### Video Transcript

Find the parametric equations of the straight line three π₯ minus seven over negative nine equals eight π¦ minus three over four which equals negative eight minus six π§ over negative nine.

Weβve been given an equation in Cartesian form. Now, in general, if a line passes through a point π nought which has coordinates π₯ nought, π¦ nought, π§ nought and is parallel to the vector πππ. That its parametric equations are π₯ equals π₯ nought plus ππ‘, π¦ equals π¦ nought plus ππ‘, and π§ equals π§ nought plus ππ‘. In Cartesian form, we can compare this and say that the equation would be π₯ minus π₯ nought over π equals π¦ minus π¦ nought over π which equals π§ minus π§ nought over π. So, to write parametric equations, weβre going to begin by comparing our Cartesian equation to the general form.

We begin by noticing that the coefficient of π₯, π¦, and π§ throughout is simply positive one. To achieve this in the first part of our equation, weβre going to divide both the numerator and denominator of the fraction by three. Now, remember, as long as we do the same to both the numerator and the denominator, weβre just creating an equivalent fraction. So, it doesnβt change the size. Three π₯ divided by three is π₯. Negative seven divided by three is negative seven-thirds and negative nine divided by three is negative three.

Now, we have the first part of the equation in the right form. That is, the coefficient of π₯ is one. Weβre going to repeat this process for the second part of the equation. This time, we want the coefficient of π¦ two to be one. So, weβre going to divide both the numerator and the denominator of our fraction by eight. Eight π¦ divided by eight is π¦. Negative three divided by eight is negative three-eighths. And four divided by eight is one-half.

We repeat this process one more time. This time, we want the coefficient of π§ to be positive one. So, weβre going to have to divide the numerator and denominator of this fraction by negative six. Eight divided by negative six is four-thirds. Negative six π§ divided by negative six is π§. And negative nine divided by negative six is three over two. Well, we write this last bit ever so slightly so it compares quite nicely to our original equation. And of course, we know that these three expressions are equal.

Letβs compare π₯ minus seven over three over negative three to π₯ minus π₯ nought over π. We can let π₯ nought be equal to seven over three here and π be equal to negative three. Using the parametric form π₯ equals π₯ nought plus ππ‘, we can say that the parametric equation for π₯ is π₯ equals seven-thirds minus three π‘. Letβs repeat this process for the second part. π¦ nought here is going to be equal to three-eighths, and π is equal to one-half. And so, using the parametric form π¦ equals π¦ nought plus ππ‘, we find π¦ equals three-eighths plus one-half π‘.

Weβll repeat this process for the final part. This time, we could say that π§ nought is equal to negative four-thirds and π is equal to three over two. Once again, we compare this to the parametric form, when we get π§ equals negative four-thirds plus three over two π‘. And so, by comparing our equation to the general form and manipulating it somewhat, we found the parametric equations of our straight line are π₯ equals seven-thirds minus three π‘, π¦ equals three-eighths plus a half π‘, and π§ equals negative four-thirds plus three over two π‘.