PATRONES LOCOMOTORES ELEMENTALES

We illustrate the above discussion of coordinate changes in the line and plane by various examples.

Example 1 Find a transformation on the line which changes the point A(3) to A′_{(5) and}

B(−1) to B′_(1).

Use your answer to find out the new coordinate of the point C(3). Also, if D(d) goes to D′_{(−5) what is d?}

Answer:

8_{Note. The calculation shows that the transformations of the type z}′ _{= pz have a very special}

form. The transformation z′_{= pz is also very special and you can work out its form. The complex}

84 CHAPTER 4. INTRODUCTION TO ANALYTIC GEOMETRY.

• We assume that the transformation is x′ _{= px + q and try to find p, q.}

From the information about the point A, we see that 5 = p(3) + q = 3p + q. The information about B says:

1 = p(−1) + q = −p + q.

We need to solve these two equations together. Of course, we can use any of the techniques that we know, but here, it is easiest to subtract the second from the first to get:

4 = 3p + q − (−p + q) = 4p.

So p = 1. Plugging the value of p into the first equation, we see that: 5 = 3 + q so q = 2.

Thus the transformation is:

x′ = x + 2.

• To find the image of C(3), we set x = 3 and deduce that x′ _{= 3 + 2 = 5.}

So, C(3) becomes C′_(5).

• To find the point D, we set x′ _{= −5 and solve −5 = x + 2 to get x = −7.}

So d = −7.

Example 2 Find a plane transformation using only translations and flips which sends P (1, 1) to P′_{(4, 1), Q(2, 3) to Q}′_{(5, −1).}

Determine the new image of the origin. Also determine the point which goes to the new origin.

What point R goes to R′_{(3, 3)?}

Answer:

• Assume that the transformation is given by x′ _{= ux − p, y}′_{= vy − q.}

We have the equations for point P :

4.5. GENERAL CHANGE OF COORDINATES. 85

Similarly for the point Q, we get:

5 = 2u − p , − 1 = 3v − q.

Solve the equations

u − p = 4 and 2u − p = 5 to deduce u = 1, p = −3.

Similarly, solve the equations

v − q = 1 and 3v − q = −1 to deduce v = −1, q = −2.

Thus the transformation is:

x′ = x + 3 , y′ _{= −y + 2.}

• Clearly, the origin O(0, 0) goes to O′_{(3, 2).}

• If x′ _{= 0 and y}′ _{= 0 then clearly, x = −3 and y = 2, so the point S(−3, 2)}

becomes the new origin!

• If x′ _{= 3 and y}′ _{= 3, then x = 0 and y = −1, so R(0, −1) is the desired}

Chapter 5

Equations of lines in the plane.

You are familiar with the equations of lines in the plane of the form y = mx + c. In this chapter, we introduce a different set of equations to describe a line, called the parametric equations. These have distinct advantages over the usual equations in many situations and the reader is advised to try and master them for speed of calculation and better understanding of concepts.

5.1

Parametric equations of lines.

Consider a point P (a, b) in the plane. If t is any real number, we can see that the point (at, bt) lies on the line joining P to the origin. Define Q(t) to be the point (at, bt). We may simply write Q for Q(t) to shorten our notation.

It is easy to see that the line joining P to the origin is filled by points Q(t) as t takes all possible real values. (See the picture below.) 1

1_{The formal proof, however, is not so easy since we are assuming that we know what a line is,}

without ever defining it formally! An algebraist resolves the problem by declaring that a line is simply the set of all points satisfying a linear equation! Then he gets lines defined even for other number fields. A geometer has to simply rely on a geometric intuition and a picture; both can be misleading from time to time. The problem of formalizing and proving the concepts of Geometry are tackled in higher courses on College Geometry.

88 CHAPTER 5. EQUATIONS OF LINES IN THE PLANE.

Origin

Q(at,bt) P(a,b)

Thus, if we write O for the origin, then we can describe our formula as −→

OQ = (at, bt) = t(a, b) = (t)(−→OP ) for some real t .

More generally, a line joining a point A(a1, b1) to B(a2, b2) can then be described as

all points Q(x, y) such that−→AQ = (t)(−→_{AB), i.e. (x − a}1, y − b1) = t(a2 − a1, b2− b1)

and this gives us the:

Parametric two point form: x = a1+ t(a2− a1) , y = b1+ t(b2− b1).

Q(–1+t,1+t)

A(–1,1)

B(0,2)

Here is a concrete example where we take points A(−1, 1) and B(0, 2). (See the picture above.)

5.1. PARAMETRIC EQUATIONS OF LINES. 89

The parametric equation then comes out as

x = −1 + t(0 − (−1)) = −1 + t and y = 1 + t(2 − 1) = 1 + t. We have marked this point as Q in the picture.

Even though, we have a good formula in hand, we find it convenient to rewrite it in different forms for better understanding. Here are some variations:

• The shift −→AB can be conveniently denoted as B − A, where we think of the points as pairs of coordinates and perform natural term by term operations on them. Now, our formula can be conveniently written as:

Compact parametric two point form:

Q(x, y) = A + t(B − A) or simply (x, y) = A + t(B − A).

For example, for our A(−1, 1) and B(0, 2) we get:

A+t(B−A) = (−1, 1)+t ((0, 2) − (−1, 1)) = (−1, 1)+t(1, 1) = (−1+t, 1+t) as before.

• Note that this formula can also be rewritten as: Q(x, y) = Q = (1 − t)A + t(B). Here, the calculation gets a bit simpler:

(1 − t)(−1, 1) + t(0, 2) = (−1 + t, 1 − t) + (0, 2t) = (−1 + t, 1 + t).

In this form, the parameter t carries more useful information which we investigate next.

Examples of parametric lines:

1. Find a parametric form of the equations of a line passing through A(2, −3) and B(1, 5).

Answer: The two point form gives:

x = 2 + t(1 − 2) , y = −3 + t(5 − (−3)) or x = 2 − t , y = −3 + 8t.

The compact form gives the same equations:

90 CHAPTER 5. EQUATIONS OF LINES IN THE PLANE.

2. Verify that the points P (1, 5) and Q(4, −19) lie on the line given above. Answer: We want to see if some value of t gives the point P . This means:

Can we solve? 1 = 2 − t , 5 = −3 + 8t.

A clear answer is yes, t = 1 does the job. Similarly, t = −2 solves: 4 = 2 − t, −19 = −3 + 8t.

Thus P and Q are points of the above line through A, B.

3. Determine the parametric form of the line joining P, Q. Note that we must get the same line! Compare with the original form.

Answer: The compact form would be given by:

(x, y) = P + t(Q − P ) = (1, 5) + t((4, −19) − (1, 5)) = (1, 5) + t(3, −24).

The original compact form of the line was:

(x, y) = (2, −3) + t(−1, 8).

An observation.

Note that the pair of coefficients multiplying t were (−1, 8) for the original equation and are (3, −24) for the new one. Thus the second pair is −3 times the first pair.

This is always the case! Given two parametric forms of equations of the same line, the pair of coefficients of t in one set is always a non zero multiple of the pair of coefficients in the other set.

Thus, it makes sense to make a:

Definition: Direction numbers of a line. The pair of multipliers of t in a parametric form of the equation of a line are called (a pair of ) direction numbers of that line. Any non zero multiple of the pair of direction numbers is also a pair of direction numbers for the same line.

The pair of direction numbers can also be understood as the shift −→P Q for some two points on the line!

In addition, if we set the parameter t = 0, then we must get some point on the line.

In turn, if we have two lines with one common point and proportional direction numbers, then they must be the same line!