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. 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

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 The point of intersection of the x-axis and the y-axis is the origin O. The two axes divide the plane into four quadrants, labeled I, II, III, and IV here.

Origin & Quadrants The points on the coordinate axes are not assigned to any quadrant.

Ordered Pair Any point P in the coordinate plane can be located by a unique ordered pair of numbers (a, b). The first number a is called the x-coordinate of P. The second number b is called the y-coordinate of P.

Coordinates We can think of the coordinates of P as its “address.” They specify its location in the plane.

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

E.g. 1—Graphing Regions in the Coordinate Plane Describe and sketch the regions given by each set. (a) {(x, y) | x ≥ 0} (b) {(x, y) | y = 1} (c) {(x, y) | |y| < 1}

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. Example (a)

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. Example (b)

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. 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. Example (c)

The Distance and Midpoint Formulas

The Distance Formula We now find a formula for the distance d(A, B) between two points A(x 1, y 1 ) and B(x 2, y 2 ) in the plane.

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 Formula So, from the figure, we see that: The distance between the points A(x 1, y 1 ) and C(x 2, y 1 ) on a horizontal line must be |x 2 – x 1 |. The distance between B(x 2, y 2 ) and C(x 2, y 1 ) on a vertical line must be |y 2 – y 1 |.

The Distance Formula Triangle ABC is a right triangle. So, the Pythagorean Theorem gives:

Distance Formula The distance between the points A(x 1, y 1 ) and B(x 2, y 2 ) in the plane is:

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 Formula This shows that d(P, A) < d(Q, A) So, P is closer to A.

The Midpoint Formula Now, let’s find the coordinates (x, y) of the midpoint M of the line segment that joins the point A(x 1, y 1 ) to the point B(x 2, y 2 ).

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 Formula It follows that d(A, P) = d(M, Q). So, x – x 1 = x 2 – x

The Midpoint Formula Solving that equation for x, we get: 2x = x 1 + x 2 Thus, x = (x 1 + x 2 )/2 Similarly, y = (y 1 + y 2 )/2

Midpoint Formula The midpoint of the line segment from A(x 1, y 1 ) to B(x 2, y 2 ) is:

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 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 Formula Thus, each diagonal bisects the other. A theorem from elementary geometry states that the quadrilateral is therefore a parallelogram.

Graphs of Equations in Two Variables

Equation in Two Variables An equation in two variables, such as y = x 2 + 1, expresses a relationship between two quantities.

Graph of an Equation in Two Variables A point (x, y) satisfies the 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 = x because 10 = However, the point (1, 3) does not, because 3 ≠

The Graph of an Equation The graph of 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 Equation The graph of an equation is a curve. So, to graph an equation, we: 1.Plot as many points as we can. 2.Connect them by a smooth curve.

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

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 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 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 Sketch the graph of the equation y = x 2 – 2

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 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 Sketch the graph of the equation y = |x|

E.g. 6—Graphing an Absolute Value Equation Again, we make a table of values.

E.g. 6—Graphing an Absolute Value Equation We plot these points and use them to sketch the graph of the equation.

Intercepts

x-intercepts The x-coordinates of the points where a graph intersects the x-axis are called the x-intercepts of the graph. They are obtained by setting y = 0 in the equation of the graph.

y-intercepts The y-coordinates of the points where a graph intersects the y-axis are called the y-intercepts of the graph. They are obtained by setting x = 0 in the equation of the graph.

E.g. 7—Finding Intercepts Find the x- and y-intercepts of the graph of the equation y = x 2 – 2

E.g. 7—Finding Intercepts To find the x-intercepts, we set y = 0 and solve for x. Thus, 0 = x 2 – 2 x 2 = 2 (Add 2 to each side) (Take the sq. root) The x-intercepts are and.

E.g. 7—Finding Intercepts To find the y-intercepts, we set x = 0 and solve for y. Thus, y = 0 2 – 2 y = –2 The y-intercept is –2.

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

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.

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.

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.

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

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

Circles From the distance formula, we have: (Square each side) This is the desired equation.

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 = r 2 This is called the standard form for the equation of the circle.

Equation of a Circle If the center of the circle is the origin (0, 0), then the equation is: x 2 + y 2 = r 2

E.g. 8—Graphing a Circle Graph each equation. (a) x 2 + y 2 = 25 (b) (x – 2) 2 + (y + 1) 2 = 25

E.g. 8—Graphing a Circle Rewriting the equation as x 2 + y 2 = 5 2, we see that that this is an equation of: The circle of radius 5 centered at the origin. Example (a)

E.g. 8—Graphing a Circle Rewriting the equation as (x – 2) 2 + (y + 1) 2 = 5 2, we see that this is an equation of: The circle of radius 5 centered at (2, –1). Example (b)

E.g. 9—Finding an Equation of a Circle (a)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 Using the equation of a circle with r = 3, h = 2, and k = –5, we obtain: (x – 2) 2 + (y + 5) 2 = 9 Example (a)

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: 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, r 2 = (3 – 1) 2 + (1 – 8) 2 = (–7) 2 = 53 Example (b)

E.g. 9—Equation of a Circle Hence, the equation of the circle is: (x – 3) 2 + (y – 1) 2 = 53 Example (b)

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) x 2 – 6x y 2 – 2y + 1 = 53 (Expand the squares) x 2 – 6x +y 2 – 2y = 43 (Subtract 10 to get the expanded form)

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 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 x 2 – 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 Show that the equation x 2 + y 2 + 2x – 6y + 7 = 0 represents a circle. Find the center and radius of the circle.

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 x 2 + 2x by adding (½ ∙ 2) 2 = 1. We complete the square for y 2 – 6y by adding [½ ∙ (–6)] 2 = 9.

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

The figure shows the graph of y = x 2 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.

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 In this situation, we say the graph is symmetric with respect to the y-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 A graph is symmetric with respect to the origin if, whenever (x, y) is on the graph, so is (–x, –y).

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 Test the equation x = y 2 for symmetry and sketch the graph.

E.g. 11—Using Symmetry to Sketch a Graph If y is replaced by –y in the equation x = y 2, we get: x = (–y) 2 (Replace y by –y) x = y 2 (Simplify) So, the equation is unchanged. Thus, the graph is symmetric about the x-axis.

E.g. 11—Using Symmetry to Sketch a Graph However, changing x to –x gives the equation –x = y 2 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 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 Graph Then, we reflect the graph in the x-axis.

E.g. 12—Using Symmetry to Sketch a Graph Test the equation y = x 3 – 9x for symmetry and sketch its graph.

E.g. 12—Using Symmetry to Sketch a Graph If we replace x by –x and y by –y, we get: –y = (–x 3 ) – 9(–x) –y = –x 3 + 9x (Simplify) y = x 3 – 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 Graph First, we plot points for x > 0.

E.g. 12—Using Symmetry to Sketch a Graph Then, we use symmetry about the origin.