### Video Transcript

Write a pair of parametric equations with parameter π describing the shown line.

In this question, weβre given a line. And we need to use this given diagram to determine a pair of parametric equations which represents this line, where we call the parameter π. To do this, letβs start by recalling how we find a pair of parametric equations representing a line. We can recall the parametric equations of a line take the form π₯ is equal to π₯ sub zero plus π times π and π¦ is equal to π¦ sub zero plus π times π, where the point with coordinates π₯ sub zero, π¦ sub zero lies on the line and the vector π, π is parallel to the line and is a nonzero vector.

In particular, itβs worth noting we can choose any point on the line to be our point π₯ sub zero, π¦ sub zero. And we can choose any nonzero vector parallel to the line to be the vector π, π. Any choice of this point or vector will give us a different set of parametric equations for the line. However, theyβre all the same parametric equations. We can note that weβre given a point which lies on the line already. The point two, three lies on the line. So in our parametric equations for this line, weβll set π₯ sub zero equal to two and π¦ sub zero equal to three. All we need to do now is determine a nonzero vector which is parallel to the line. And thereβs many different ways of doing this.

One way of doing this is to note that weβre given the angle that the line makes with the π₯-axis. And we can recall the tangent of this angle will be equal to the slope of the line. π is equal to the tangent of 30 degrees. And this is one of our special angles. We can recall the tangent of 30 degrees is the square root of three divided by three. And since the slope of the line tells us the displacement up or down in the line for every unit we travel to the right, we could note that our line with slope π will be parallel to the vector one π. Since our value of π is root three over three, this line must be parallel to the vector one, root three over three. For every unit we move to the right, the line moves root three over three units upwards.

We can use these values to find one set of parametric equations for this line. We substitute π₯ sub zero is equal to two, π¦ sub zero is equal to three, π is equal to one, and π is equal to root three over three into the parametric equations. We get π₯ is equal to two plus π and π¦ is equal to three plus root three over three multiplied by π. And this is one way of answering this question. However, there are many different ways of finding different pairs of parametric equations which also represent the line. For example, instead of using the tan of 30 degrees to determine the slope of the line, we could have instead calculated the sin and cos of 30 degrees.

For example, we can sketch the following right triangle with angle 30 degrees. We know the sin of 30 degrees is equal to one-half, and the sine function represents the ratio between the side opposite the angle and the hypotenuse. So the ratio of these two lengths needs to be equal to one-half. One way of representing this is to say that the side opposite this angle is length one-half, and the hypotenuse is length one. Similarly, the cosine ratio tells us the ratio of the length of the side adjacent to the angle and the hypotenuse. And we can recall the cos of 30 degrees is equal to the square root of three divided by two. And this is a second method for determining a nonzero vector parallel to the line.

For every root three over two units, we move to the right, we move half a unit up. So we can find another pair of parametric equations representing the line by using π₯ sub zero is two, π¦ sub zero is three, π is equal to root three over two, and π is equal to one-half. This then gives us that π₯ is equal to two plus π times root three over two and π¦ is equal to three plus π divided by two. However, there are many other different possible variations of these parametric equations.

Therefore, we were able to show a number of different methods for determining pairs of parametric equations representing the line shown. And one possible option was π₯ is equal to two plus π root three over two and π¦ is equal to three plus π over two.