The Geometry Mathematics 15: Lecture 20 Dan Sloughter Furman University November 6, 2006 Dan Sloughter (Furman University) The Geometry November 6, 2006 1 / 18
René Descartes 1596-1650 Dan Sloughter (Furman University) The Geometry November 6, 2006 2 / 18
René Descartes 1596-1650 Central figure in the move from medieval to modern mathematics, as well as from medieval to modern philosophy Dan Sloughter (Furman University) The Geometry November 6, 2006 2 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Wrote Al-jabr wa l muqabalah Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Wrote Al-jabr wa l muqabalah Our word algorithm comes from his name Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Wrote Al-jabr wa l muqabalah Our word algorithm comes from his name As an example of techniques found in Al-Khowarizmi s work, consider the problem of solving the quadratic equation x 2 + 10x = 39. Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Wrote Al-jabr wa l muqabalah Our word algorithm comes from his name As an example of techniques found in Al-Khowarizmi s work, consider the problem of solving the quadratic equation x 2 + 10x = 39. Al-Khowarizmi constructed the following diagram: 5 5x 25 x x 2 x 5x 5 Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra Al-Khowarizmi (780? - 850?) Wrote Al-jabr wa l muqabalah Our word algorithm comes from his name As an example of techniques found in Al-Khowarizmi s work, consider the problem of solving the quadratic equation x 2 + 10x = 39. Al-Khowarizmi constructed the following diagram: 5 5x 25 x x 2 x 5x 5 Note: he has completed the square by adding the square of area 25 to the figure which has area x 2 + 10x, which, in the given equation, is supposed to have area 39. Dan Sloughter (Furman University) The Geometry November 6, 2006 3 / 18
Geometry and algebra (cont d) That is, we have (x + 5) 2 = x 2 + 5x + 5x + 25 = x 2 + 10x + 25 = 39 + 25 = 64. Dan Sloughter (Furman University) The Geometry November 6, 2006 4 / 18
Geometry and algebra (cont d) That is, we have (x + 5) 2 = x 2 + 5x + 5x + 25 = x 2 + 10x + 25 = 39 + 25 = 64. Hence x + 5 = 8, and so x = 3. Dan Sloughter (Furman University) The Geometry November 6, 2006 4 / 18
Geometry and algebra (cont d) That is, we have (x + 5) 2 = x 2 + 5x + 5x + 25 = x 2 + 10x + 25 = 39 + 25 = 64. Hence x + 5 = 8, and so x = 3. Note: Al-Khowarizmi did not consider negative solutions. Dan Sloughter (Furman University) The Geometry November 6, 2006 4 / 18
Geometric algebra of Descartes Multiplying numbers as lines (see page 240): Dan Sloughter (Furman University) The Geometry November 6, 2006 5 / 18
Geometric algebra of Descartes Multiplying numbers as lines (see page 240): BE BD = BC BA Dan Sloughter (Furman University) The Geometry November 6, 2006 5 / 18
Geometric algebra of Descartes Multiplying numbers as lines (see page 240): BE BD = BC BA Since BA = 1, we have either BE = BD BC or, equivalently, BE BD = BC. Dan Sloughter (Furman University) The Geometry November 6, 2006 5 / 18
Geometric algebra of Descartes (cont d) Finding a square root (see page 240): Dan Sloughter (Furman University) The Geometry November 6, 2006 6 / 18
Geometric algebra of Descartes (cont d) Finding a square root (see page 240): (GI ) 2 = (KI ) 2 (GK) 2 = (KI GK)(KI + GK) = KH + GK = GH Dan Sloughter (Furman University) The Geometry November 6, 2006 6 / 18
Geometric algebra of Descartes (cont d) Finding a square root (see page 240): (GI ) 2 = (KI ) 2 (GK) 2 = (KI GK)(KI + GK) = KH + GK = GH So GI is the square root of GH. Dan Sloughter (Furman University) The Geometry November 6, 2006 6 / 18
Geometric algebra of Descartes(cont d) Solving the quadratic equation z 2 = az + b 2 (see page 248): Dan Sloughter (Furman University) The Geometry November 6, 2006 7 / 18
Geometric algebra of Descartes(cont d) Solving the quadratic equation z 2 = az + b 2 (see page 248): OM LM = LM PM, so OM PM = (LM)2. Dan Sloughter (Furman University) The Geometry November 6, 2006 7 / 18
Geometric algebra of Descartes(cont d) Solving the quadratic equation z 2 = az + b 2 (see page 248): OM LM = LM PM, so OM PM = (LM)2. That is, z(z a) = b 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 7 / 18
Geometric algebra of Descartes(cont d) Solving the quadratic equation z 2 = az + b 2 (see page 248): OM LM = LM PM, so OM PM = (LM)2. That is, z(z a) = b 2. Equivalently, z 2 = az + b 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 7 / 18
Geometric algebra of Descartes(cont d) Solving the quadratic equation z 2 = az + b 2 (see page 248): OM LM = LM PM, so OM PM = (LM)2. That is, z(z a) = b 2. Equivalently, z 2 = az + b 2. Then z = ON + NM = 1 2 a + 1 4 a2 + b 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 7 / 18
The Cartesian plane Cartesian plane: identify every point in the plane, uniquely, with an ordered pair of numbers which give the horizontal and vertical distances of the point from two perpendicular axes y (x, y) x Dan Sloughter (Furman University) The Geometry November 6, 2006 8 / 18
The Cartesian plane Cartesian plane: identify every point in the plane, uniquely, with an ordered pair of numbers which give the horizontal and vertical distances of the point from two perpendicular axes y (x, y) x Note: the first coordinate of a point to the left of the vertical axis is negative, as is the second coordinate of a point which lies below the horizontal axis. Dan Sloughter (Furman University) The Geometry November 6, 2006 8 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Example: If P = (x 1, y 1 ) and Q = (x 2, y 2 ) are two points in the plane, then Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Example: If P = (x 1, y 1 ) and Q = (x 2, y 2 ) are two points in the plane, then x 2 x 1 represents the horizontal displacement of Q from P, Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Example: If P = (x 1, y 1 ) and Q = (x 2, y 2 ) are two points in the plane, then x 2 x 1 represents the horizontal displacement of Q from P, y 2 y 1 represents the vertical displacement of Q from P, and, Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Example: If P = (x 1, y 1 ) and Q = (x 2, y 2 ) are two points in the plane, then x 2 x 1 represents the horizontal displacement of Q from P, y 2 y 1 represents the vertical displacement of Q from P, and, by the Pythagorean theorem, PQ = (x 2 x 1 ) 2 + (y 2 y 1 ) 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
The Cartesian plane (cont d) Important idea: certain geometric relationships may be expressed in the language of algebra. Example: If P = (x 1, y 1 ) and Q = (x 2, y 2 ) are two points in the plane, then x 2 x 1 represents the horizontal displacement of Q from P, y 2 y 1 represents the vertical displacement of Q from P, and, by the Pythagorean theorem, PQ = (x 2 x 1 ) 2 + (y 2 y 1 ) 2. Example: The distance between the points ( 1, 2) and (3, 4)is (3 ( 1)) 2 + (4 ( 2)) 2 = 16 + 36 = 52 = 4 13 = 2 13. Dan Sloughter (Furman University) The Geometry November 6, 2006 9 / 18
Circles Note: curves may now be expressed by compact algebraic expressions. Dan Sloughter (Furman University) The Geometry November 6, 2006 10 / 18
Circles Note: curves may now be expressed by compact algebraic expressions. Example: A circle of radius r centered at the point (a, b) consists of the set of all points (x, y) which satisfy (x a) 2 + (y b) 2 = r 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 10 / 18
Example (x 2) 2 + (y 3) 2 = 4 is the equation of a circle of radius 2 centered at (2, 3). 5 4 3 2 1 1 2 3 4 5 Dan Sloughter (Furman University) The Geometry November 6, 2006 11 / 18
Lines Example Dan Sloughter (Furman University) The Geometry November 6, 2006 12 / 18
Lines Example Consider a line through the origin with slope m: 1 m x y Dan Sloughter (Furman University) The Geometry November 6, 2006 12 / 18
Lines Example Consider a line through the origin with slope m: 1 m x y That is, for every change of one unit in the x direction, the line changes by m units in the y direction. Dan Sloughter (Furman University) The Geometry November 6, 2006 12 / 18
Lines Example Consider a line through the origin with slope m: 1 m x y That is, for every change of one unit in the x direction, the line changes by m units in the y direction. Then if (x, y) is a point on the line, we must have, by similar triangles, m 1 = y x. Dan Sloughter (Furman University) The Geometry November 6, 2006 12 / 18
Lines Example Consider a line through the origin with slope m: 1 m x y That is, for every change of one unit in the x direction, the line changes by m units in the y direction. Then if (x, y) is a point on the line, we must have, by similar triangles, Hence y = mx. m 1 = y x. Dan Sloughter (Furman University) The Geometry November 6, 2006 12 / 18
Lines (cont d) Note: in general, y = mx + b is the equation of a line with slope m which passes through (0, b). Dan Sloughter (Furman University) The Geometry November 6, 2006 13 / 18
Lines (cont d) Note: in general, y = mx + b is the equation of a line with slope m which passes through (0, b). Example: y = 2x + 3 is the equation of a line which passes through the point (0, 3) with slope 2. 8 6 4 2-3 -2-1 1 2 3-2 Dan Sloughter (Furman University) The Geometry November 6, 2006 13 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Let C be the parabola with focus at (0, d) and directrix the line y = d. Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Let C be the parabola with focus at (0, d) and directrix the line y = d. Then a point P = (x, y) is on C if and only if the distance from (x, y) to (0, d) is the same as the distance from (x, y) to (x, d). Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Let C be the parabola with focus at (0, d) and directrix the line y = d. Then a point P = (x, y) is on C if and only if the distance from (x, y) to (0, d) is the same as the distance from (x, y) to (x, d). That is, (x, y) is on C if and only if x 2 + (y d) 2 = y + d. Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Let C be the parabola with focus at (0, d) and directrix the line y = d. Then a point P = (x, y) is on C if and only if the distance from (x, y) to (0, d) is the same as the distance from (x, y) to (x, d). That is, (x, y) is on C if and only if x 2 + (y d) 2 = y + d. Hence x 2 + (y d) 2 = (y + d) 2, and so x 2 + y 2 2dy + d 2 = y 2 + 2dy + d 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Parabolas Geometric definition: given a line l, called the directrix, and a point F, called the focus, a parabola is the set of all points P such that the distance from P to F is the same as the distance from P to l. Let C be the parabola with focus at (0, d) and directrix the line y = d. Then a point P = (x, y) is on C if and only if the distance from (x, y) to (0, d) is the same as the distance from (x, y) to (x, d). That is, (x, y) is on C if and only if x 2 + (y d) 2 = y + d. Hence x 2 + (y d) 2 = (y + d) 2, and so x 2 + y 2 2dy + d 2 = y 2 + 2dy + d 2. It follows that (x, y) is on C if and only if y = 1 4d x 2. Dan Sloughter (Furman University) The Geometry November 6, 2006 14 / 18
Example The curve with equation y = x 2 is a parabola with focus at ( 0, 1 ) 4 and directrix y = 1 4. 2 1.5 1 0.5-2 -1 1 2 Dan Sloughter (Furman University) The Geometry November 6, 2006 15 / 18
Geometry to algebra With the introduction of rectangular coordinate systems by Descartes and Fermat (1601-1665), it is possible to describe curves algebraically. Dan Sloughter (Furman University) The Geometry November 6, 2006 16 / 18
Geometry to algebra With the introduction of rectangular coordinate systems by Descartes and Fermat (1601-1665), it is possible to describe curves algebraically. At first, only curves described geometrically are accepted as legitimate, but, with time, it is accepted that a curve may be defined first by an algebraic equation. Dan Sloughter (Furman University) The Geometry November 6, 2006 16 / 18
Problems 1. Use the geometric method of Al-Khowarizmi to find solutions for the following equations. a. x 2 + 12x = 108 b. x 2 + 8x = 20 2. Suppose p > 0 and q > 0. a. Using the geometric method of Al-Khowarizmi, show that one solution of x 2 + 2px = q is given by x = p 2 + q p. b. Use the quadratic formula to verify this result. 3. Plot the points (2, 1), ( 2, 3), ( 4, 1), and (5, 3) in the Cartesian plane. Dan Sloughter (Furman University) The Geometry November 6, 2006 17 / 18
Problems (cont d) 4. Draw the circles in the Cartesian plane with the following equations: a. x 2 + y 2 = 25 b. (x 1) 2 + (y 1) 2 = 1 c. (x + 2) 2 + (y 3) 2 = 4 5. Draw the lines in the Cartesian plane with the following equations: a. y = 3x b. y = x + 4 c. y = 5 x d. y = 2x 4 e. y = 4 f. x = 2 6. Identify the directrix and focus of each of the following parabolas. a. y = 4x 2 b. y = 1 4 x 2 c. y = 1 16 x 2 d. y = 12x 2 Dan Sloughter (Furman University) The Geometry November 6, 2006 18 / 18