

Then if we divide both sides by three, we get our value for 𝑦. So 𝑦 is equal to negative four. So brilliant! If we check both our answers, we found that 𝑦 is equal to negative four. And we’ve got that both; we’ve got it graphically and algebraically. So that’s the correct answer.