# Video: EG19M2-ALGANDGEO-Q16

Find the different forms of the equation of the straight line whose Cartesian equation is (π₯ β 3)/4 = (π§ + 6)/3, π¦ = 4. Then determine a point that lies on this straight line.

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### 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, weβre 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 ππ‘. π¦ equals π¦ one plus ππ‘. And π§ equals π§ one plus ππ‘.

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 thatβs because weβre 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 thereβs 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. Iβve chosen three, four, negative six as the point on our line. But as weβve seen, there are a whole host of options we could have chosen to find a point on this line.