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1.E: Analytic Geometry (Exercises)


These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here.

1.1: Lines

Ex 1.1.1 Find the equation of the line through ((1,1)) and ((-5, -3)) in the form (y=mx+b). (answer)

Ex 1.1.2 Find the equation of the line through ((-1,2)) with slope (-2) in the form (y=mx+b). (answer)

Ex 1.1.3 Find the equation of the line through ((-1,1)) and ((5, -3)) in the form (y=mx+b). (answer)

Ex 1.1.4 Change the equation (y-2x=2) to the form (y=mx+b), graph the line, and find the (y)-intercept and (x)-intercept. (answer)

Ex 1.1.5 Change the equation (x+y=6) to the form (y=mx+b), graph the line, and find the (y)-intercept and (x)-intercept. (answer)

Ex 1.1.6 Change the equation (x=2y-1) to the form (y=mx+b), graph the line, and find the (y)-intercept and (x)-intercept. (answer)

Ex 1.1.7 Change the equation (3=2y) to the form (y=mx+b), graph the line, and find the (y)-intercept and (x)-intercept. (answer)

Ex 1.1.8 Change the equation (2x+3y+6=0) to the form (y=mx+b), graph the line, and find the (y)-intercept and (x)-intercept. (answer)

Ex 1.1.9 Determine whether the lines (3x+6y=7) and (2x+4y=5) are parallel. (answer)

Ex 1.1.10 Suppose a triangle in the (x,y)--plane has vertices ((-1,0)), ((1,0)) and ((0,2)). Find the equations of the three lines that lie along the sides of the triangle in (y=mx+b) form. (answer)

Ex 1.1.11 Suppose that you are driving to Seattle at constant speed. After you have been traveling for an hour you pass a sign saying it is 130 miles to Seattle, and after driving another 20 minutes you pass a sign saying it is 105 miles to Seattle. Using the horizontal axis for the time (t) and the vertical axis for the distance (y) from your starting point, graph and find the equation (y=mt+b) for your distance from your starting point. How long does the trip to Seattle take? (answer)

Ex 1.1.12 Let (x) stand for temperature in degrees Celsius (centigrade), and let (y) stand for temperature in degrees Fahrenheit. A temperature of (0^circ) C corresponds to (32^circ ) F, and a temperature of (100^circ)C corresponds to (212^circ)F. Find the equation of the line that relates temperature Fahrenheit (y) to temperature Celsius (x) in the form (y=mx+b). Graph the line, and find the point at which this line intersects (y=x). What is the practical meaning of this point? (answer)

Ex 1.1.13 A car rental firm has the following charges for a certain type of car: $25 per day with 100 free miles included, $0.15 per mile for more than 100 miles. Suppose you want to rent a car for one day, and you know you'll use it for more than 100 miles. What is the equation relating the cost (y) to the number of miles (x) that you drive the car? (answer)

Ex 1.1.14 A photocopy store advertises the following prices: 5cents per copy for the first 20 copies, 4cents per copy for the 21st through 100th copy, and 3cents per copy after the 100th copy. Let (x) be the number of copies, and let (y) be the total cost of photocopying. (a) Graph the cost as (x) goes from 0 to 200 copies. (b) Find the equation in the form (y=mx+b) that tells you the cost of making (x) copies when (x) is more than 100. (answer)

Ex 1.1.15 In the Kingdom of Xyg the tax system works as follows. Someone who earns less than 100 gold coins per month pays no tax. Someone who earns between 100 and 1000 gold coins pays tax equal to 10% of the amount over 100 gold coins that he or she earns. Someone who earns over 1000 gold coins must hand over to the King all of the money earned over 1000 in addition to the tax on the first 1000. (a) Draw a graph of the tax paid (y) versus the money earned (x), and give formulas for (y) in terms of (x) in each of the regions (0le xle 100), (100le xle 1000), and (xge 1000). (b) Suppose that the King of Xyg decides to use the second of these line segments (for (100le xle 1000)) for (xle 100) as well. Explain in practical terms what the King is doing, and what the meaning is of the (y)-intercept. (answer)

Ex 1.1.16 The tax for a single taxpayer is described in the figure 1.1.3. Use this information to graph tax versus taxable income (i.e., (x) is the amount on Form 1040, line 37, and (y) is the amount on Form 1040, line 38). Find the slope and (y)-intercept of each line that makes up the polygonal graph, up to (x=97620). (answer)

1990 Tax Rate Schedules
Schedule X—Use if your filing status is Single
If the amount on Form 1040 line 37 is over:But not over:Enter on Form 1040 line 38of the amount over:

$0$19,45015%$0
19,45047,050$2,917.50+28%19,450
47,05097,620$10,645.50+33%47,050

97,620......Use Worksheet below to figure your tax
Schedule Z—Use if your filing status is Head of household
If the amount on Form 1040 line 37 is over:But not over:Enter on Form 1040 line 38of the amount over:

$0 $26,050 15% $0
$26,050 67,200 $3,907.50+28% 26,050
67,200 134,930 $15,429.50+33% 67,200

134,930......Use Worksheet below to figure your tax

Figure 1.1.3. Tax Schedule.

Ex 1.1.17 Market research tells you that if you set the price of an item at $1.50, you will be able to sell 5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Let (x) be the number of items you can sell, and let (P) be the price of an item. (a) Express (P) linearly in terms of (x), in other words, express (P) in the form (P=mx+b). (b) Express (x) linearly in terms of (P). (answer)

Ex 1.1.18 An instructor gives a 100-point final exam, and decides that a score 90 or above will be a grade of 4.0, a score of 40 or below will be a grade of 0.0, and between 40 and 90 the grading will be linear. Let (x) be the exam score, and let (y) be the corresponding grade. Find a formula of the form (y=mx+b) which applies to scores (x) between 40 and 90. (answer)

1.2: Distance Between Two Points; Circles

Ex 1.2.1Find the equation of the circle of radius 3 centered at:

a) ((0,0))d) ((0,3))
b) ((5,6))e) ((0,-3))
c) ((-5,-6))f) ((3,0))

(answer)

Ex 1.2.2 For each pair of points (A(x_1,y_1)) and (B(x_2,y_2)) find (i) (Delta x) and (Delta y) in going from (A) to (B), (ii) the slope of the line joining (A) and (B), (iii) the equation of the line joining (A) and (B) in the form (y=mx+b), (iv) the distance from (A) to (B), and (v) an equation of the circle with center at (A) that goes through (B).

a) (A(2,0)), (B(4,3))d) (A(-2,3)), (B(4,3))
b) (A(1,-1)), (B(0,2))e) (A(-3,-2)), (B(0,0))
c) (A(0,0)), (B(-2,-2))f) (A(0.01,-0.01)), (B(-0.01,0.05))

( (b) (Delta x=-1), (Delta y = 3), (m=-3), (y=-3x+2), (sqrt{10})

(c) (Delta x=-2), (Delta y = -2), (m=1), (y=x), (sqrt{8})">answer

)

Ex 1.2.3 Graph the circle (x^2+y^2+10y=0).

Ex 1.2.4 Graph the circle (x^2-10x+y^2=24).

Ex 1.2.5 Graph the circle (x^2-6x+y^2-8y=0).

Ex 1.2.6 Find the standard equation of the circle passing through ((-2,1)) and tangent to the line (3x-2y =6) at the point ((4,3)). Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer)

1.3: Functions

Find the domain of each of the following functions:

Ex 1.3.1 ( y=f(x)=sqrt{2x-3}) (answer)

Ex 1.3.2 (y=f(x)=1/(x+1)) (answer)

Ex 1.3.3 (y=f(x)=1/(x^2-1)) (answer)

Ex 1.3.4 (y=f(x)=sqrt{-1/x}) (answer)

Ex 1.3.5 (y=f(x)={ oot 3 of x}) (answer)

Ex 1.3.6 (y=f(x)={ oot 4 of x}) (answer)

Ex 1.3.7 (y=f(x)=sqrt{r^2-(x-h)^2 }), where (r) and (h) are positive constants. (answer)

Ex 1.3.8 (y=f(x)=sqrt{1-(1/x)}) (answer)

Ex 1.3.9 (y=f(x)=1/sqrt{1-(3x)^2}) (answer)

Ex 1.3.10 (y=f(x)=sqrt{x}+1/(x-1)) (answer)

Ex 1.3.11 (y=f(x)=1/(sqrt{x}-1)) (answer)

Ex 1.3.12 Find the domain of (h(x) = cases{ (x^2-9)/(x-3)& x eq 3cr 6& if (x=3).cr}) (answer)

Ex 1.3.13 Suppose (f(x) = 3x-9) and ( g(x) = sqrt{x}). What is the domain of the composition ((gcirc f)(x))? (Recall that composition is defined as ((gcirc f)(x) = g(f(x))).) What is the domain of ((fcirc g)(x))? (answer)

Ex 1.3.14 A farmer wants to build a fence along a river. He has 500 feet of fencing and wants to enclose a rectangular pen on three sides (with the river providing the fourth side). If (x) is the length of the side perpendicular to the river, determine the area of the pen as a function of (x). What is the domain of this function? (answer)

Ex 1.3.15 A can in the shape of a cylinder is to be made with a total of 100 square centimeters of material in the side, top, and bottom; the manufacturer wants the can to hold the maximum possible volume. Write the volume as a function of the radius (r) of the can; find the domain of the function. (answer)

Ex 1.3.16 A can in the shape of a cylinder is to be made to hold a volume of one liter (1000 cubic centimeters). The manufacturer wants to use the least possible material for the can. Write the surface area of the can (total of the top, bottom, and side) as a function of the radius (r) of the can; find the domain of the function. (answer)

1.4: Shifts and Dilations

Starting with the graph of ( y=sqrt{x}), the graph of ( y=1/x), and the graph of ( y=sqrt{1-x^2}) (the upper unit semicircle), sketch the graph of each of the following functions:

Ex 1.4.1 (f(x)=sqrt{x-2})

Ex 1.4.2 (f(x)=-1-1/(x+2))

Ex 1.4.3 (f(x)=4+sqrt{x+2})

Ex 1.4.4 (y=f(x)=x/(1-x))

Ex 1.4.5 ( y=f(x)=-sqrt{-x})

Ex 1.4.6 ( f(x)=2+sqrt{1-(x-1)^2})

Ex 1.4.7 (f(x)=-4+sqrt{-(x-2)})

Ex 1.4.8 (f(x)=2sqrt{1-(x/3)^2})

Ex 1.4.9 (f(x)=1/(x+1))

Ex 1.4.10 (f(x)=4+2sqrt{1-(x-5)^2/9})

Ex 1.4.11 (f(x)=1+1/(x-1))

Ex 1.4.12 (f(x)=sqrt{100-25(x-1)^2}+2)

The graph of (f(x)) is shown below. Sketch the graphs of the following functions.

Ex 1.4.13 ( y=f(x-1))

Ex 1.4.14 (y=1+f(x+2))

Ex 1.4.15 (y=1+2f(x))

Ex 1.4.16 (y=2f(3x))

Ex 1.4.17 (y=2f(3(x-2))+1)

Ex 1.4.18 (y=(1/2)f(3x-3))

Ex 1.4.19 (y=f(1+x/3)+2)


Textbook and Homework

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Part 2: XAS calculations

To compute the absorption spectra, download or copy the input file bellow to the working directory. It is a general input that needs to be edited depending on which system you are working with.

To compute the absorption spectra for the bulk MgS, first rename the input file changing the X to S . It can be done by typing in the terminal:

Now change all the X s in the input file to S s. Move the new input file to the correct working directory. The next step is to add the optimized coordinates of the system, that you can find them in the .xyz file written by the program after the geometry optimization. Use the last iteration step values, and write them in the &COORD subsection. The final step is to add the correct values for the lattice vectors. You can copy it from the geometry optimization input file.

To run this calculation proceed as you did before.

This calculation should take longer than the geometry optimization to run. Once it is finished, check the number of warnings and if the calculation converged. Sometimes it does not converge within the maximum number of iterations we set in the input file. If this is the case, you can increase the number using the keyword MAX_SCF .

You can check in the working directory that some files were created. The absorption energies and intensities (oscillator strength) are written in the files named MgS-xas_at1_st1.spectrum and MgS-xas_at2_st1.spectrum , where the first one corresponds to the atom 1 in your input file, and the second one to atom number 2.

To convolute the spectra with gaussian functions, download the files lib_tools.zip and extract them in the same directory as the output files. Now run the script typing in the terminal:

As an output you are going to get two files: spectrum.inp and spectrum.out . The first one contains the same information as the Mgs-xas_at1_st1.spectrum file, and in the second one you will find you absorption spectrum for atom 1. Change the name of the files to S_K-edge.inp and S_K-edge.out , for example. You can now plot both absorption intensities from the file S_K-edge.inp and the convoluted spectrum from the file S_K-edge.out . From the first one only the second and sixth columns need to be plotted.

In order to obtain the spectrum for atom 2, you can open the file get_average_spectrum.sh , and replace at1 by at2 in the line for i in $(ls $

/*xas_at2*spectrum) . Run the script again and you will obtain the same two files again, but now with the absorption intensities and spectrum of atom 2. Change their names to Mg_K-edge.inp and Mg_K-edge.out , and plot the absorption spectrum.


Question 1.
A bridge has a parabolic arch that is 10m high in the centre and 30m wide at the bottom. Find the height of the arch 6m from the centre, on either sides.
Solution:
From the diagram, equation of the parabolic arch

∴ The required height =10 – y1 = 10 – 1.6 = 8.4m.

Question 2.
A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16m, and the height at the edge of the road must be sufficient for a truck 4m high to clear if the highest point of the opening is to be 5m approximately. How wide must the opening be?
Solution:
From the diagram,
AA’ = 16 m, OA = 8m, OB = 5m
∴ Equation of the ellipse is

∴ The required wide for the opening is 2y1 = 2(4.8) = 9.6 m

Question 3.
At a water fountain, water attains a maximum height of 4m at horizontal distance of 0.5m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75m from the point of origin.
Solution:
From the diagram
Equation of the path of water

The refined height = 4 – y1 = 4 – 1 = 3 m

Question 4.
An engineer designs a satellite dish with a parabolic cross section. The dish is 5m wide at the opening, and the focus is placed 1.2m from the vertex
(a) Position a coordinate system with the origin at the vertex and the x-axis on the parabola’s axis of symmetry and find an equation of the parabola.
(b) Find the depth of the satellite dish at the vertex.
Solution:
From the diagram,

(a) Consider the satellite dish is open rightward parabola
y 2 = 4 ax ……….. (1)
Clearly a = 1.2m
(1) ⇒ y 2 = 4(1.2)
y 2 = 4.8x
(b) Use the point (x1, 2.5) in (1)
(2.5) 2 = 4(1.2)x1
(frac<(2.5)^<2>><4(1.2)>) = y1
x1 = 1.3 m
∴ The depth of the satellite dish at vertex is 1.3 m

Question 5.
Parabolic cable of a 60m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every 6m along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.
Solution:
From the diagram,

Equation of the suspension bridge
(x – h) 2 = 4a(y – k)
But V (0, 3)
x 2 = 4a (y – 3)
Use the point (30, 16) in (1)
30 2 = 4a (16 – 3) ⇒ 900 = 13 × 4a
(frac<900><13 imes 4>) = a

(i) The length of the first vertical cable from the vertex is
Use (6, y1) in (2)
(2) ⇒ (6) 2 = (frac<900><13>) (y1 – 3)
(frac<36 imes 13><900>) = y1 – 3
0.52 = y1 – 3
y1 = 3.52 m
(ii) The length of the second vertical cable from the vertex is
Use the point (12, y2) in (2)
(2) ⇒ (12) 2 = (frac<900><13>) (y2 – 3)
(frac<144 imes 13><900>) = y2 – 3
0.52 = y2 – 3
y2 = 3.52 m

Question 6.
Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation (frac><30^<2>>-frac><44^<2>>) = 1. The tower is 150m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
Solution:
From the diagram,equation of hyperbola is

Question 7.
A rod of length 1.2m moves with its ends always touching the coordinate axes. The locus of a point P on the rod, which is 0.3m from the end in contact with x-axis is an ellipse. Find the eccentricity.
Solution:
From the diagram,
(i) ∆ le OAB be a right angle triangle.
(ii) ∠APD and ∠PBC are corresponding angles, so corresponding angles are equal.

Question 8.
Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line of the pipe, the flow of water has curved outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground?
Solution:
From the diagram,

Equation of the water path is
x 2 = – 4 ay
Use the point (3, – 2.5) in (1)
(3) 2 = – 4a(- 2.5)
9 = 10a
a = (frac<9><10>) substituting in (1)
(1) ⇒ x 2 = -4(frac<9><10>)y …………. (2)
Use the point (x1, -7.5) in (2)
(2) ⇒ x1 2 = -4 (frac<9><10>)(-7.5) ⇒ x1 2 = 30((frac<9><10>))
x1 = (sqrt<3 imes 9>)
x1 = (3 sqrt<3>) m
∴ The water strikes the ground (3 sqrt<3>) m beyond the vertical line.

Question 9.
On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4m when it is 6m away from the point of projection. Finally it reaches the ground 12m away from the starting point. Find the angle of projection?
Solution:
From the diagram,

Equation of the parabolic path is
x 2 = -4ay
Use the point (6, -4) in (1)
(1) ⇒ (6) 2 = 16a
(frac<36><16>) = 10a
substitute a = (frac<9><14>) in (1)
(1) ⇒ x 2 = -4(left(frac<9><4> ight))y
x 2 = -9y ……….. (2)
Differentiate with respect to ‘x’

Question 10.
Points A and B are 10km apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is 6km closer to A than B. Show that the location of the explosion is restricted to a particular curve and find an equation of it.
Solution:
From the diagram,

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.5 Additional Problems

Question 1.
If a parabolic reflector is 20cm in diameter and 5cm deep, find the distance of the focus from the centre of the reflector.
Solution:

By the property of the parabolic reflector, the position of the bulb should be placed at the focus.
By taking the vertex at the origin the equation of the reflector is y 2 = 4ax.
Let PQ be the diameter of the reflector P = (5, 10)
Since P (5, 10) lies on the parabola,
10 2 = 4a × 5
ie., 100 = 20a ⇒ a = 5
So the focus is at a distance of 5cm from the vertex and focus is (5, 0).

Question 2.
The focus of a parabolic mirror is at a distance of 8cm from its centre (vertex). If the mirror 25cm deep, find the diameter of the mirror.
Solution:

Let the vertex be at the origin.
VF = a = 8cm
The equation of the parabola is
Y 2 = 4ax = 4(8)x = 32x
Depth of the mirror = x1 = 25cm.
So, radius is 0.
⇒ y 2 = 32(25) = 800
y = (sqrt<800>=10 sqrt<8>=10 imes 2 sqrt<2>=20 sqrt<2>) = Radius of the mirror
∴ Diameter of the mirror = 2 × 20(sqrt<2>) = 40 (sqrt<2>) cm of the mirror.

Question 3.
A cable of a suspension bridge is in the form of a parabola whose span is 40 mts. The roadway is 5 mts below the lowest point of the cable. If an extra support is provided across the cable 30 mts above the ground level, find the length of the support if the height of the pillars are 55 mts.
Solution:

The lowest point on the cable is taken as the vertex and it is taken as the origin.
Let AB, CD be the pillars.
Span of parabola = 40 mts = distance between AB and CD
C’V = VA’ = 20 mts
Height of each pillar = 55 mts ⇒ AB = 55 mts
So, A’B = 55 – 5 = 50 mts
Thus, the point B is (20, 50).
Equation of the parabola is x 2 = 4ay
Here, B is a point on the parabola, x 2 = 4ay
(20) 2 = 4a (50) ⇒ 4a = (frac<20 imes 20><50>) = 8
∴ The equation is x 2 = 8y
Let PQ be the length of the extra support RQ.
RQ = 30, RR’ = 5 ⇒ R’Q = 25
Let VR’ be x1 ∴ Q is (x1, 25).
Q is a point on parabola
x1 2 = 8 x 25 = 200
x1 = (sqrt<200>=10 sqrt<2>)
The entire length, PQ = 2 x1 = 20(sqrt<2>).mts.

Question 4.
A kho-kho player in a practice session while running realises that the sum of the distances from the kho-kho poles from him is always 8m. Find the equation of the path traced by him if the distance between the poles is 6m.
Solution:

Give FP + F’P = 4
ie., 2a = 8 and
FF’ = 2ae = 6
ie., a = 4 and ae = 3
e = (frac=frac<3><4>)
b 2 = a 2 (1 – e 2 ) = 16(1 – (frac<9><16>)) = 7
So the equation of the path is an ellipse whose equation is (frac><16>+frac><7>) = 1.


Hyperbola

8.1 DEFINITION

A hyperbola is defined as the locus of a point that moves in a plane such that its distance from a fixed point is always e times (e > 1) its distance from a fixed line. The fixed point is called the focus of the hyperbola. The fixed straight line is called the directrix and the constant e is called the eccentricity of the hyperbola.

8.2 STANDARD EQUATION

Let S be the focus and the line l be the directrix. Draw SX perpendicular to the directrix. Divide SX internally and externally in the ratio e : 1 (e > 1). Let A and A′ be the point of division. Since and the points A and A′ lie on the curve.

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A Symbolic Dynamic Geometry System Using the Analytical Geometry Method

A symbolic geometry system such as Geometry Expressions can generate symbolic measurements in terms of indeterminate inputs from a geometric figure. It has elements of dynamic geometry system and elements of automated theorem prover. Geometry Expressions is based on the analytical geometry method. We describe the method in the style used by expositions of semi-synthetic theorem provers such as the area method. The analytical geometry method differs in that it considers geometry from a traditional Euclidean/Cartesian perspective. To the extent that theorems are proved, they are only proved for figures sufficiently close to the given figure. This clearly has theoretical disadvantages, however they are balanced by the practical advantage that the geometrical model used is familiar to students and engineers. The method decouples constructions from geometrical measurements, and thus admits a broad variety of measurement types and construction types. An algorithm is presented for automatically deriving simple forms for angle expressions and is shown to be equivalent to a class of traditional proofs. A semi-automated proof system comprises the symbolic geometry system, a CAS and the user. The user’s inclusion in the hybrid system is a key pedagogic advantage. A number of examples are presented to illustrate the breadth of applicability of such a system and the user’s role in proof.


1.E: Analytic Geometry (Exercises)

3.3.3 Isometry for the Analytic Euclidean Plane Model Printout
In mathematics the art of asking questions is more valuable than solving problems.
Georg Cantor (1845–1918)

A natural question arises, "What is the form of the matrix of an isometry that is an affine transformation of the Euclidean plane?" We investigate that question. The matrix of an affine transformation of the Euclidean plane has the form . What restrictions need to be placed on the values aij for A to be a matrix of an isometry? For two points X and Y, let X' = AX and Y' = AY. Then

If we assume A is an isometry, then d( X, Y) = d(X', Y'). Hence

For the first and last expressions to be equal, we must have

Assume a11 = 0. Then by (1), a21 = ±1. And by (3), a22 = 0. Finally, by (2), a12 = ±1.
Assume a11 is nonzero. Then by (3), Substitute into (2), Hence, by (1), a22 = ± a11. If a22 = a11, then (3) implies a12 = –a21. If a22 = –a11, then (3) implies a12 = a21.

Since , there is a real number such that and Further, note that there are no restrictions on a13 and a23.
Thus, we summarize the results in the following proposition, which only has the converse left to be proven.

Proposition 3.7. An affine transformation of the Euclidean plane is an isometry if and only if the matrix representation is

( direct isometry ) or ( indirect isometry ).

Corollary to Proposition 3.7. The determinant of a direct isometry is 1 and the determinant of an indirect isometry is –1.

Examples.Which is a direct isometry? Which is an indirect isometry? Note the positions of the triangles. What happens with the measures of the angles between the sides? Use the definition of the measure of the angle between two lines to check your conjectures. Investigate further by looking at the animation video clips. (See below between the two examples for the links.)

Click here for an animation of the graphic examples: example above or example below.

Proposition 3.8. The product of the matrices of two affine direct or two affine indirect isometries of the Euclidean plane is the matrix of an affine direct isometry. Further, the product of an affine direct and an affine indirect isometry of the Euclidean plane is an affine indirect isometry of the Euclidean plane.

Proposition 3.9. The set of affine direct isometries of the Euclidean plane is a group.

Proposition 3.10. The set of affine isometries of the Euclidean plane is a group.

We partially examine the questions asked before the illustrations above. Notice that the first diagram illustrates a direct isometry whereas, the second diagram is an indirect isometry. If we label the vertices in the original triangle clockwise as A, B, and C, what happens with the vertices in the image figure for each diagram? In the first diagram, the vertices of the image remain in the same clockwise order but reverse to a counter-clockwise order in the second diagram. It appears that a direct isometry keeps the orientation the same, and an indirect isometry reverses the orientation.
Examine this further by computing the measures of the angles between the lines determined by the sides for both diagrams. The angle between the lines l[1, –1, 0] and m[1, –3, 2] measures approximately –0.464, where (Check the computations determining the two lines and the measure of the angle.) The measure of the angle between the two image lines l'[1, –3.085, –4.322] and m'[1, 6.655, 8.210], in the first diagram, is approximately –0.464. The angles between the lines l and m and the lines l' and m' measure the same. The measure of the angle between the two image lines l'[1, 0.325, –1.424] and m'[1, 0.985, –1.751], in the second diagram, is approximately 0.464. The measure of the angle between the image lines l' and m' has the opposite sign of the measure of the angle between the lines l and m. Compute the values for the other two angles, and .
The observations in the above examples lead us to conjecture the next two propositions.

Proposition 3.11. For an affine direct isometry of the Euclidean plane, the measure of the angle between two lines equals the measure of the angle between the two image lines.

Proof. Let the lines p' and q' be the images of lines p and q under a direct isometry with matrix A. Let B be the inverse of the matrix A. By Proposition 3.9, B is the matrix of a direct isometry. By Proposition 3.6, there are nonzero real numbers k1 and k2 such that k1 p' = pB and k2q' = qB. We use the results of the previous two sentences together with Proposition 3.7 to compute . The line q' may be expressed in a similar form. Compute the measure of the angle between p' and q' where the angle is not a right angle. (The case for a right angle is left for you to verify.)

Proposition 3.12. For an affine indirect isometry of the Euclidean plane, the measure of the angle between the two image lines has the opposite sign of the measure of the angle between the two lines.

Exercise 3.37. Let A(0, 0, 1), B(1, 0, 1), C(0, 1, 1), D(1, 1, 1), E(2, 1, 1), and F(1, 2, 1). Show the sets <A, B, C> and <D, E, F> are congruent. (Two sets of points are said to be congruent provided there is an isometry where one set is the image of the other set.)

Exercise 3.38. An affine transformation maps X(5, 0, 1) to X'(4, 6, 1) and Y(0, 0, 1) to Y'(1, 2, 1). (a) Show d(X, Y) = d(X', Y') and show the transformation may not be an isometry. (b) Find a direct isometry for the transformation. (c) Find an indirect isometry for the transformation. (d) Find the image of Z(3, 10, 1) for the isometries you obtained in parts (b) and (c).

Exercise 3.39. Complete the proof of Proposition 3.7.

Exercise 3.40. Prove Proposition 3.8.

Exercise 3.41. Prove Proposition 3.9.

Exercise 3.42. Prove Proposition 3.10.

Exercise 3.43. Fill in the missing steps of the two computations in the proof of Proposition 3.11.

Exercise 3.44. Prove the inverse of an affine indirect isometry of the Euclidean plane is an affine indirect isometry of the Euclidean plane.

Exercise 3.45. Prove Proposition 3.12. (Note Exercise 3.44.)


1.E: Analytic Geometry (Exercises)

Let two perpendicular lines pass through the orthocenter $H$ of $Delta ABC.$ Assume they meet the sides $AB,$ $AC,$ and $BC$ in $C_1,$ $B_1,$ $A_1$ and $C_2,$ $B_2,$ $A_2,$ respectively. Let $tin [0,1].$ Define $M_1= tA_1+(1-t)A_2,$ $M_2=tB_1+(1-t)B_2,$ and $M_3=tC_1+(1-t)C_2.$

Prove that $M_<1>,$ $M_<2>$ and $M_3$ are collinear.

Solution

As has already been done previously, we choose to place the origin of the coordinate system at the orthocenter and the axes along the two given lines. We thus may specify the coordinates of the points involved: $A_<1>(a,0),$ $A_<2>(0,b),$ $B_<1>(c,0),$ $B_<2>(0,d),$ $C_<1>(e,0),$ $C_<2>(0,f).$ From here, $M_<1>(ta,(1-t)b),$ $M_<2>(tb,(1-t)d),$ $M(te,(1-t)f).$

The three points are collinear iff

$left|egin ta & (1-t)b & 1 tc & (1-t)d & 1 te & (1-t)f & 1 end ight|= t(1-t)left|egin a & b & 1 c & d & 1 e & f & 1 end ight|=0, $

We may exclude cases where $t=0$ or $t=1$ as trivial because then the three midpoints all lie on one of the given lines and are, therefore, automatically collinear:

Thus the collinearity of the three points is equivalent to the determinant identity:

$left|egin a & b & 1 c & d & 1 e & f & 1 end ight|=0, $

Note that segment $A_1A_2$ lies on the side line $BC$ of $Delta ABC,$ and, therefore, the latter has the slope $-b/a.$ Let $A=(x_<1>,y_<1>).$ Then since $AHperp BC,$

Now, again, since $A_1A_2$ and $BC$ define the same line, its equation is

In particular, $bx_<2>+ay_<2>=ab$ and $bx_<3>+ay_<3>=ab.$ Thus we can solve (2) and (4) for $x_2$ and $y_2,$ whereas from (3) and (4) we can obtain for $x_3$ and $y_3.$ Let's focus on $x_2$ and $y_2:$

But we can also use the fact that $B$ lies on $AB,$ i.e., $C_1C_2.$ Since the equation of the latter is $fx+ey=ef,$ we have $fx_<2>+ey_<2>=dc.$ Solving this together with (2) gives,

Equating the two expressions for either $x_2$ or $y_2$ we obtain

The latter expression is equivalent to

Proceeding similarly for $A$ and $C$ yields additional identities $ef(ac+bd)=cd(ae+bf)$ and $cd(ae+bf)=ab(ce+df).$ Between them, the three identities hold three equal quantities. We'll need just (7'). Let's return to the determinant in (0):

$left|egin a & b & 1 c & d & 1 e & f & 1 end ight|= left|egin a & b & 1 c & d & 1 e-a & f-b & 0 end ight|. $

We may assume that none of the six quantities $a,b,c,d,e,f$ is ,$ for, otherwise, one of the vertices lies on one of the given lines, making it an altitude. For example, if $b=0$ the second line is the altitude through $A,$ whereas the first is parallel to $BC:$

In this case, $M_2M_3parallel BC,$ $M_1$ is the point at infinity of the pencil of lines parallel to $BC,$ making the three points collinear. We thus assume that $abcdef e 0$ and proceed to modify the determinant in (8). With (7') in mind:

$ abcdefleft|egin a & b & 1 c & d & 1 e-a & f-b & 0 end ight| = left|egin abdf & abce & 1 bcdf & acded & 1 bdf(e-a) & ace(f-b) & 0 end ight|$

We continue with the determinant on the right. Invoking (7') and (7):

$egin left|egin abdf & abce & 1 bcdf & acded & 1 bdf(e-a) & ace(f-b) & 0 end ight| &= left|egin abdf & abce+abdf & 1 bcdf & acde+bcdf & 1 bdf(e-a) & 0 & 0 end ight| &=bdf(e-a)[abce+abdf-acde-bcdf] &=bdf(e-a)[ab(ce+df)-cd(ae+bf)] &=0. end$

Which shows the required collinearity.

Acknowledgment

The proof is a slight modification of the one supplied by Leonard Giugiuc (Romania). A synthetic proof for the case of $t=1/2$ could be found on a separate page.


1.E: Analytic Geometry (Exercises)

David Catlin: Monday, 10:30 pm - 11:20 pm. Thursday, 1:30 pm - 2:20 pm.

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Differential Calculus

1.2.5 Analytic functions

(I) P ower series Let E be a normed vector space with norm |.| and suppose that F is a Hausdorff quasi-complete locally convex space (see Remark 1.1 for the case where F is normable). With the notation of section 1.2.1 , let S E F be the K -vector space of formal power series S = ∑pSp, where Sp = cp.X p and c p ∈ ℒ p , s E F . Let (|.|γ)γ ∈ Γ be a family of semi-norms which induces the topology of F and let r > 0 we write that ‖Sγ, r = ∑pr pcpγ and

The set S E F is a K -vector space called the space of convergent power series. If S ∈ S E F , we say that ρ S ≔ inf r > 0 : S ∈ S r E F is the radius of convergence of S. Suppose that ρ (S) > 0 if we replace the indeterminate X with an element hE such that | h | < ρ (S), then the family cp.h p is summable ([P2], section 3.2.1 (III)), as can be seen by adapting the proof of ([P2], section 3.4.1 (I), Theorem 3.41), and the mapping S : SS (h) is continuous in the open set | h | < ρ (S).If F is a Banach space and ρ (S) > 0, then the power series S is absolutely convergent in | h | < ρ (S) and normally convergent in | h | ≤ r′ for every r′ such that 0 < r′ < ρ (S) ([P2], section 4.3.2 (I)).

(II) A nalytic functions Let A be a non-empty open subset of E. We say that a function f from A into F is analytic (or is a mapping of class C ω ) if, for each point aA, there exists a convergent series S ∈ S E F , denoted by fa, such that f (a + h) = fa (h) for every hE with sufficiently small norm. This definition generalizes ([P2], section 4.3.2 (I), Definition 4.74). Write C ω A F for the K -vector space of analytic functions from A into F. If K = ℝ and f ∈ C ω A F , then f is of class C ∞ in A, and so is each of its differentials D p f (p ≥ 1). Every mapping f ∈ C ω A F admits the following Taylor series expansion at the point a, which converges in | h | < ρ(fa):

If | a | < ρ(fa), then the radius of convergence of the Taylor expansion of f at the point a is greater than or equal to ρ (fa) − | a |. If A = E and ρ (fa) = +∞, then the function f is said to be entire.

Let E, F be Banach spaces, G a quasi-complete locally convex space, A an open subset of E, f : AF an analytic function, B an open subset of F containing f (A) and g : BG an analytic function. Then, gf is analytic (exercise) see ( [BOU 82a] , 3.2.7), ( [WHI 65] , p. 1079).

The principle of analytic continuation ([P2], section 4.3.2 , Theorem 4.76) can be generalized as follows (exercise: see [WHI 65] , p. 1080): let E and F be two Banach spaces, Ω a connected open subset of E and f, g two analytic functions from Ω into F. If f and g coincide in any non-empty open subset of Ω, then they must be equal.

Let E be a Banach space and writefor the subset of invertible operators in ℒ E . Let ℐ : ℌ → ℌ : u ↦ u − 1 . The mappingis analytic and satisfies D ℐ u 0 . h = − u 0 − 1 . h . u 0 − 1 for every u 0 ∈ ℌ .

We know that ℌ is open in ℒ E ([P2], section 3.4.1 (II), Corollary 3.49 ). Let u 0 ∈ ℌ and s ∈ ℒ E . We have u0 + s = u0 (1Ev), where v = − u − 1 s. If || v || < 1, then 1Ev is invertible with inverse Σn ≥ 0 v n (ibid.). Hence, if s < 1 u 0 , u0 + s has inverse ∑n ≥ 0(− u0 − 1 . s) n u0 − 1 , which shows that ℐ is analytic. Furthermore, ∑n ≥ 0(− u0 − 1 . s) n u0 − 1 = u0 − 1 − u0 − 1 . s. u0 − 1 + o(‖s‖).

(III) H olomorphic functions Let E be a normed complex vector space with norm |.|, A some non-empty open subset of E, and F a complex quasi-complete locally convex space. Goursat’s theorem ([P2], section 4.2.4 , Proposition 4.56) can be generalized as follows ( [BOU 82a] , 3.1.1): the function f : AF is analytic if and only if it is holomorphic (i.e. complex-differentiable). If this condition is satisfied for E = E1 × … × En, let a =(a1,…, an) ∈ A, r = (r1, …, rn), where ri > 0, and c α = 1 α ! D α f a , where α is the multi-index (α1,…, αn). The Cauchy inequalities ([P2], section 4.3.2 (II), Lemma-Definition 4.78(2)) can be generalized as follows (exercise): for r αr1 α1rn αn ,

Hence ([P2], section 4.3.2 (II), Theorem-Definition 4.81(3)), if f is entire in E and bounded in F, then it must be constant (Liouville’s theorem). The statement of Hartogs’ theorem ([P2], section 4.3.2 (II), Corollary 4.80) also holds, mutatis mutandis, for a function f : AF, where A is an open subset of E1 × … × En, and each Ei is a complex normed vector space: any such function is analytic if and only if it is analytic in each of its variables when the others are held fixed.

Theorem 1.25

(maximum modulus) Let E (respectively F) be a complex Banach space (respectively quasi-complete Hausdorff locally convex space), A some connected non-empty open subset of E and f : EF a holomorphic function. Let |.|γ be a continuous semi-norm on F. If the function | f |γ : x ↦ | f (x)|γ is not constant, then it does not have a maximum in A.

Let us begin by showing the result by contradiction when E = F = ℂ . Suppose that f has a maximum in A. By translation, we may assume that 0 ∈ A and that this maximum is attained at 0. Let c0 = f (0). If f is not constant, then there exists bm ≠ 0 such that f (z) = c0 (1 + bmz m + z m .h (z)), where h is holomorphic in A and satisfies h (0) = 0. Choose r > 0 such that | z | ≤ r implies zA and h z ≤ 1 2 b m . Let t ∈ ℝ be such that e mit = b m b m . For z = re it , we have

In the case where E = ℂ , we can similarly argue by contradiction by assuming that there exist z0, z1A such that | f (z) |γ ≤ | f (z0)|γ for all zA and | f (z1)|γ < | f (z0)|γ. Let V = λ . f z 0 : λ ∈ ℂ and η : V → ℂ : λ . f z 0 ↦ λ . f z 0 γ . Then, | η |γ = 1, where η γ ≔ sup y ∈ F , y γ ≤ 1 η y . By the Hahn–Banach theorem ([P2], section 3.3.4 (II), Theorem 3.25), there exists a continuous linear form ξ ∈ F ∨ extending η such that | ξ |γ = 1. Therefore, for all xA, | ξf (z)| ≤ | f (z)|γ ≤ | f (z0)|γ = | ξf (z0)|, so ξf is constant by (1). Hence, | ξf (z1) = | ξf (z0)| = | f (z0)|γ and | ξf (z1)| ≤ | f (z1)|γ < | f (z0)|γ, contradiction.

In the general case, let g (ξ) = f (z0 + ξ (zz0)) and suppose that | f (z)|γ ≤ | f (z0)|γ for all zA. Then, g is holomorphic in Ω = ξ ∈ ℂ : ξ < 1 + r for sufficiently small r > 0 and z sufficiently close to z0. Therefore, | g (ξ)|γ ≤ | f (z0)|γ = | g (0)|γ, and g is constant in Ω by (2). Thus, g (0) = g (1), so f (z) = f (z0). The set of zA satisfying this condition is non-empty, open and closed in A, and so must be equal to A ([P2], section 2.3.8 ).


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