Hi. This lesson will involve a few examples and applications of the parametric form of a straight line.

By the way, I have used (and might use again) ‘*straight line*‘ in place of ‘*equation of the straight line*‘. It doesn’t matter much. The two phrases ‘*A line 3x + y = 5…*‘ and ‘*A line whose equation is 3x + y = 5…*‘ can be used interchangeably; they mean the same.

Coming back to the examples…

**Q1. **Find the coordinates of the point(s) on the line 3x – 4y + 1 = 0 at a distance of 5 units from the point A(1,1).

**Sol. **First of all, let me illustrate a method of finding out the required coordinates without using the parametric form.

Let the coordinates of the required point be (x_{1},y_{1}).

Two variables, so we need two equations, which will be obtained using the given conditions:

(i) The point lies on the given line, therefore 3x_{1 }– 4y_{1 }+ 1 = 0.

(ii) It is at a distance 5 units from (1,1), therefore (x_{1} – 1)^{2} + (y_{1} – 1)^{2} = 25.

Substituting the value of x_{1} from the first equation in the second, we get two values of y_{1}, 4 and -2, and correspondingly, two values of x_{1}, 5 and -3.

Therefore, we have two points which satisfy the given conditions: **(5, 4) **and** (-3, -2). **(Note that A is the midpoint of these two points. It has to be ! )

Here’s another method.

Since the point A lies on the given line, we can use the parametric form of its equation. (You may ask what would have happened in case the point didn’t lie on the line. I’ll discuss this in a subsequent example.)

The slope of the line (tanθ) is 4/3. This makes cosθ=3/5, and sinθ=4/5. The equation therefore becomes \(\frac{x-1}{3/5}=\frac{y-1}{4/5}=r\).

This means that any point on this line has the coordinates (1+ 3r/5, 1+4r/5). (where r denotes the distance of the point from (1,1) ).

What now? Simple! Put r =±5, and we get the two points effortlessly as (1+3, 1+4) and (1-3,1-4) or **(5, 4) **and** (-3, -2) **

Here is a figure to illustrate the situation..

In case you’re confused about what just happened, go back and review the parametric form of the equation.

**Q2. **Find the coordinates of the point on the line 2x+y=2, which lies at a distance of 1 unit from the origin.

**Sol. **This is a similar problem. The difference here is that the point, from which the distance is given, does not lie on the given line.

We can still use the parametric equation. Here’s how..

The parametric equation of the red line is x=0 + rcosθ, y = 0 + rsinθ. (where r is the distance from the point (0,0)).

Or, any point on the red line is (rcosθ, rsinθ).

We are interested in that particular point where r=1, and also the point should lie on the line 2x + y = 2.

Putting r = 1, and substituting the resulting coordinates in the equation 2x+y=2, we have 2cosθ + sinθ=2.

This gives two values of θ, 0 and arctan(4/3).

Geometrically, this means there will be two such points on the given line. Also, there will be a case when there is exactly one point satisfying the condition, and another in which no such point would exist. Try to figure that out yourself.

And we’re done ! The required coordinates will be obtained by substituting the values of θ and r.

The required points are **(1, 0)** and **(4/5, 3/5)**.

Note that you could have used the first method as was used in the solution of the previous problem. Please try it yourself and compare the two solutions to get an idea about which one is shorter/efficient.

One more..

**Q3.** In what direction must the line passing through P(1, 2) be drawn, so that its point of intersection with the line x + y = 4, is at a distance 1/\(\sqrt{2}\) units from P?

**Sol. **Another similar problem. Let the slope (referred to as direction) of the line be tanθ.

Equation of this line in the parametric form becomes \(\frac{x-1}{cosθ}=\frac{y-2}{sinθ}=r\).

Putting r = 1/\(\sqrt{2}\), and substituting the coordinates (the values of x and y) in the given line, we have cosθ + sinθ=\(\sqrt{2}\), which gives the value of **θ = 45°**.

Here’s a figure to illustrate.

Okay. This is too much for you to digest at one go. I’ll therefore stop here, and continue with a few more examples in the next lesson.