Web Homework College Algebra

Chapter 1: Algebra and Geometry Review

1.1 Algebra

Algebra is a part of mathematics which solves problems by representing quantities by symbols (often called variables), expressing the relationships between the quantities as equations or inequalities, and manipulating these expressions according to well defined rules in order to find additional properties of the quantities and solve the problem.

This course course assumes that you have already had considerable experience in using algebra, and that the material in this chapter is, for the most part, simply review. The remainder of this section will deal with rules for manipulating algebraic expressions, for solving linear and quadratic equations in one unknown, for solving systems of equations in more than one unknown, and for solving inequalities.

1.1.1 Simplifying Expressions

We begin with a review of the basic rules for simplifying algebraic expressions. You probably already know hundreds of these, and the purpose here is point out those that we think are the most important. This will reduce the number of such rules down to a few dozen; in subsequent chapters, we will see that even this short list can be pared down to a few fundamentals from which all the rest follow.

Some really fundamental rules are:

  1. Commutativity of Addition and Multiplication: images/review1.png , and images/review2.png . Note that subtraction and division are NOT commutative.
  2. Associativity of Addition and Multiplication: images/review3.png , and images/review4.png . Because of the Associative law, one often simply omits the parentheses: we write a + b + c because the value of the expression is the same regardless of how it is parenthesized.
  3. Multiplication distributes over addition: images/review5.png , and images/review6.png . Does addition distribute over multiplication?
  4. Additive and Multiplicative Identities: images/review7.png and images/review8.png .
  5. Inverses: For every number a there is a number denoted -a such that images/review9.png . Furthermore, if images/review10.png , then there is a number denoted images/review11.png such that images/review12.png .

The following examples show how these three fundamental rules can be combined to simplify more complicated expressions:

Example 1: You can distribute over longer sums by applying the distributive law multiple times:

images/review13.png

Example 2: One can multiply out products of arbitrary sums using the Example 1:

images/review14.png

Operator Precedence: In the above expressions we have been omitting parentheses. For example, we write ab + ac which might be interpreted as a(b + a)c, but we know that the intention is (ab) + (bd). We know this because of the following precedence rules:

  1. Parenthesized expressions are computed first.
  2. Powers are computed next.
  3. Unary negation is applied next.
  4. Multiplication and Division are next in precedence.
  5. Addition and subtraction have least precedence and are done last.
  6. Within the same precedence level, binary operations are done left to right except for exponentiation. Successive exponentiations should be completely parenthesized to avoid any possibility of ambiguity.

Note: Depending on the context, multiple exponentiations without parentheses might be interpreted as being grouped left to right, righjt to left, or as simply being syntactically incorrect. The best practice is to always use an explicit parenthesization to avoid any possible misinterpretation.

Example 3: Be careful to evaluate operators of equal precedence from left to right. For example, images/review15.png and not 3/8.

Negatives: These are the main properties of the negation operator (unary minus):

PropertyExample
images/review16.png images/review17.png
images/review18.png images/review19.png
images/review20.png images/review21.png
images/review22.png images/review23.png

Fractions: These are the main properties of fractions. In all the formulas, one assumes that the quantities in the denominators are all non-zero.

PropertyExample
images/review24.png if and only if images/review25.png
images/review26.png images/review27.png
images/review28.png images/review29.png
images/review30.png images/review31.png
images/review32.png images/review33.png
images/review34.png images/review35.png
images/review36.png images/review37.png

Powers: One can raise arbitrary real numbers images/review38.png to integer powers. One defines images/review39.png and, by induction, images/review40.png . For negative integers images/review41.png , one defines images/review42.png . For non-negative real numbers images/review43.png and positive integers images/review44.png , one can define the root images/review45.png to be unique real number images/review46.png such that images/review47.png . This allows one to define rational powers of non-negative numbers images/review48.png by images/review49.png . The following properties are true for these rational powers:

PropertyExample
images/review50.png images/review51.png
images/review52.png images/review53.png
images/review54.png images/review55.png

1.1.2 Solving Equations

The last section was concerned with the problem:

Given the values of some quantities, how do you calculate expressions involving those quantities?
Most of the course will be concerned with the inverse problem:
Given the value of some expressions involving quantities, how do you find the values of the quantities.

The two principal arithmetic operations are addition and multiplication. Here is a problem of the second type:

Problem 1: Find two numbers given the values of their sum and product.

The solution of this problem was already known to the Babylonians, and is one of the most important algebra problems known to them. How can we solve it?

First let use represent the two numbers by the letters x and y and represent their sum and product by the letters a and b. The problem can then be expressed as: Given a and b, find x and y such that

images/review56.png

Now there are lots of pairs x and y such that x + y = a. One possibility is to take two equal numbers, this would give x = y and our equation becomes 2x = 2y = a. So, x = y = a/2. Now, if images/review57.png , we would have our solution.

On the other hand, if this weren't the case, we would not have the solution. This would be the case where the two numbers x and y are not equal. We can think of this is x is different from a/2 and so x = a/2 + t for some number t. Since we still want x + y = a, increasing x by t means that we have to decrease y by the same amount, i.e. y = a/2 - t. Now, let's try this as the solution by putting it in the second equation:

images/review58.png

Although it was not clear before how to choose the exact value for t, this last equality tells us that images/review59.png and so images/review60.png . We know the square of the number t we need, and so images/review61.png . But then images/review62.png and images/review63.png . Another solution is obtained by letting t be its second possible value, this simply switches the values of x and y.

Problem 1 is the most important algebra problem solved in antiquity. The approach we have taken is that of Diophantus of Alexandria. The presentation was very quick; so let us go back and comment on a number of important points:

  1. From the statement of the problem, one might have assumed that there was a single answer -- in fact, we seem to have come up with multiple answers, both of which are (hopefully) equally valid.
  2. Logically, we started off by assuming that there was a solution given by x and y and then deduced what the values of x and y must be. We know that the problem can't have any other solutiosn, but we do not yet know that the values of x and y are solutions. So, check this:

    images/review64.png

    The verification for the second solution is analogous -- or one could note that it follows by commutativity of addition and multiplication.

  3. But is this enough? We had been computing without being concerned about whether or not the operations we were doing actually made any sense. In some cases, they may not: notably, we found the square root of images/review65.png . When images/review66.png , there is a problem. There is no real number whose square is negative. There are two ways of resolving this:
    1. At that point in the argument, note that if images/review67.png , then there is no t of the required type and so there can be no solution to the problem.
    2. Extend our number system to include numbers whose squares are negative. The extended number system is called the complex numbers.
    Both of these approaches will be used. It is convenient to have a notion of number in which we can proceed to take square roots with abandon, i.e. without first checking to see if the number is non-negative. A portion of the course will be devoted to showing that one can indeed extend the domain of numbers in this manner. On the other hand, one needs to distinguish the types of numbers -- although one computes with the complex numbers, the solution may be required to be a real number, or may have even more stringent conditions, like it may be required to be a positive integer.
  4. So, we typically have two values of t or none. There is also the possibility that there be exactly one value of t -- this would occur if images/review68.png . So, even though the expressions for (x, y) given for the two values of t appeared different, they might actually correspond to the same numbers.
  5. Let's go back to our original problem: Given a and b, find x and y such that

    images/review69.png

    One could have interpreted the first equation as x = a - y and substituted into the second equation to get images/review70.png . This can be rewritten as images/review71.png . Once one has y, it is easy to obtain x as x = a - y and so our problem is reduced to

    Problem 2: Given a and b, find y such that

    images/review72.png

    Our original problem is symmetric in x and y. So, we could just as well have solved for y = a - x and substituted into the second equation and seen that x satisfied exactly the same quadratic equation as we have seen y satisfies.

    On the other hand, suppose that y is a solution of Problem 2. Then we could set x = a - y. We would then have x + y = a and images/review73.png . So, we see that solving Problem 2 is equivalent to solving Problem 1. In other words, solving a system of simultaneous linear equations (Problem 1) is the same as solving a quadratic equation in one variable (Problem 2). You may have already recognized this as our solution is exactly what one would expect from the quadratic formula applied to Problem 2.

  6. Again, consider the quadratic equation images/review74.png . What we have seen is that the coefficients a and b have an interpretation -- a is the sum of the roots and b is the product of the roots.

1.1.3 Solving Equations in one Variable

Solving systems of equations was the topic of the last section. Let's concentrate in this section on the special case in which there is only one equation and only one variable. The next section will return to the more general case.

There are two basic ideas in working with equations:

  1. If two quantities are equal, then applying the same procedure to each of the quantities yields equal results.
  2. If a product is zero, then so is at least one of the factors.
In other words:

Proposition 1: For all real numbers images/review75.png , images/review76.png , and images/review77.png :

  1. If images/review78.png , then images/review79.png and images/review80.png .
  2. If images/review81.png , then either images/review82.png or images/review83.png . Conversely, one has images/review84.png .

Example 4: To solve the general linear equation ax + b = c, we apply the first principle. Assuming that both sides are equal, we can add -b to both sides to get (ax + b) + (-b) = c + (-b) or ax = c - b. If a is non-zero, then one can then multiply both sides by images/review85.png to obtain images/review86.png , which simplifies to images/review87.png .

Doing the same operations with numbers, one can start with 2x + 3 = 5. Assuming that both sides are equal, we can add -3 to both sides to get (2x + 3) + (-3) = 5 + (-3) or 2x = 5 - 3. Since the coefficient 2 is non-zero, one can multiply both sides by images/review88.png to obtain images/review89.png , or images/review90.png . Now, substitute 1 for x in the original equation to check to see that it is indeed a solution.

If you want to solve: images/review91.png for x, then, assuming that both sides of the equations are equal, one can multiply them by images/review92.png to get

images/review93.png

So, the solution is x = 1. Again, one should substitute this back into the original equation to check that it is indeed a solution.

In each example, we started out by assuming that we had a solution, solved to find out what the solution must have been, and then checked to see that the value actually worked. This last step is NOT just a check for errors in algebra, but is a NECESSARY step in the procedure. For example, suppose you want to solve: images/review94.png . Assuming that you have a solution, we can proceed exactly as in the last example:

images/review95.png

So, the solution can only be x = 2. But, when you try to substitute this value back into the original equation, you see that the denominator is zero. So, x = 2 is NOT a solution. This means that there are NO solutions to the original equation.

Example 5: Suppose you want to solve images/review96.png . Assume that x is a solution to the equation. The left side can be factored to get images/review97.png . Using the second part of Proposition 1, we know that either x - 2 = 0 or x - 4 = 0. So, x = 2 or x = 4. Substituting each of these values back into the original equation verifies that each of two possibilities is indeed a solution of the original equation.

Quadratic Formula: Suppose we want to solve the general quadratic equation: images/review98.png .

If a = 0, then the equation is bx + c = 0. This is a simple linear equation. If b is not zero, then it has a single solution x = -c/b. If b = 0 and c is not zero, then there are no solutions. Finally, if both b and c are zero, then every real number x satisfies the equation.

If a is not zero, then we can divide through by a to put the equation in the form images/review99.png . This is of the form of Problem 2 of the last section. The solutions obtained there were:

images/review100.png

(provided that all the operations make sense, i.e. a is non-zero and images/review101.png ).

Completing the square: This is another approach to solving a quadratic equation, and it is a method we will use in many other problems as well.

Let us assume that we wish to solve the equation images/review102.png . If we could rewrite the equation in the form images/review103.png , then it would be easy to solve for x. This can be accomplished by the operation of completing the square. To see how to do it, just expand out this last expression to get images/review104.png . If this is to be the same as images/review105.png , then corresponding coefficients in the two equations must be equal, i.e. we must have:

images/review106.png

But this system of equation is easy to solve for d and e. We have images/review107.png and images/review108.png .

Let's rework our last example using this method:

Example 5: Solve images/review109.png .

We want to express this in the form images/review110.png . To do this, we take d = a/2 = -6/2 = -3. So, we would get images/review111.png . Rather than trying to remember the formula for e, we can just multiply out the square to get images/review112.png . Comparing the constant terms, we need 9 + e = 8 and so e = -1, i.e. images/review113.png . We can now solve for x, we get images/review114.png and so images/review115.png . So images/review116.png and so the solutions are x = 4 or x = 2. It is easy to verify that these are both solutions of the original equation.

1.1.4 Solving Systems of Equations

When there is more than one variable, one often has more than one equation which the values of the variables are required to simultaneously satisfy. For example, in completing the square, we needed to find all solutions of the system of equations: images/review117.png and images/review118.png where the variables were d and e, and a, b, and c were constants. We were not interested in pairs of numbers (d, e) which satisfied just one of the two equations; they needed to satisfy both equations.

Another example is the general system of 2 linear equations in two unknowns x and y: ax + by = e, cx + dy = f. Now, in the special case of the system in the last paragraph, it was easy to solve for the variables because one of the equations involved only one of the variables. We could use it to solve for that variable. Having its value, we could substitute its value into the other equation and obtain an equation involving only the second variable; this equation was then solved and we had all the possible solutions of the system.

What makes the general system of 2 linear equations look more difficult is that both equations involve both variables. There are two approaches:

  1. One could solve the first equation for one of the variables in terms of the other. Then substitute this into the second equation giving an equation in only the second variable. Then proceed as in the easy case.
  2. One could subtract an appropriately chosen multiple of one equation from the other in order to obtain an equation involving only one variable.
As before, one needs to check that the possible solutions one obtains do indeed satisfy the original equations.

Example 6: Consider the system: 2x + 3y = 5, 4x - 7y = -3. Assume that x and y are satisfy both equations. One can proceed using either method:

  1. Solving for x using the first equation, gives x = (5/2) - (3/2)y. Substituting this into the second equation gives 4((5/2) - (3/2)y) - 7y = -3, which can be used to find y = 1. Substituting this back into our expression for x yields x = 1. One then checks that the values x = 1, y = 1 do indeed satisfy the original equations.
  2. If one subtracts twice the first equation from the second equation, one gets (4x - 7y) - 2(2x + 3y) = -3 - 2(5) or -13y = -13. This gives y = 1 and substituting this value back into the first equation gives 2x + 3 = 5 or x = 1. As before, we need to check that x = 1, y = 1 does indeed satisfy the original two equations.

Example 7: Find all the solutions of the system of equations: images/review119.png , images/review120.png . Assume that one has a solution x, y. Solving the second equation for y, one gets a value y = 1 -x which when substituted into the firs equation gives: images/review121.png or images/review122.png . Collecting terms, we get a quadratic images/review123.png . Factoring and using the second part of Proposition 1, gives x = 0 or x = 1. Substituting these values into our expression for y, gives two possible solutions (x,y) = (0, 1) and (x,y) = (1, 0). Substituting each of these into the original equations, verifies that both of these pairs are solutions of the original system of equations.

If we have more than two variables and more equations, we can apply the same basic strategies. For example, if you have three linear equations in three unknowns, you can use one of them to solve for one variable in terms of the other two. Substituting this expression into the two remaining equations gives two equations in two unknowns. This system can be solved by the method we just described. Then the solutions can be substituted back into the expression for the first variable to find all possible solutions. When you have these, substitute each triple of numbers into the original equations to see which of the possibilities are really solutions. One can also use the second approach as is illustrated by the next example.

Example 8: Solve the system of equations: x + y + z = 0, x + 2y + 2z = 2, x - 2y + 2z = 4. Assume that (x, y, z) is a solution. Subtracting the first equation from each of the other two equations gives y + z = 2, -3y + z = 4. Now subtracting the first of these from the second gives -4y = 2 or y = -1/2. Substituting this into the y + z = 2 gives z = 5/2. Finally, substituting these into the first of the original equations gives x = -2. So the only possible solution is (x, y, z) = (-2, -1/2, 5/2). Substituting these values into the original equations shows that this possible solution is, in fact, a solution of the original system.

1.1.5 Inequalities

In addition to the four basic algebraic operations on real numbers, there is also an order relation. The basic properties are:

Proposition 2: Let a, b, and c be real numbers.

  1. (Trichotomy) For any pair a, b of real numbers, exactly one of these conditions holds true: a < b, a = b, or a > b.
  2. (Transitivity) If a < b and b < c, then a < c.
  3. If a < b, then a + c < b + c.
  4. If a < b and c > 0, then images/review124.png .
  5. If a < b and c < 0, then images/review125.png .

Just as Proposition 1 allowed one to operate with equations, Proposition 2 allows one to work with inequalities. The main difference is that multiplication tends to complicate things as there are two cases depending on whether the multiplier is positive or negative.

Example 9: Solve the general linear inequality ax + b < 0. Using Proposition 2, this implies ax < - b. If a > 0, Proposition 2 gives x < -b/a. On the other hand, if a < 0, then x > -b/a. Finally, if a = 0, then there are no possible solutions if images/review126.png or every real number is a solution if images/review127.png and images/review128.png .

We still need to check that our possible solutions are indeed solutions. This is somewhat more difficult because we do not simply have a small number of values of substitute:

  • Case 1: If a > 0 and x < -b/a, then Proposition 2 shows that ax < -b and so ax + b < 0. So all the x smaller than -b/a are solutions of the inequality.
  • Case 2: If a <0 and x > -b/a, then Proposition 2 shows that ax < -b and so ax + b < 0. So all the x larger than -b/a are solutions of the inequality.
  • Case 3: If a = 0 and images/review129.png , there are no possible solutions. On the other hand, if images/review130.png , then any number x clearly satisfies images/review131.png .

The absolute value |r| of a real number r is defined to be r or -r depending on whether or not r is non-negative or not. For example, |2| = 2, but |-2| = -(-2) = 2.

Example 10: Find all solutions of |2x - 3| > 5. There are two cases:

  • Case 1: If images/review132.png , then |2x - 3| = 2x - 3 and so the inequality becomes 2x - 3 > 5. Solving the inequality gives x > 4. Such x satisfy both images/review133.png and |2x - 3| > 5.
  • Case 2: If images/review134.png , then |2x - 3| = -(x - 3) and so the inequality is -(2x - 3) > 5. Solving this inequality gives x < -1. Such x satisfy both 2x - 3 < 0 and |2x - 3| > 5.
So, the solutions are all real numbers x that are either less than -1 or bigger than 4.

1.2 Geometry

Geometry is the mathematical study of the relationships between collections of points, curves, angles, surfaces and solid objects including the measurement of these objects and the distances between them. This review will discuss the correspondence between the real numbers and the points of a line, the general idea of analytic geometry, as well as the equations of lines, circles, and parabolas as well as the measurement of distance between points.

1.2.1 The Pythagorean Theorem

The most important result of classical synthetic geometry is:

Theorem 1: If images/review135.png is a right triangle with hypotenuse of length c and legs of length a and b, then images/review136.png .

To understand its proof, we need to know

Proposition 3: The sum of the angles of any triangle is 180 degrees.

Proof: Let images/review137.png be an arbitrary triangle. Construct a line DE parallel to the base BC and through the third vertex A as shown in the diagram below.

Since BA is transversal to the pair of parallel lines BC and DE, one has images/review138.png . Similarly, CA is transversal to the pair of parallel lines BC and DE. So, images/review139.png . Since DAE is a straight line, we have images/review140.png . From the diagram, we see that

images/review141.png

as was to be shown.

The proof of the Pythagorean Theorem will also require some simple facts about the areas of squares and triangles. For future reference, let's state these and other results as

Proposition 4:(Areas and Perimeters)

  1. The area of a rectangle (or even a parallelogram) is the product of the lengths of its base and its height.
  2. The area of a triangle is half the product of the lengths of its base and its height.
  3. The area of a circle is images/review142.png where r is its radius. The circumference (perimeter) of of a circle is images/review143.png .

Now it is easy to see why the Pythagorean Theorem is true. Start with an arbitrary right triangle images/review144.png where C is the right angle and the opposite side (the hypotenuse) is of length c. Let a and b be the lengths of the legs which are opposite to the angles A and B respectively.

  1. Construct the figure below where the square in the middle has sides of length c. On each of its sides, place a triangle congruent to images/review145.png using the side of the square as the hypotenuse.
  2. The interior angles of the square are, of course, right angles and the sum of the angles opposite the legs of each of the triangles is images/review146.png by Proposition 3. It follows that the legs of the triangles at each of the vertices of the square actually form a straight line segment (because the sum of the three angles there is precisely 180 degrees. So our figure is actually a large square with sides of length a + b.
  3. Now the large square can also be considered to be the union of the smaller square with side c and four congruent right triangles with base a and height b. So we can calculate the area of the figure in two ways (using Proposition 4) and equate the answers: images/review147.png . When you multiply out and simplify this equation, one finds images/review148.png , as was to be proved.

There are at least hundreds of extant proofs of the Pythagorean Theorem. The proof given here is quite different from the proof appearing in Euclid's Elements.

Corollary 1: i. The length of the diagonal of a rectangle with sides of lengths a and b is precisely images/review149.png .

ii. Let T be a triangle images/review150.png with sides of length a, b, and c opposite angles A, B, and C respectively. Then T is a right triangle with right angle C if images/review151.png .

Proof: i. The first assertion is an obvious consequence of Theorem 1.

ii. Let T' be a triangle images/review152.png where angle C' is a right angle and the sides opposite angles A' and B' are of length a and b respectively. By Theorem 1, the length of the side opposite angle C' is of length images/review153.png . Then corresponding sides of the triangles T and T' are of equal length; so the two triangles are congruent. But then angle C must be equal to angle C' and so angle C is a right angle.

We will need to know about similar triangles. Two triangles images/review154.png and images/review155.png are defined to be similar if images/review156.png , images/review157.png , and images/review158.png . By Proposition 3, if two of these equations hold, then so does the third.

Proposition 5: If the two triangles images/review159.png and images/review160.png are similar, then the ratio of the lengths of a pair of sides of one triangle is equal to the ratio of the lengths of the corresponding pair of sides of the other triangle, e.g. images/review161.png .

1.2.2 Analytic Geometry

The rough idea of analytic geometry is to model the plane as the set of all pairs of real numbers. A curve is then the set of solutions of some algebraic equation. In order to solve a geometric problem, one first translates it into an algebra problem about the sets of algebraic equations. Then one uses algebra to solve this problem, and finally translates the answer back into geometric terms.

In the case of a line, one can model it as the set of real numbers.

  1. Given the line L, choose two points called 0 and 1. Make the distance between 0 and 1 our unit length, so the distance between 0 and 1 is one unit. From 1, mark off another unit distance to determine a new point which we will call 2. Repeat the process to get 3, 4, etc. Then mark off -1, -2, etc. in the opposite direction. Since we know the distance between 0 and 1, one can construct line segments of length 1/n for
  2. For every positive integer n, one can construct a segment of length 1/n. To do this, construct a right triangle images/review162.png with one leg AC of length 1, with right angle at C, and with hypotenuse of length n. Mark off a point D on the hypotenuse such that the distance from A to D is 1 unit. Drop a perpendicular from D to the leg AC. The point of intersection of the perpendicular and AC is denoted E. Then images/review163.png is similar to images/review164.png . Using Proposition 5, one sees that AE is of length 1/n.
  3. Again starting from 0, one can mark off points 1/n, 2/n, etc. as well as points -1/n, -2/n, etc. just as we labeled the integer points. Since we can do this for all positive integers n, we now have all the rational points labeled.
  4. Any real number is determined by its (infinite) decimal expansion. By taking more and more digits of the decimal expansion we get rational numbers which approach the real number. By taking the limiting point of the corresponding points on the line, we get the point on the line which corresponds to the given real number. This completes our mapping of the real numbers to the points of the line L.

To handle the plane, start with two perpendicular lines in the plane. Their point of intersection is denoted 0. The first line is called the x-axis and the second line is called the y-axis. Using the same unit distance as before, one can map the real numbers onto each of the two lines. With any pair (a, b) of real numbers, we can associate a point in the plane: From the point marked a on the x-axis, erect a line perpendicular to the x-axis. Similarly, from the point marked b on the y-axis, erect a line perpendicular to the y-axis. The point of intersection of these two lines is the point associated with (a, b).

Assumption 1: The above mapping is a 1-1 and onto mapping between the set of all pairs (a,b) of real numbers and the points of the plane.

Because of this assumption, we will usually not distinguish between the ordered pair (a, b) and its corresponding point in the plane. In particular, we will refer to the point as being the ordered pair (a,b).

Remark: By using three pairwise perpendicular lines intersecting at a point 0, one can name points in three dimensional space as simply triples (a, b, c) of real numbers. Of course, one could then define n-dimensional space, as simply being the set of ordered n-tuples of numbers.

Definition 1: The distance between the points P = (a,b) and Q = (c, d) is defined to be images/review165.png . This is called the distance formula.

Although the formula is a bit complicated, this is the obvious definition as you can see by examining the diagram below in which PQ is simply the hypotenuse of a right triangle whose legs are of length |c - a| and |d - b| respectively. In light of the Pythagorean Theorem the formula for the length of this hypotenuse is the value given in the definition.

Example 11: The points on either axis labeled a and b are of distance |b - a| from each other -- where distance is calculated by Definition 1.

Example 12: The distance between the points (3, 5) and (2,7) is images/review166.png .

Translation of Graphs If the point (x,y) satisfies the equation f(x, y) = 0, and if A and B are real numbers, then the point (x + A, y + B) satisfies the equation f(x - A, y - B) = 0 (because f((x+A) - A, (y+B) - B) = f(x, y) = 0). Another way of saying this is: if the graph of a function y = f(x) is moved A units to the right and B units to the right, then one obtains the graph of y - B = f(x - A). This amounts to the simple rules:

MotionChange in Equation
Move Graph A units to right Replace x with x - A
Move Graph A units to left Replace x with x + A
Move Graph B units upwards Replace y with y - B
Move Graph B units down Replace y with y + B

Graph Compression and Expansion If the point (x,y) satisfies the equation f(x, y) = 0, and if A and B are non-zero real numbers, then the point (Ax, By) satisifes the equation f(x/A, y/B) = 0 (because f((Ax)/A, (By)/B) = f(x, y) = 0). Another way of saying this is: if the graph of a function y = f(x) is expanded by a factor of A horizontally and by a factor of B vertically, then the result is the graph of y = Bf(x/A). Again, you can give simple rules:

Expansion or CompressionChange in Equation
Expand Graph horizontally by a factor A Replace x with x/A
Compress Graph horizontally by a factor A Replace x with Ax
Expand Graph vertically by a factor B Replace y with y/B
Compress Graph vertically by a factor B Replace y with By

Reflecting a graph across an axis If the point (x, y) satisfies the equation f(x, y) = 0, then the point (-x, y) obtained by reflecting the point (x, y) across the y-axis satisfies f(-x, y) = 0. Similarly, the point (x, -y) obtained by reflecting the point (x, y) across the x-axis satsifes the equation f(x, -y) = 0.

1.2.3 Lines, Circles, and Parabolas

Lines, circles, and other curves like parabolas should simply be the sets of solutions of certain algebraic equations. This appears to be the case. For example, we can define a vertical line to be the set of solutions of some equation x = a, where a is a constant. Similarly, a horizontal line is the set of solutions of an equation of the form images/review167.png for some constant b.

Definition 2: A line is either a horizontal or vertical line or else it is the set of solutions of an equation of the form y = mx + b where m and b are constants.

The equation y = mx + b (or x = a) is called the equation of the line which is the set of solutions of the equation. Note that this is a 1-1 correspondence between the set of lines and the set of equations of lines. In particular, different equations correspond to different lines; in fact, you can easily convince yourself that through any two points there is precisely one line. The constant m is called the slope of the line and the constant b is called the y-intercept of the line. Note that the y-intercept is the y-coordinate of the point where the line intersects the y-axis. The slope and y-intercept of a vertical line is not defined. There are many algebraically equivalent forms of the equation of a line; the particular one y = mx + b is called the slope-intercept form of the equation of the line.

If images/review168.png for i = 1, 2 are two distinct points on the line y = mx + b, then one has images/review169.png for i = 1, 2 and, taking differences, one gets

images/review170.png

Solving for m gives:

Proposition 6: The slope of the non-vertical line through images/review171.png and images/review172.png is given by images/review173.png .

Example 13: The equation of the line through the two points images/review174.png and images/review175.png is given by

images/review176.png

This is the so-called 2 point form of the equation of the line. It is proved by substituting in the two points.

Example 14: The equation of the line with slope m which passes through the point images/review177.png is images/review178.png . This is called the point-slope form of the equation of the line.

Example 15: Consider the line through the two points (2, 3) and (5,7). Its slope is images/review179.png . Its equation in point-slope form is images/review180.png (or equivalently images/review181.png ). To find the y-intercept, just substitute x = 0 into either of these equations and solve for y; the y-intercept is 1/3. So the slope-intercept form of the equation of the line is y = (4/3)x + 1/3. To find points on this line, just substitute in arbitrarily chosen values of x and solve for the y-coordinate.

Two lines which have no point of intersection are said to be parallel.

Example 16: Two distinct lines are parallel if and only if they are either both vertical or else they both have the same slope.

Suppose the two lines are not vertical. Let images/review182.png and images/review183.png be their equations and let's assume (x, y) is a point of intersection. Then it satisfies both equations. Equating them, we get images/review184.png and so images/review185.png . If the slopes were unequal, the coefficient of x would be non-zero and we could solve for images/review186.png and substitute this back into either of the original equations to obtain the value for y. If the slopes are not equal, it is easy to then verify that this pair is indeed a point of intersection, i.e. satisfies both equations. On the other hand, if the slopes were equal, then we would have images/review187.png which would mean that the lines were not distinct (because they have the same equation).

In case the two lines are vertical, the equations are of the form images/review188.png and images/review189.png where images/review190.png . Clearly, no pair (x, y) could satisfy both of these equations; so distinct vertical lines are parallel.

Finally, if one line has the equation images/review191.png and the other the equation images/review192.png , then the point (a, ma + b) is a point of intersection. This completes the proof of the assertion.

We say that two lines are perpendicular if they intersect at right angles to each other.

Example 17: Two lines are pendendicular if and only if one of the following is true:

  1. One line has slope 0 and the other is vertical.
  2. Neither line is vertical and the product of the slopes of the lines is -1.

It is easy to verify the result in case either of the two lines is vertical, either of the two lines is horizontal, or if the lines are are parallel or coincident. So, assume that none of these conditions are true. Then the two lines have slopes and their equations can be written images/review193.png for i = 1, 2 where images/review194.png is the point of intersection of the two lines. The lines have points A and B with x-coordinates, say, a + 1. Using the equations of the lines, we see that the points are images/review195.png and images/review196.png . By the Pythagorean Theorem, the lines are perpendicular if and only if images/review197.png . Using the distance formula, this amounts to

images/review198.png

If you multiply out and simplify this expression, you will see that it is equivalent to images/review199.png .

Remark: The last example shows both the power and danger of analytic geometry. There was no understanding involved, the result came from brute algebraic computation.

A circle with center (a, b) and radius r is the set of points (x,y) whose distance from (a, b) is precisely r. By the distance formula, this circle is the set of solutions of

images/review200.png

The set of solutions of an equation of the form images/review201.png where images/review202.png , b, and c are constants is called a parabola. The following properties are easy to verify:

  1. If a > 0, the parabola opens upward; if a < 0, it opens downward.
  2. If a > 0, the lowest point on the parabola (i.e. the one with smallest y-coordinate) is at (b, c). This point is called the vertex of the parabola.
  3. The parabola is symmetric about its axis x = b (i.e. For every real number z, the points on the parabola with x-coordinates images/review203.png have the same y-coordinates).
If you interchange the x and y variables, one obtains the equations of parabolas which open to the right and left instead of up and down.

Example 18: Find the equation of the line which passes through the two points of intersection of the circle centered at the origin of radius 1 and the circle centered at (1,1) with radius 1.

The equations of the two circles are images/review204.png and images/review205.png . If (x, y) is a point of intersection, it must satisfy both equations. Expanding out the second equation, one gets images/review206.png . Subtracting the first equation and simplifying gives the equation x + y = 1. If we knew that there were two points of intersection, then they would both satisfy this equation and it is the equation of a line; so it must be the desired line. (In order to verify that there are indeed two points of intersection, finish solving the system of equations and verify that the two solutions satisfy the original equations.)

1.2.4 Conic Sections

Let L be a non-vertical line through the origin. By rotating the line L around the y-axis, one obtains a pair of infinite cones with vertex at the origin. By intersecting this pair of cones with various planes, one obtains curves called ellipses, parabolas, and hyperbolas as well as some degenerate cases such as a single point or a single line. These intersections were called sections and so ellipses, parabolas, and hyperbolas were referred to as conic sections, i.e. sections of a cone.

Nowadays the importance of conic sections is not that they arise by intersecting a cone with a plane, but rather that they can be used to categorize all curves whose equations are polynomials of degree 2. So, lines are the curves represented by equations which are polynomials of degree 1, and conic sections are the curves represented by equations which are polynomials of degree 2.

In this section, we will define ellipses, parabolas, and hyperbolas without reference to sections of cones, and obtain equations for each in some special cases.

1.2.4.1 Parabolas

We have already defined parabolas to be the graphs of functions of the form images/review207.png . Let us give a more traditional definition:

Definition 3: Let F be a point and D be a line in the plane which does not contain F. Then the parabola with focus F and directrix D is the set of points P = (x, y) such that the distance from P to F is the same as the distance from P to D. (By the distance from P to the line D, we mean the shortest distance from P to any point of D, i.e. the length of the line segment obtained by dropping a line through P perpendicular to D.)

Now, let's look at a special case in which the vertex F = (0, d) and the line D is y = -d. If P = (x,y) is on the parabola with focus F and directrix D, then the distance formula tells us that

images/review208.png

When you multiply this out and collect terms, one sees that this is just images/review209.png or images/review210.png .

Given the parabola, images/review211.png , we see that a = 1/(4d) and so the focus is at (0, d) = (0, 1/(4a)) and the directrix is y = -d = -1/(4a). Where is the focus and directrix of images/review212.png ?

1.2.4.2 Ellipses

One could define ellipses and hyperbolas in a manner exactly analogous to Definition 3, viz.

Definition 3a: Let F be a point and D be a line in the plane which does not contain F and e be a positive real number. Then the conic section with focus F, directrix D, and eccentricity e is the set of points P = (x, y) such that the distance from P to F is equal to e times the distance from P to D. If e = 1, the conic section is called a parabola. The conic section is called an ellipse if e < 1 and a hyperbola if e > 1.

Such a definition leads to rather complicated formulas, and so, instead, we will define

Definition 4: Let images/review213.png and images/review214.png be two not necessarily distinct points in the plane and let a be a positive real number. Then the ellipse with foci images/review215.png and images/review216.png and semimajor axis a is the set of points P = (x, y) such that the sum of the distances from P to the foci is exactly 2a.

Clearly, if a is less than half the distance between the foci, then the ellipse with semimajor axis a is the empty set.

Again, let us consider a special case: Let images/review217.png and images/review218.png where images/review219.png is a real number. Let a be a real number greater than c. Then if P = (x, y) is a point of the ellipse with these points as foci and with semimajor axis a, then we have by the distance formula:

images/review220.png

This is the equation of the ellipse, but it is better to simplify it a bit. To do so, subtract the second square root from both sides and then square both sides to get:

images/review221.png

Moving all the terms except for the square root from the right side to the left side and simplifying gives

images/review222.png

Dividing both sides by 4 and squaring again gives

images/review223.png

which can be re-arranged to get

images/review224.png

Finally, since a is larger than c and both are positive, we can let b be a positive number such that images/review225.png . Substituting this into our last formula and dividing both sides by images/review226.png gives us the standard form of the equation of the ellipse:

images/review227.png

The quantity e = c/a is called the eccentricity of the ellipse. The quantity b is called the semiminor axis of the ellipse. In particular, if the eccentricity of the ellipse is 0, then the two foci coincide, the semimajor and semiminor axes are equal, and the ellipse is a circle whose radius is precisely the semimajor axis.

Of course, one can also have ellipses with a vertical semimajor axis. If we simply interchange the roles of x and y in the above calculation, the final formula is the same, except that quantities a and b are interchanged. You distinguish the two cases by looking to see which of the two is the larger. You can also translate the graph of the equation to obtain ellipses centered at points (a, b) instead of at the origin (0, 0). As usual, the formulas look the same except that x is replaced with x - a and y is replace by y - b. Expansion and compression either horizontally or vertically just expands or contracts the values of a and b.

1.2.4.3 Hyperbolas

In analogy with Definition 4, we have

Definition 5: Let images/review228.png and images/review229.png be two not necessarily distinct points in the plane and let a be a positive real number. Then the ellipse with foci images/review230.png and images/review231.png and semimajor axis a is the set of points P = (x, y) such that the difference of the distances from P to the foci is exactly 2a. (Note that we have to subtract the smaller of two distances from the larger one in order to get 2a.)

Taking the foci (0, -c) and (0, c) as before, it is easy to use the distance formula to write down the equation of the hyperbola. You should go through steps just as with the ellipse. When you are done, you will see that the equation of the hyperbola reduces down to

images/review232.png

where b is a positive number such that images/review233.png . As before, we call images/review234.png the eccentricity of the hyperbola. Note that e > 1 for hyperbolas and less than 1 for ellipses.

1.2.5 Trigonometry

Angle Measurement: Angles are measured in radians. Start with the unit circle images/review235.png centered at the origin (0, 0). Make one side of the angle the positive x-axis. The other side of the angle is another ray from the origin, say 0A in the diagram below. The magnitude of the radian measure of the angle is equal to the length of the arc of the circle swept out as the positive x-axis is rotated through the angle to the ray 0A. The radian measure is positive if the arc is swept out in a counterclockwise direction and is negative otherwise. If we take instead of the unit circle, the circle centered at the origin of radius r, then the arc swept out by the same action is of length images/review236.png where images/review237.png is the radian measure of the angle.

Again referring to the diagram above, if A is the point on the unit circle corresponding to the second ray of the angle, then its coordinates are by definition the cosine and sine of the angle images/review238.png , i.e. images/review239.png . Since A lies on the unit circle, we have the identity:

images/review240.png

. Further, it is clear by the symmetry of the circle that we have:

images/review241.png

If B is the point on the ray 0A where the ray intersects the circle of radius r: images/review242.png , then images/review243.png is equal to the ratio of the x-coordinate of B and the the hypotenuse r (because of similar triangles). This explains the common definition of the cosine as being the ratio of the adjacent side to the hypotenuse; when using this, be careful, that that the adjacent side may be either a positive or negative number. Similarly, the sine of the same angle is the quotient of the opposite side over the hypotenuse (where the opposite side might be negative). These definitions allow one to create tables of trigonometric functions for practical use in solving right triangles. For example, Ptolemy computed tables in half degree increments.

The remaining trigonometric functions are:

images/review244.png
images/review245.png
images/review246.png
images/review247.png

Example 19: Dividing the identity images/review248.png , by images/review249.png and using the above definitions, one obtains the identity:

images/review250.png

Similarly, by dividing by images/review251.png , one obtains the identity:

images/review252.png

Example 20: Let images/review253.png be the angle opposite the side of length 3 in a right triangle whose sides are of length 3, 4, and 5. Then the trigonometric functions of images/review254.png are:

images/review255.png
images/review256.png
images/review257.png
images/review258.png
images/review259.png
images/review260.png

The values of the trigonometric functions at a number of commonly occurring angles are given in the table below:

DegreesRadianssinecosinetangentcotangentsecantcosecant
00010undefined1undefined
30images/review261.png 1/2images/review262.png images/review263.png images/review264.png images/review265.png 2
45images/review266.png images/review267.png images/review268.png 11images/review269.png images/review270.png
60images/review271.png images/review272.png 1/2images/review273.png images/review274.png 2images/review275.png
90images/review276.png 10undefined0undefined1

Rather than memorize the table, it is easier to keep in mind the figure above used to define the trig functions and reconstruct the two standard triangles shown below.

There are many more important results from trigonometry. The most important are:

Proposition 7:(Addition Formulas) For arbitrary angles images/review277.png and images/review278.png , one has:

  1. images/review279.png .
  2. images/review280.png .

and

Proposition 8: Let images/review281.png be any triangle and a, b, c be the lengths of the sides opposite angles A, B, and C respectively. Then

  1. ( Law of Cosines) images/review282.png .
  2. (Law of Sines) images/review283.png .

You will have an opportunity to prove these results in the exercises.

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Revised: July 12, 2001
All contents © copyright 2001 K. K. Kubota. All rights reserved