Geometry

 

1.     the Pythagorean Theorem

2.     Circles

3.     Straight Lines

slope

point-slope and slope-intercept forms

4.     Parabolas

5.     Other Conic Sections

 

 

1.     The Pythagorean Theorem

 

Perhaps the single most significant theorem covered by secondary math courses is the Pythagorean Theorem. It is the basis for much analytic geometry, force diagrams in Physics and Civil Engineering, and Number Theory in Mathematics.

 

It applies only to right triangles and says that if the legs have lengths a and b, and the hypotenuse c, then c2 = a2 + b2.

 

Examples of triangles exhibiting the theorem are

 

 

 

 

 

These triangles have integers as sides; most do not but the algebra is all the same.

 

The negation of the Pythagorean Theorem is also of interest: if c2 ≠ a2 + b2.

then the triangle is not a right triangle.

 

Problems:  in each case below, the three sides of a triangle are given. Decide which ones are right triangles.

1.  6,8,12

2.  5,10,12

3.  8,15,17

4.  8,6, 10

5.  1, √3 , 2

 

A repeatedly useful skill is “completing a right triangle”. This means that if you are given only 2 sides, you can find the third side. 

 

For example, if the hypotenuse is 7 and one leg is 5, what is the other leg?

 

solution: letting x denote the length of the unknown leg, we must have that

 

52  + x2 = 72       or   25 + x2 = 49    so  x2 = 24  and x =

 

Distance and the Pythagorean Theorem:

 

If we use coordinate axes then the theorem can be used to find the distance between two points. The key reason behind this is that the axes are perpendicular to each other. Suppose we want to find the distance from the point  (1,1)  to  (7,9). Plotted these look like:

and distance translates into the length of the green , sloped line. We can base a right triangle on this situation:

 

T

 

The length of the horizontal, red line is 6 since we go from (1,1) to (7,1). The length of the vertical, red line is 8 since we then go from (7,1) to (7,9).  Thus we have a right triangle with legs of 6 and 8 and hypotenuse being the number we want.  By the Pythagorean Theorem, its length is the square root of  62 + 82 or  which is 10.

 

 

This can be easily generalized to provide a formula for the distance between any two points. Suppose one point is  (x1,y1)  and the other is  (x2,y2). A diagram might look like

 

 

 

 

 

The hypotenuse is the distance, d, that we are after.  The horizontal leg is the amount x changes by in going from the first point to the second and the vertical leg is the change in y.  Thus by the Pythagorean Theorem

 

                   (x2 – x1)2   + (y2 – y1)2 = d2

 

or, equivalently,

 

This is sometimes called the distance formula but it is really just the Pythagorean Theorem adapted to a particular situation.           

 

Example:   how far apart are  (1, 3 )    and  (13, 8)  ?

Solution:    by the above discussion, the distance is

 

Comments:

 

1)      these are “textbook” problems designed to have “nice” answers. In reality, most answers will have a square root in them.

2)      It does not matter which point you treat as the first one and which as the second one. If you exchange them you will pick up a minus sign inside the parentheses but once it is squared, it vanishes.

3)      If the points have any negative coordinates in them, you must be careful not to make any sign errors. Parentheses are good !  Don’t skip steps. See below.

 

Example 2:  How far apart are  (-2,-5)  and  ( 7,-1)?

Solution:

 

Straight Lines

            Slope

                 If we have two points in the x-y plane on a line and we compute the ratio of the vertical change to horizontal change it is called the slope of the line. It provides a measure of how steep the line goes up (or down).

 

If the points are  (2,1)   and  (5,7) then the line through them has slope (7 – 1)/(5-2) = 2.

 

This can be interpreted in the following way:  for every unit you move to the right, you will go up two units.  If you move 5 units to the right, you will go up 10 units. If you move 3 units to the left, you will go down 6 units.

 

 

 

 

 

 

            Slope-Intercept Form

 

            Point Slope Form

 

 

 

Circles

 

Having developed the capability to measure distance analytically, the next obvious thing is to look at circles. Recall the definition of a circle:

 

A circle is all points the same distance from a fixed point (center). That distance is called the radius.

 

Let (x0, y0)   be the center , (x,y) be an arbitrary point on the circle, and  r be the radius.  Then the distance formula becomes

 

 

If we square both sides then this becomes the more often seen version

 

Example:  determine the equation of the circle of radius 5 centered at  (4,-1)

 

Solution: 

 

                       

or

                       

 

Now what are the coordinates of the points at the top, bottom, left and right of the circle?

 

solution:

            beginning at (4,-1) we move 5 unit in the appropriate direction, so

 

            (4, -1+5)  =  (4,4)  at the top

            (4, -1 -5) = (4,-6)  at the bottom

            (4 -5, -1) = (-1,-1) at the left

(4 + 5, -1) = (9,-1)  at the right.

 

three of  these points appear on the circle as:

 

 

 

 

 

Tangent Lines to Circles

 

Calculus I is the study of derivatives and tangent lines.  The circle is the first instance of a place where we run into tangent lines.  The following graph shows one:

 

 

 

The red line is the tangent line to the circle at the point P.  This means all of the following:

 

 

Example:  for the circle centered at the origin of radius 10,

a)      confirm that the point (6,8) is on the circle

b)      determine the equation of the tangent line to the circle at (6,8)

 

Solution:

 

            a)    so the point (6,8) is 10 units from the center putting it on the circle

 

b)   the slope of the line from (0,0) to (6,8)  is m=8/6 = 4/3.   The line passes through (6,8)  so if

 

y = (-3/4)x + b    for the line perpendicular to this

 

            then b is picked to satisfy    8 = (-3/4)(6) + b    or  b = 8 + 9/2 = 25/2

 

and the equation of the tangent line is  y = (-3/4)x + 25/2.

 

 

Determining the Circle from the Equation.

 

This is the reverse of what we did in the last section. We are given the equation of the circle and wish to know its circle and radius.  It takes some algebra to do this.

 

Example:  a circle has the equation

 

                   x2  + 2x + y2 + 10x  =  10

 

Where  is it’s center and what is it’s radius?

 

Solution:

we use the algebraic technique of “completing the square” by adding both 1 and 25 to both sides as follows:

 

            x2  + 2x + 1+ y2 + 10y + 25   =  10+1+25

Since we added the same thing to both sides, the equation is still valid. But why?

 

The 1 makes a perfect square of the first three terms while the 25 makes a perfect square of the last three:

 

                        (x+1)2  +  (y+5)2  =  36

 

(the reader is urged to expand (x+1)2  as well as   (y+5)2 and check all this ). So the center is at  (-1, -5)  and the radius is 6. 

 

Why 1 and 25??  Here is the rule for completing a square of  x2  + bx  

 

a)      compute half of b

b)      square it

c)      add your result from the previous step to both sides of the equation

d)      form  (x + b/2)2

Example:  determine the center and radius of the circle whose equation is

 

x2  + 6x + y2 -8y  =  -9

 

Solution:

 

            we add both 32  and  (-4)2  to both sides:

 

x2  + 6x + 9 +  y2 - 8y + 16  =  -9 + 9 + 16

 

(x+3)2  +  (y-4)2  = 16

 

so the circle has its center at (-3,4)  and radius of  4  (not 16…)

 

 

 

 

 

           

 

 

Problem   Describe the circle whose equation is   x2 + y2 + 8y = 0.

 

Solution: we only need to complete the square on y, by adding 16 to both sides:

 

 

x2 + y2 + 8y + 16 = 0 + 16.

 

or    x2 + (y + 4)2 = 42.   This means it has the center at (0, -4)  and radius of 4.  The graph, while we are at it ,looks like