Parametric Form

Hello. I’ll go a little off the track to cover what I had left behind – the parametric form of the equation of a straight line.

This is an important form of the equation, which looks quite different from the rest of the forms. It is useful in a wide variety of problems, where it leads to much shorter methods as compared to using the other forms. We’ll observe this during examples.

To the derivation..

Consider a line which has slope tanθ and passes through the point A(x1, y1).

Let P(x, y) be a point on the line which is at a distance r from the point A. Observe the following figure.

This looks similar to what we used while deriving the point-slope form of the equation.

But here, we’re not interested in finding out the relation between the x and y coordinates of P. (We already know that relation, which is (y – y1)=m(x – x1). )

Instead, we’re going to relate the coordinates of P (x and y) with r. That is, given the value of the ‘parameter’ r, we should be able to get the coordinates of P.

This is what the parametric equation does. Instead of relating x and y with each other directly, it relates both of them with a third parameter (thereby relating them together again, but indirectly). And given the parameter’s value, we are able get the coordinates of a point on the curve.

In this case the parameter (r) is the distance between the fixed point A and the point P. The parameter can be anything, and the parametric form of curve is not unique. (You needn’t worry too much about this.)

Now moving back to the figure. We have, cosθ = AQ/AP = (x-x1)/r and sinθ = PQ/AP= (y-y1)/r

This gives the coordinates of P as (x1 + rcosθ, y+ rsinθ).

And this is the parametric form of the equation of a straight line:  x = x+ rcosθ,  y = y+ rsinθ. (Looks a little different, as I told earlier.)

This can also be written in a fancy way as $$\frac{x-x_1}{cosθ} = \frac{y-y_1}{sinθ} = r$$

To find the relation between x and y, we should eliminate the parameter from the two equations. (This will lead us to the point-slope form. Try doing that yourself as an exercise.)

To repeat what I said, r being a parameter means that its value can change, and each value of r corresponds to a unique point on the line.

As an example, let A be (1, 2) and θ=30°. Then the parametric equation of this line looks like $$\frac{x-1}{cos30°} = \frac{y-2}{sin30°} = r$$.

Or, any point the line will have the coordinates of the form (1 + rcos30°, 2 + rsin30°), where r denotes the distance between this point and the point (1,2).

Putting the value of r equal to 2, we get the coordinates of the point as P(1+$$\sqrt{3}$$, 2+1). These are the coordinates of the point (on the given line) which is at a distance 2 unit from the point (1, 2).

There will also be another point at the same distance from this point, whose coordinates will be obtained by putting r = -1, which is Q(1-$$\sqrt{3}$$, 2-1)

This doesn’t mean that the distance is negative, only that the is on the opposite direction from the point where r = +1.

Here’s how things look..

That’s all for now. Things will become a little more clear when I discuss a few examples.

Lesson Summary

1. The parametric equation of a straight line passing through (x1, y1) and making an angle θ with the positive X-axis is given by $$\frac{x-x_1}{cosθ} = \frac{y-y_1}{sinθ} = r$$, where r is a parameter, which denotes the distance between (x,y) and (x1, y1)

In the next lesson, I’ll discuss a few related examples. See you there !