Video: Finding Parametric Equations of Straight Lines

Find the parametric equations of the straight line that makes an angle of 135Β° with the positive π‘₯-axis and passes through the point (1, βˆ’15).

03:51

Video Transcript

Find the parametric equations of the straight line that makes an angle of 135 degrees with the positive π‘₯-axis and passes through the point one, negative 15.

The line makes an angle of 135 degrees with the positive π‘₯-axis, so let’s draw in the positive π‘₯-axis. The line in question presumably passes through the π‘₯-axis at some point, and we’re told that it does so at an angle of 135 degrees, which by convention is measured counterclockwise. And of course this line continues infinitely in both directions. This angle tells us about the gradient of this line, but of course there are infinitely many lines which have this gradient, so we’re going to need some more information to work out exactly which line we’re dealing with.

And that information in this case comes in the form of a point through which the line passes, so let’s mark this point with coordinates one and negative 15. And having chosen where this point lies, we can also draw in our 𝑦-axis. We need to find the parametric equations of this line, and this involves writing the coordinates of points on this line, π‘₯ and 𝑦, as functions of a parameter π‘˜. We can choose that when the π‘˜ is zero, we’re at this special point that we were given, one, negative 15. And as π‘˜ varies from negative infinity to infinity, it draws the line of course passing through this point when π‘˜ reaches zero.

When π‘˜ is zero, π‘₯ is one and 𝑦 is negative 15. So 𝑓 of zero is one, and 𝑔 of zero is negative 15. We now have several choices to make about π‘˜. We know that as π‘˜ varies from a negative infinity to infinity, this line should be drawn, and we can make a choice that the line is drawn from left to right, so with π‘₯ increasing. So π‘˜ equals one corresponds to some point slightly further along the line, and say π‘˜ equals negative nine corresponds to some point further up the line.

The easiest way to do this, and again and this is a choice, is that the value of π‘˜ is the change of π‘₯ between the special point, one and negative 15, and the point in question. So π‘˜ being one means that the change in π‘₯ from the point one and negative 15 is one, and so the π‘₯-coordinate to this point is two. And similarly, when π‘˜ is negative nine, the change in π‘₯ from the point of one negative, 15 is negative nine, and so the π‘₯-coordinate of this point is a negative eight.

In general then, π‘₯ is one plus π‘˜. To find the equivalent equation for 𝑦, we can either do some trigonometry. We can notice that an angle of 135 degrees to the positive π‘₯-axis means an angle of 45 degrees to the negative π‘₯-axis, and therefore we can recognize this line is having a gradient of negative one. What does this mean? Well it means that the change in 𝑦 over the change of π‘₯ is negative one, and as the change in π‘₯ is π‘˜, we can see that the change and 𝑦 must be negative π‘˜.

As we’ve already seen when π‘˜ is zero, 𝑦 is negative 15. And as π‘˜ varies from that value, the change in 𝑦 is equal to negative π‘˜. And so we get the parametric equations π‘₯ is equal to one plus π‘˜ and 𝑦 is equal to negative 15 minus π‘˜. One important thing to realize here is that there were lots of choices that went into finding these parametric equations. And as a result, there are lots of different parametric equations which represent the same line.

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