### Video Transcript

If a line that passes through the points 𝐴 at coordinate six, zero and 𝐵 at coordinate four, negative six is perpendicular to the line passing through the points 𝐶 coordinate negative nine, 19 and 𝐷 at coordinate 𝑥, 15. What is the value of 𝑥?

Let’s start by reminding ourselves that the word “perpendicular” means that they cross at right angles or at 90 degrees. So here we have two lines, a line that goes through 𝐴 and 𝐵 and a line that goes through 𝐶 and 𝐷. But point 𝐷 has an unknown 𝑥-coordinate, which we need to work out. So there are two methods that we can potentially solve this question.

The first one is by solving graphically. And by that, we could create a coordinate grid and plot our points. We don’t know the complete coordinate for point 𝐷. But we could in theory work backwards after finding a line that’s perpendicular to 𝐴𝐵 and goes through point 𝐶. However, there are some disadvantages to this.

We can see that there are large values, which means that we’d have to have very large grid. We’d also need to be very accurate. So let’s see if we can find another method. In this option, we’re going to solve it algebraically. In this case, we’re going to use the equations for finding the slope of a line and facts that we know about perpendicular lines. It can be helpful to draw a quick sketch of our coordinates and the lines.

Let’s start by reminding ourselves what the slope or the gradient of a line is. The slope of a line is defined as the rise over the run. And it’s a measure of how steep a line is. If we have two coordinates to find as 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, then we can calculate the numerical value for the slope by calculating 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. That is, the changes in the rise values divided by the changes in the run values.

Let’s start by calculating the slope of our line 𝐴𝐵, since we know the coordinates for 𝐴 and 𝐵. We can use coordinate 𝐴 as our 𝑥 one, 𝑦 one values and coordinate 𝐵 for our 𝑥 two, 𝑦 two values. But it doesn’t matter if we use them the other way round. So Substituting these values into our slope equation would give us negative six take away zero, since our 𝑦 two value is negative six and our 𝑦 one value is zero. Our denominator values would give us four minus six, since 𝑥 two is equal to four and 𝑥 one is equal to six.

Simplifying the numerator and denominator would give us negative six over negative two. This is equal to six over two. And since six over two is the same as six divided by two, this means that the slope of the line going through 𝐴𝐵 is three.

And now since we have an unknown value in our 𝐷-coordinate, we can’t work out the slope of the line going through 𝐶 and 𝐷. But we can use an important fact about perpendicular lines to help us get towards that point. Our key fact about perpendicular lines is that if we have two perpendicular lines, then the values for their slope will multiply to give negative one. Since we know that this slope of the line going through 𝐴 and 𝐵 is three, this means we have the calculation three times what gives us negative one.

To find our missing value then, we would simply divide both sides of this by three. So our missing slope value must be negative one-third. So now we find that the slope of the line going through 𝐶𝐷 is equal to negative one-third.

Let’s go back and use our equation with our 𝑥- and 𝑦-values to see if we can find the missing 𝑥-coordinate for coordinate 𝐷. We can use coordinate 𝐶 for our 𝑥 one, 𝑦 one values and coordinate 𝐷 for our 𝑥 two, 𝑦 two values. Substituting in our values will give us negative one-third, since that’s the slope we’ve calculated. Then on the numerator, we have 15 minus 19, since that’s 𝑦 two minus 𝑦 one. And on the denominator, we have 𝑥 minus negative nine, since we have 𝑥 two subtract 𝑥 one and our 𝑥 one is already a negative value. Simplifying this will give us negative one-third equals negative four over 𝑥 plus nine, since our 𝑥 subtract negative nine is equivalent to saying 𝑥 plus nine.

And now we want to take our 𝑥 out of the denominator. So we’re going to multiply both sides of our equation by 𝑥 plus nine. This will give us negative 𝑥 plus nine over three equals negative four. And at this point, we can drop the negatives from both sides of our equation to give us 𝑥 plus nine over three equals four.

To continue rearranging then, we can multiply both sides by three to remove the three on the denominator on the left-hand side. This will give us 𝑥 plus nine equals 12, since, on the right-hand side, we had four times three, which is 12. So we have 𝑥 equals three. This is our final answer. And it means that the coordinate 𝐷 must have been equal to three, 15.