### Video Transcript

Find the different forms of the equation of the straight line whose Cartesian equation is π₯ minus three over four equal π§ plus six over three, π¦ equals four. Then determine a point that lies on this straight line.

By different forms, weβre being asked to find the vector form and the parametric form of the equation of our straight line. Remember, for a line with the Cartesian equation π₯ minus π₯ one over π equals π¦ minus π¦ one over π, which equals π§ minus π§ one over π, its vector form can be written as shown. Where the line passes through a point π΄ with position vector π₯ one, π¦ one, π§ one and it has a direction vector π, π, π. And we can finally write this in parametric form as π₯ equals π₯ one plus ππ‘. π¦ equals π¦ one plus ππ‘. And π§ equals π§ one plus ππ‘.

We could work out the values of the various unknowns by comparing the general form to our Cartesian equation. We can see that π₯ one is equal to three, π is equal to four, π§ one must be equal to negative six so that π§ minus π§ one becomes π§ plus six. And π must be equal to three. And then here we do need to be a little bit careful, as our equation does look slightly different to the general form. And thatβs because weβre told that π¦ is equal to four.

We can, therefore, say that the value of π¦ will remain unchanged, no matter the value of π‘. And this means that π¦ one is equal to four. But the π¦-component of our direction vector, π, must be equal to zero. This means the position vector of π is three, four, negative six. And the direction vector for our line is four, zero, three. And so the vector equation of our line is equal to three, four, negative six plus π‘ multiplied by four, zero, three.

And in parametric form, we can say that π₯ must be equal to three plus four π‘. π¦ is equal to four plus zero π‘ or just four. And π§ is equal to negative six plus three π‘. Now, we simply need to determine a point that lies on this straight line. Earlier, we said that our line passes through a point π΄ which has a position vector of π₯ one, π¦ one, π§ one. This means we know that thereβs a point on the line with a position vector of three, four, negative six or a coordinate three, four, negative six.

However, we can find any other point that lies on the line by changing our value of π‘. When π‘ is equal to one, π₯ is three plus four, which is seven. π¦ is four plus zero, which is four. And π§ is negative six plus three, which is negative three. And when π‘ is two, π₯ is three plus eight, which is 11. π¦ is once again four. And π§ is negative six plus two times three, which is zero. Iβve chosen three, four, negative six as the point on our line. But as weβve seen, there are a whole host of options we could have chosen to find a point on this line.