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Precalculus Fifth Edition Mathematics for Calculus James Stewart  Lothar Redlin  Saleem Watson. Fundamentals. 1. Coordinate Geometry. 1.8. Coordinate Geometry. The coordinate plane is the link between algebra and geometry.

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Precalculus fifth edition mathematics for calculus james stewart lothar redlin saleem watson

Precalculus

Fifth Edition

Mathematics for Calculus

James StewartLothar RedlinSaleem Watson




Coordinate geometry
Coordinate Geometry

  • The coordinate plane is the link between algebra and geometry.

    • In the coordinate plane, we can draw graphs of algebraic equations.

    • In turn, the graphs allow us to “see” the relationship between the variables in the equation.



The coordinate plane
The Coordinate Plane

  • Points on a line can be identified with real numbers to form the coordinate line.

  • Similarly, points in a plane can be identified with ordered pairs of numbers to form the coordinate plane or Cartesian plane.


Axes

  • To do this, we draw two perpendicular real lines that intersect at 0 on each line.

  • Usually,

    • One line is horizontal with positive direction to the right and is called the x-axis.

    • The other line is vertical with positive direction upward and is called the y-axis.


Origin quadrants
Origin & Quadrants

  • The point of intersection of the x-axis and the y-axis is the origin O.

  • Thetwo axes divide the plane into four quadrants, labeled I, II, III, and IV here.


Origin quadrants1
Origin & Quadrants

  • The points onthe coordinate axes are not assigned to any quadrant.


Ordered pair
Ordered Pair

  • Any point P in the coordinate plane can be located by a unique ordered pairof numbers (a, b).

    • The first number a is called the x-coordinateof P.

    • The second number b is called the y-coordinateof P.


Coordinates
Coordinates

  • We can think of the coordinates of P as its “address.”

    • They specify its location in the plane.


Coordinates1
Coordinates

  • Several points are labeled with their coordinates in this figure.


E g 1 graphing regions in the coordinate plane
E.g. 1—Graphing Regions in the Coordinate Plane

  • Describe and sketch the regions given by each set.

  • {(x, y) | x≥ 0}

  • {(x, y) | y = 1}

  • {(x, y) | |y| < 1}


E g 1 regions in the coord plane

Example (a)

E.g. 1—Regions in the Coord. Plane

  • The points whose x-coordinates are 0 or positive lie on the y-axis or to the right of it.


E g 1 regions in the coord plane1

Example (b)

E.g. 1—Regions in the Coord. Plane

  • The set of all points with y-coordinate 1 is a horizontal line one unit above the x-axis.


E g 1 regions in the coord plane2

Example (c)

E.g. 1—Regions in the Coord. Plane

  • Recall from Section 1-7 that |y| < 1 if and only if –1 < y < 1

    • So, the given region consists of those points in the plane whose y-coordinates lie between –1 and 1.

    • Thus, the region consists of all points that lie between (but not on) the horizontal lines y = 1 and y = –1.


E g 1 regions in the coord plane3

Example (c)

E.g. 1—Regions in the Coord. Plane

  • These lines are shown as broken lines here to indicate that the points on these lines do not lie in the set.



The distance formula
The Distance Formula

  • We now find a formula for the distance d(A, B) between two points A(x1, y1) and B(x2, y2) in the plane.


The distance formula1
The Distance Formula

  • Recall from Section 1-1 that the distance between points a and b on a number line is: d(a, b) = |b – a|


The distance formula2
The Distance Formula

  • So, from the figure, we see that:

    • The distance between the points A(x1, y1) and C(x2, y1) on a horizontal line must be |x2 – x1|.

    • The distance between B(x2, y2) and C(x2, y1) on a vertical line must be |y2 – y1|.


The distance formula3
The Distance Formula

  • Triangle ABC is a right triangle.

  • So, the Pythagorean Theorem gives:


Distance formula
Distance Formula

  • The distance between the points A(x1, y1) and B(x2, y2) in the plane is:


E g 2 applying the distance formula
E.g. 2—Applying the Distance Formula

  • Which of the points P(1, –2) or Q(8, 9) is closer to the point A(5, 3)?

    • By the Distance Formula, we have:


E g 2 applying the distance formula1
E.g. 2—Applying the Distance Formula

  • This shows that d(P, A) < d(Q, A)

  • So, P is closer to A.


The midpoint formula
The Midpoint Formula

  • Now, let’s find the coordinates (x, y) of the midpoint M of the line segment that joins the point A(x1, y1) to the point B(x2, y2).


The midpoint formula1
The Midpoint Formula

  • In the figure, notice that triangles APM and MQB are congruent because:

    • d(A, M) = d(M, B)

    • The corresponding angles are equal.


The midpoint formula2
The Midpoint Formula

  • It follows that d(A, P) = d(M, Q).

  • So, x – x1 = x2 – x


The midpoint formula3
The Midpoint Formula

  • Solving that equation for x, we get: 2x = x1 + x2

  • Thus, x = (x1 + x2)/2

    • Similarly, y = (y1 + y2)/2


Midpoint formula
Midpoint Formula

  • The midpoint of the line segment from A(x1, y1) to B(x2, y2) is:


E g 3 applying the midpoint formula
E.g. 3—Applying the Midpoint Formula

  • Show that the quadrilateral with vertices P(1, 2), Q(4, 4), R(5, 9), and S(2, 7) is a parallelogram by proving that its two diagonals bisect each other.


E g 3 applying the midpoint formula1
E.g. 3—Applying the Midpoint Formula

  • If the two diagonals have the same midpoint, they must bisect each other.

    • The midpoint of the diagonal PR is:

    • The midpoint of the diagonal QS is:


E g 3 applying the midpoint formula2
E.g. 3—Applying the Midpoint Formula

  • Thus, each diagonal bisects the other.

    • A theorem from elementary geometry states that the quadrilateral is therefore a parallelogram.



Equation in two variables
Equation in Two Variables

  • An equation in two variables, such as y = x2 + 1, expresses a relationship between two quantities.


Graph of an equation in two variables
Graph of an Equation in Two Variables

  • A point (x, y) satisfiesthe equation if it makes the equation true when the values for x and y are substituted into the equation.

    • For example, the point (3, 10) satisfies the equation y = x2 + 1 because 10 = 32 + 1.

    • However, the point (1, 3) does not, because 3 ≠ 12 + 1.


The graph of an equation
The Graph of an Equation

  • The graphof an equation in x and y is:

    • The set of all points (x, y) in the coordinate plane that satisfy the equation.


The graph of an equation1
The Graph of an Equation

  • The graph of an equation is a curve.

  • So, to graph an equation, we:

    • Plot as many points as we can.

    • Connect them by a smooth curve.


E g 4 sketching a graph by plotting points
E.g. 4—Sketching a Graph by Plotting Points

  • Sketch the graph of the equation

  • 2x – y = 3

    • We first solve the given equation for y to get: y = 2x – 3


E g 4 sketching a graph by plotting points1
E.g. 4—Sketching a Graph by Plotting Points

  • This helps us calculate the y-coordinates in this table.


E g 4 sketching a graph by plotting points2
E.g. 4—Sketching a Graph by Plotting Points

  • Of course, there are infinitely many points on the graph—and it is impossible to plot all of them.

    • Still, the more points we plot, the better we can imagine what the graph represented by the equation looks like.


E g 4 sketching a graph by plotting points3
E.g. 4—Sketching a Graph by Plotting Points

  • We plot the points we found.

    • As they appear to lie on a line, we complete the graph by joining the points by a line.


E g 4 sketching a graph by plotting points4
E.g. 4—Sketching a Graph by Plotting Points

  • In Section 1-10, we verify that the graph of this equation is indeed a line.


E g 5 sketching a graph by plotting points
E.g. 5—Sketching a Graph by Plotting Points

  • Sketch the graph of the equation y = x2 – 2


E g 5 sketching a graph by plotting points1
E.g. 5—Sketching a Graph by Plotting Points

  • We find some of the points that satisfy the equation in this table.


E g 5 sketching a graph by plotting points2
E.g. 5—Sketching a Graph by Plotting Points

  • We plot these points and then connect them by a smooth curve.

    • A curve with this shape is called a parabola.


E g 6 graphing an absolute value equation
E.g. 6—Graphing an Absolute Value Equation

  • Sketch the graph of the equation y = |x|


E g 6 graphing an absolute value equation1
E.g. 6—Graphing an Absolute Value Equation

  • Again, we make a table of values.


E g 6 graphing an absolute value equation2
E.g. 6—Graphing an Absolute Value Equation

  • We plot these points and use them to sketch the graph of the equation.



X intercepts
x-intercepts

  • The x-coordinates of the points where a graph intersects the x-axis are called the x-interceptsof the graph.

    • They are obtained by setting y = 0 in the equation of the graph.


Y intercepts
y-intercepts

  • The y-coordinates of the points where a graph intersects the y-axis are called the y-interceptsof the graph.

    • They are obtained by setting x = 0 in the equation of the graph.


E g 7 finding intercepts
E.g. 7—Finding Intercepts

  • Find the x- and y-intercepts of the graph of the equation y = x2 – 2


E g 7 finding intercepts1
E.g. 7—Finding Intercepts

  • To find the x-intercepts, we set y = 0 and solve for x.

    • Thus, 0 = x2 – 2 x2 = 2 (Add 2 to each side) (Take the sq. root)

    • The x-intercepts are and .


E g 7 finding intercepts2
E.g. 7—Finding Intercepts

  • To find the y-intercepts, we set x = 0 and solve for y.

    • Thus, y = 02 – 2 y = –2

    • The y-intercept is –2.


E g 7 finding intercepts3
E.g. 7—Finding Intercepts

  • The graph of this equation was sketched in Example 5.

    • It is repeated here with the x- and y-intercepts labeled.



Circles
Circles

  • So far, we have discussed how to find the graph of an equation in x and y.

  • The converse problem is to find an equation of a graph—an equation that represents a given curve in the xy-plane.


Circles1
Circles

  • Such an equation is satisfied by the coordinates of the points on the curve and by no other point.

    • This is the other half of the fundamental principle of analytic geometry as formulated by Descartes and Fermat.


Circles2
Circles

  • The idea is that:

    • If a geometric curve can be represented by an algebraic equation, then the rules of algebra can be used to analyze the curve.


Circles3
Circles

  • As an example of this type of problem, let’s find the equation of a circle with radius r and center (h, k).


Circles4
Circles

  • By definition, the circle is the set of all points P(x, y) whose distance from the center C(h, k) is r.

    • Thus, P is on the circle if and only if d(P, C) = r


Circles5
Circles

  • From the distance formula, we have: (Square each side)

    • This is the desired equation.


Equation of a circle standard form
Equation of a Circle—Standard Form

  • An equation of the circle with center (h, k) and radius r is: (x – h)2 + (y – k)2 = r2

    • This is called the standard formfor the equation of the circle.


Equation of a circle
Equation of a Circle

  • If the center of the circle is the origin (0, 0), then the equation is:x2 + y2 = r2


E g 8 graphing a circle
E.g. 8—Graphing a Circle

  • Graph each equation.

  • x2 + y2 = 25

  • (x – 2)2 + (y + 1)2 = 25


E g 8 graphing a circle1

Example (a)

E.g. 8—Graphing a Circle

  • Rewriting the equation as x2 + y2 = 52, we see that that this is an equation of:

    • The circle of radius 5 centered at the origin.


E g 8 graphing a circle2

Example (b)

E.g. 8—Graphing a Circle

  • Rewriting the equation as (x – 2)2 + (y + 1)2 = 52, wesee that this is an equation of:

    • The circle of radius 5 centered at (2, –1).


E g 9 finding an equation of a circle
E.g. 9—Finding an Equation of a Circle

  • Find an equation of the circle with radius 3 and center (2, –5).

  • (b) Find an equation of the circle that has the points P(1, 8) and Q(5, –6) as the endpoints of a diameter.


E g 9 equation of a circle

Example (a)

E.g. 9—Equation of a Circle

  • Using the equation of a circle with r = 3, h = 2, and k = –5, we obtain: (x – 2)2 + (y + 5)2 = 9


E g 9 equation of a circle1

Example (b)

E.g. 9—Equation of a Circle

  • We first observe that the center is the midpoint of the diameter PQ.

    • So,by the Midpoint Formula, the center is:


E g 9 equation of a circle2

Example (b)

E.g. 9—Equation of a Circle

  • The radius r is the distance from P to the center.

    • So, by the Distance Formula,r2 = (3 – 1)2 + (1 – 8)2 = 22 + (–7)2 = 53


E g 9 equation of a circle3

Example (b)

E.g. 9—Equation of a Circle

  • Hence, the equation of the circle is: (x – 3)2 + (y – 1)2 = 53


Equation of a circle1
Equation of a Circle

  • Let’s expand the equation of the circle in the preceding example.

  • (x – 3)2 + (y – 1)2 = 53 (Standard form)

  • x2 – 6x + 9 + y2 – 2y + 1 = 53 (Expand the squares)

  • x2 – 6x +y2 – 2y = 43 (Subtract 10 to get the expanded form)


Equation of a circle2
Equation of a Circle

  • Suppose we are given the equation of a circle in expanded form.

    • Then, to find its center and radius, we must put the equation back in standard form.


Equation of a circle3
Equation of a Circle

  • That means we must reverse the steps in the preceding calculation.

    • To do that, we need to know what to add to an expression like x2 – 6x to make it a perfect square.

    • That is, we need to complete the square—as in the next example.


E g 10 identifying an equation of a circle
E.g. 10—Identifying an Equation of a Circle

  • Show that the equation x2 + y2 + 2x – 6y + 7 = 0 represents a circle.

  • Find the center and radius of the circle.


E g 10 identifying an equation of a circle1
E.g. 10—Identifying an Equation of a Circle

  • First, we group the x-terms and y-terms.

  • Then, we complete the square within each grouping.

    • We complete the square for x2 + 2x by adding (½ ∙ 2)2 = 1.

    • We complete the square for y2 – 6y by adding [½ ∙ (–6)]2 = 9.



E g 10 identifying an equation of a circle3
E.g. 10—Identifying an Equation of a Circle

  • Comparing this equation with the standard equation of a circle, we see that: h = –1, k = 3, r =

    • So, the given equation represents a circle with center (–1, 3) and radius .



Symmetry
Symmetry

  • The figure shows the graph of y = x2

    • Notice that the part of the graph to the left of the y-axis is the mirror image of the part to the right of the y-axis.


Symmetry1
Symmetry

  • The reason is that, if the point (x, y) is on the graph, then so is (–x, y), and these points are reflections of each other about the y-axis.


Symmetric with respect to y axis
Symmetric with Respect to y-axis

  • In this situation, we say the graph is symmetric with respect to the y-axis.


Symmetric with respect to x axis
Symmetric with Respect to x-axis

  • Similarly, we say a graph is symmetric with respect to the x-axis if, whenever the point (x, y) is on the graph, then so is (x, –y).


Symmetric with respect to origin
Symmetric with Respect to Origin

  • A graph is symmetric with respect to the originif, whenever (x, y) is on the graph, so is (–x, –y).


Using symmetry to sketch a graph
Using Symmetry to Sketch a Graph

  • The remaining examples in this section show how symmetry helps us sketch the graphs of equations.


E g 11 using symmetry to sketch a graph
E.g. 11—Using Symmetry to Sketch a Graph

  • Test the equation x = y2for symmetry and sketch the graph.


E g 11 using symmetry to sketch a graph1
E.g. 11—Using Symmetry to Sketch a Graph

  • If y is replaced by –y in the equation x = y2, we get: x = (–y)2(Replace y by –y)x = y2(Simplify)

    • So, the equation is unchanged.

    • Thus, the graph is symmetric about the x-axis.


E g 11 using symmetry to sketch a graph2
E.g. 11—Using Symmetry to Sketch a Graph

  • However, changing x to –x gives the equation –x = y2

    • This is not the same as the original equation.

    • So, the graph is not symmetric about the y-axis.


E g 11 using symmetry to sketch a graph3
E.g. 11—Using Symmetry to Sketch a Graph

  • We use the symmetry about the x-axis to sketch the graph.

  • First, we plot points just for y > 0.


E g 11 using symmetry to sketch a graph4
E.g. 11—Using Symmetry to Sketch a Graph

  • Then, we reflect the graph in the x-axis.


E g 12 using symmetry to sketch a graph
E.g. 12—Using Symmetry to Sketch a Graph

  • Test the equation y = x3 – 9xfor symmetry and sketch its graph.


E g 12 using symmetry to sketch a graph1
E.g. 12—Using Symmetry to Sketch a Graph

  • If we replace x by –x and y by –y, we get: –y = (–x3) – 9(–x) –y = –x3 + 9x(Simplify)y = x3 – 9x(Multiply by –1)

    • So, the equation is unchanged.

    • This means that the graph is symmetric with respect to the origin.


E g 12 using symmetry to sketch a graph2
E.g. 12—Using Symmetry to Sketch a Graph

  • First, we plot points for x > 0.


E g 12 using symmetry to sketch a graph3
E.g. 12—Using Symmetry to Sketch a Graph

  • Then, we use symmetry about the origin.


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