Analytic Geometry

Cartesian Coordinate System

A Cartesian coordinate system consists of two perpendicular real number lines, called coordinate axes, that intersect at their origins. Generally one line is horizontal and called the x-axis, and the other is vertical and called the y-axis. The axes divide the coordinate plane, or xy-plane, into four parts, called quadrants and are numbered first, second, third, and fourth, or I, II, III, and IV. Points on the axes are not in any quadrant.

One-to-One Correspondence

A one-to-one correspondence exists between ordered pairs of numbers $ (a,b) $ and points in the coordinate plane (Fig. 8-1). Thus,

  1. To each point P there corresponds an ordered pair of numbers $ (a,b) $ called the coordinates of P. a is called the x-coordinate or abscissa; b is called the y-coordinate or ordinate.

  2. To each ordered pair of numbers there corresponds a point, called the graph of the ordered pair. The graph can be indicated by a dot.

Figure 8-1

Distance between Two Points

The distance between two points $ P_{1}(x_{1},y_{1}) $ and $ P_{2}(x_{2},y_{2}) $ in a Cartesian coordinate system is given by the distance formula:

\[ d(P_{1},P_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \]

EXAMPLE 8.1 Find the distance between $ (-3,5) $ and $ (4,-1) $ .

Label $ P_{1}(x_{1},y_{1})=(-3,5) $ and $ P_{2}(x_{2},y_{2})=(4,-1) $ . Then substitute into the distance formula.

\[ \begin{aligned}d(P_{1},P_{2})&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\&=\sqrt{[4-(-3)]^{2}+[(-1)-5]^{2}}\\&=\sqrt{7^{2}+(-6)^{2}}=\sqrt{85}\\ \end{aligned} \]

Graph of an Equation

The graph of an equation in two variables is the graph of its solution set, that is, of all ordered pairs $ (a,b) $ that satisfy the equation. Since there are ordinarily an infinite number of solutions, a sketch of the graph is generally sufficient. A simple approach to finding a sketch of a graph is to find several solutions, plot them, then connect the dots with a smooth curve or line.

EXAMPLE 8.2 Sketch the graph of the equation x - 2y = 10.

Form a table of values; then plot the points and connect them. The graph is a straight line, as shown in Fig. 8-2.

x -2 0 2 4 6 8 10
y -6 -5 -4 -3 -2 -1 0

Figure 8-2

I ntercepts

The coordinates of the points where the graph of an equation crosses the x-axis and y-axis have special names:

  1. The x-coordinate of a point where the graph crosses the x-axis is called the x-intercept of the graph. To find it, set y = 0 and solve for x.

  2. The y-coordinate of a point where the graph crosses the y-axis is called the y-intercept of the graph. To find it, set x = 0 and solve for y.

EXAMPLE 8.3 In the previous example, the x-intercept of the graph is 10, since the graph crosses the x-axis at $ (10,0) $ ; and the y-intercept is -5, since the graph crosses the y-axis at $ (0,-5) $ .

EXAMPLE 8.4 Find the intercepts of the graph of the equation $ y = 4 - x^{2} $ . Set x = 0; then $ y = 4 - 0^{2} = 4 $ . Hence the y-intercept is 4.

Set y = 0. If $ 0 = 4 - x^{2} $ , then $ x^{2} = 4 $ ; thus $ x = $ . Hence 2 and -2 are the x-intercepts.

Symmetry

Symmetry is an important aid to graphing more complicated equations: A graph is

  1. Symmetric with respect to the y-axis if $ (-a,b) $ is on the graph whenever $ (a,b) $ is on the graph. (y-axis symmetry)

  2. Symmetric with respect to the x-axis if $ (a,-b) $ is on the graph whenever $ (a,b) $ is on the graph. $ (x $ -axis symmetry)

  3. Symmetric with respect to the origin if $ (-a,-b) $ is on the graph whenever $ (a,b) $ is on the graph. (origin symmetry)

  4. Symmetric with respect to the line $ y = x (b, a) (a, b) $

Tests for Symmetry

Tests for symmetry (Fig. 8-3):

  1. If substituting -x for x leads to the same equation, the graph has symmetry with respect to the y-axis.

  2. If substituting -y for y leads to the same equation, the graph has symmetry with respect to the x-axis.

  3. If simultaneously substituting -x for x and -y for y leads to the same equation, the graph has symmetry with respect to the origin.

Terminology Test Illustration
The graph is symmetric with respect to the y-axis The equation is unchanged when x is replaced by -x
The graph is symmetric with respect to the x-axis The equation is unchanged when y is replaced by -y
The graph is symmetric with respect to the origin The equation is unchanged when x is replaced by -x and y is replaced by -y
The graph is symmetric with respect to the line y = x The equation is unchanged when x and y are interchanged

Note: a graph may have none of these three symmetries, one, or all three. It is not possible for a graph to have exactly two of these three symmetries.

The fourth symmetry is less commonly tested:

  1. If interchanging the letters x and y leads to the same equation, the graph has symmetry with respect to the line y = x.

EXAMPLE 8.5 Test the equation $ y = 4 - x^{2} $ for symmetry and draw the graph.

Substitute -x for x: $ y = 4 - (-x)^{2} = 4 - x^{2} $ . Since the equation is unchanged, the graph has y-axis symmetry (see Fig. 8-4).

Substitute -y for y: $ -y = 4 - x^{2} $ ; $ y = -4 + x^{2} $ . Since the equation is changed, the graph does not have x-axis symmetry. It is not possible for the graph to have origin symmetry; see the previous note. Since the graph has y-axis symmetry, it is only necessary to find points with nonnegative values of x, and then reflect the graph through the y-axis.

x 0 1 2 3 4
y 4 3 0 -5 -12

Figure 8-4

Circle

A circle with center $ C(h,k) $ and radius r > 0 is the set of all points in the plane that are r units from C (Fig. 8-5).

Figure 8-5

Equation of a Circle

The equation of a circle with center $ C(h,k) $ and radius r > 0 can be written as (standard form)

\[ (x-h)^{2}+(y-k)^{2}=r^{2} \]

If the center of the circle is the origin $ (0,0) $ , this reduces to

\[ x^{2}+y^{2}=r^{2} \]

If r = 1 the circle is called a unit circle.

Midpoint of a Line Segment

The midpoint of a line segment with endpoints $ P_{1}(x_{1},y_{1}) $ and $ P_{2}(x_{2},y_{2}) $ is given by the midpoint formula:

Midpoint of $ P_{1}P_{2}=(,) $

SOLVED PROBLEMS

8.1. Prove the distance formula

In Fig. 8-6, $ P_{1} $ and $ P_{2} $ are shown. Introduce $ Q(x_{2}, y_{1}) $ as shown. Then the distance between $ P_{1} $ and Q is the difference in their x-coordinates, $ |x_{2} - x_{1}| $ ; similarly, the distance between Q and $ P_{2} $ is the difference in their y-coordinates, $ |y_{2} - y_{1}| $ . In the right triangle $ P_{1}P_{2}Q $ , apply the Pythagorean theorem: $ d^{2} = |x_{2} - x_{1}|^{2} + |y_{2} - y_{1}|^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} $ , since $ |a|^{2} = a^{2} $ by the properties of absolute values. Hence, taking the square root and noting that d, the distance, is always positive, $ d(P_{1}, P_{2}) = $ .

Figure 8-6

8.2. Find the distance $ d(P_{1}, P_{2}) $ given

  1. $ P_{1}(-5,-4) $ , $ P_{2}(-8,0) $ ; (b) $ P_{1}(2,2) $ , $ P_{2}(0,5) $ ; (c) $ P_{1}(x,x^{2}) $ , $ P_{2}(x+h,(x+h)^{2}) $

  2. Substitute $ x_{1} = -5 $ , $ y_{1} = -4 $ , $ x_{2} = -8 $ , $ y_{2} = 0 $ into the distance formula:

\[ \begin{aligned}d&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\&=\sqrt{[(-8)-(-5)]^{2}+[0-(-4)]^{2}}\\&=\sqrt{9+16}=\sqrt{25}=5\end{aligned} \]

  1. Substitute $ x_{1}=2 $ , $ y_{1}=2 $ , $ x_{2}=0 $ , $ y_{2}=5 $ into the distance formula:

\[ \begin{aligned}d&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\&=\sqrt{(0-2\sqrt{2})^{2}+(5\sqrt{2}-2\sqrt{2})^{2}}\\&=\sqrt{(-2\sqrt{2})^{2}+(3\sqrt{2})^{2}}\\&=\sqrt{8+18}=\sqrt{26}\end{aligned} \]

  1. Substitute $ x_{1}=x $ , $ y_{1}=x^{2} $ , $ x_{2}=x+h $ , $ y_{2}=(x+h)^{2} $ into the distance formula and simplify.

\[ \begin{aligned}d&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\&=\sqrt{(x+h-x)^{2}+[(x+h)^{2}-x^{2}]^{2}}\\&=\sqrt{h^{2}+(2xh+h^{2})^{2}}\\&=\sqrt{h^{2}+4x^{2}h^{2}+4xh^{3}+h^{4}}\\ \end{aligned} \]

8.3. Analyze intercepts and symmetry, then sketch the graph:

  1. y = 12 - 4x; (b) $ y = x^{2} + 3 $ ; (c) $ y^{2} + x = 5 $ ; (d) $ 2y = x^{3} $ .

  2. Set x = 0, then $ y = 12 - 4 = 12 $ . Hence 12 is the y-intercept.

Set y = 0, then 0 = 12 - 4x; thus x = 3. Hence 3 is the x-intercept.

Substitute -x for x: $ y = 12 - 4(-x) $ ; $ y = 12 + 4x $ . Since the equation is changed, the graph (see Fig. 8-7) does not have y-axis symmetry.

Substitute -y for y: -y = 12 - 4x; $ y = -12 + 4x $ . Since the equation is changed, the graph does not have x-axis symmetry.

Substitute -x for x and -y for y: $ -y = 12 - 4(-x) $ ; y = -12 - 4x. Since the equation is changed, the graph does not have origin symmetry.

Form a table of values; then plot the points and connect them. The graph is a straight line.

x -1 0 1 2 3 4 5
y 16 12 8 4 0 -4 -8

Figure 8-7
  1. Set x = 0, then $ y = 0^{2} + 3 = 3 $ . Hence 3 is the y-intercept.

Set y = 0, then $ 0 = x^{2} + 3 $ . This has no real solution; hence there is no x-intercept.

Substitute -x for x: $ y = (-x)^{2} + 3 = x^{2} + 3 $ . Since the equation is unchanged, the graph (Fig. 8-8) has y-axis symmetry.

Substitute -y for y: $ -y = x^{2} + 3 $ ; $ y = -x^{2} - 3 $ . Since the equation is changed, the graph does not have x-axis symmetry.

It is not possible for the graph to have origin symmetry. Since the graph has y-axis symmetry, it is only necessary to find points with nonnegative values of x, and then reflect the graph through the y-axis.

x 0 1 2 3 4
y 3 4 7 12 19

Figure 8-8
  1. Set x = 0, then $ y^{2} + 0 = 5 $ ; thus $ y = $ . Hence $ $ are the y-intercepts.

Set y = 0, then x = 5; hence 5 is the x-intercept.

Substitute -x for x: $ y^{2} - x = 5 $ . Since the equation is changed, the graph (see Fig. 8-9) does not have y-axis symmetry.

Substitute -y for y: $ (-y)^{2} + x = 5 $ ; $ y^{2} + x = 5 $ . Since the equation is unchanged, the graph has x-axis symmetry.

It is not possible for the graph to have origin symmetry. Since the graph has x-axis symmetry, it is only necessary to find points with nonnegative values of y, and then reflect the graph through the x-axis.

x 5 4 1 -4 -11
y 0 1 2 3 4
Figure 8-9

  1. Set x = 0, then $ 2y = 0^{3} $ ; thus y = 0. Hence 0 is the y-intercept.

Set y = 0, then $ 2 = x^{3} $ ; thus x = 0. Hence 0 is the x-intercept.

Substitute -x for x: $ 2y = (-x)^{3} $ ; $ 2y = -x^{3} $ . Since the equation is changed, the graph (Fig. 8-10) does not have y-axis symmetry.

Substitute -y for y: $ 2(-y) = x^{3} $ ; 2y = -x^{3}. Since the equation is changed, the graph does not have x-axis symmetry.

Substitute -x for x and -y for y: $ -2y = (-x)^{3} $ ; $ 2y = x^{3} $ . Since the equation is unchanged, the graph has origin symmetry.

From a table of values for positive x, plot the points and connect them, then reflect the graph through the origin.

x 0 1 2 3 4
y 0 1/2 4 27/2 32

Figure 8-10

8.4. Analyze intercepts and symmetry, then sketch the graph:

  1. $ y = |x| - 4 $ ; (b) $ 4x^{2} + y^{2} = 36 $ ; (c) $ |x| + |y| = 3 $ ; (d) $ x^{2}y = 12 $ .

  2. Proceeding as in the previous problem, the x-intercepts are $ $ and the y-intercept is -4. The graph has y-axis symmetry. Form a table of values for positive x, plot the points and connect them, then reflect the graph (Fig. 8-11) through the y-axis.

x 0 1 2 3 4 5 6
y -4 -3 -2 -1 0 1 2

Figure 8.11

The graph has x-axis, y-axis, and origin symmetry. Form a table of values for positive x and y, plot the points and connect them, then reflect the graph (Fig. 8-12), first through the y-axis, then through the x-axis.

x 0 1 2 3
y 6 $ \(</td><td>\) $ 0
Figure 8-12

  1. The x-intercepts are ±3 and the y-intercepts are ±3.

The graph has x-axis, y-axis, and origin symmetry. Form a table of values for positive x and y, plot the points and connect them, then reflect the graph (Fig. 8-13), first through the y-axis, then through the x-axis.

x 0 1 2 3
y 3 2 1 0
Figure 8-13

  1. There are no x- or y-intercepts.

The graph has y-axis symmetry. Form a table of values for positive x, plot the points and connect them, then reflect the graph (Fig. 8-14) through the y-axis.

x 0 1 2 3 4
y undefined 12 3 4/3 3/4

Figure 8-14

8.5. Find the center and radius for the circles with the following equations:

  1. $ x^{2} + y^{2} = 9 $ ; (b) $ (x - 3)^{2} + (y + 2)^{2} = 25 $ ; (c) $ (x + 5)^{2} + (y + )^{2} = 21 $

  2. Comparing the given equation with the form $ x^{2} + y^{2} = r^{2} $ , the center is at the origin. Since $ r^{2} = 9 $ , the radius is $ = 3 $ .

  3. Comparing the given equation with the form $ (x - h)^{2} + (y - k)^{2} = r^{2} $ , h = 3 and k = 2; hence the center is at $ (h, k) = (3, -2) $ . Since $ r^{2} = 25 $ , the radius is $ = 5 $ .

  4. Comparing the given equation with the form $ (x-h){2}+(y-k){2}=r^{2}, -h=5 $ and $ -k= $ ; hence the center is at $ (h,k)=(-5,-) $ . Since $ r^{2}=21 $ , the radius is $ $ .

8.6. Find the equations of the following circles: (a) center at origin, radius 7; (b) center at (2,-3), radius √14;

  1. center at $ (-5,0) $ , radius $ 5 $ .

  2. Substitute r = 7 into $ x^{2} + y^{2} = r^{2} $ . The equation is $ x^{2} + y^{2} = 49 $ .

  3. Substitute h = 2, k = -3, $ r = $ into $ (x - h)^{2} + (y - k)^{2} = r^{2} $ .

The equation is $ (x-2){2}+[y-(-3)]{2}=()^{2} $ or $ (x-2){2}+(y+3){2}=14 $ .

  1. Substitute $ h = -5 $ , k = 0, $ r = 5 $ into $ (x - h)^{2} + (y - k)^{2} = r^{2} $ .

The equation is $ [x - (-5)]^{2} + (y - 0)^{2} = (5)^{2} $ or $ (x + 5)^{2} + y^{2} = 50 $ .

8.7. Find the center and radius of the circle with equation $ x^{2} + y^{2} - 4x - 12y = 9 $

Complete the square on x and y.

\[ \begin{aligned}x^{2}-4x+y^{2}-12y&=9\quad&\left[\frac{1}{2}(-4)\right]^{2}&=4;\quad\left[\frac{1}{2}(-12)\right]^{2}&=36\\ x^{2}-4x+4+y^{2}-12y+36&=4+36+9\quad&Add4+36to both sides\\ (x-2)^{2}+(y-6)^{2}&=49\end{aligned} \]

Comparing this equation with the form $ (x - h)^{2} + (y - k)^{2} = r^{2} $ , the center is at $ (h, k) = (2, 6) $ and the radius is 7.

8.8. Prove the midpoint formula

In Fig. 8-15, $ P_{1}(x_{1},y_{1}) $ and $ P_{2}(x_{2},y_{2}) $ are given. Let $ (x,y) $ be the unknown coordinates of the midpoint M. Project the points $ M, P_{1}, P_{2} $ to the x-axis as shown.

Figure 8-15

From plane geometry it is known that the projected segments are in the same ratio as the original segments. Hence the distance from $ x_{1} $ to x is the same as the distance from x to $ x_{2} $ . Thus, $ x_{2} - x = x - x_{1} $ . Solving for x yields

\[ -2x=-x_{1}-x_{2} \]

\[ x=\frac{x_{1}+x_{2}}{2} \]

Similarly, it can be shown by projecting onto the y-axis that $ y = $ .

8.9. Find the midpoint M of the segment $ P_{1}P_{2} $ given $ P_{1}(3,-8) $ , $ P_{2}(-6,6) $ .

Substitute $ x_{1}=3 $ , $ y_{1}=-8 $ , $ x_{2}=-6 $ , $ y_{2}=6 $ into the midpoint formula. Then

\[ \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)=\left(\frac{3+(-6)}{2},\frac{(-8)+6}{2}\right)=\left(-\frac{3}{2},-1\right) \]

are the coordinates of M.

8.10. Find the equation of a circle given that $ (0,6) $ and $ (8,-8) $ are the endpoints of a diameter.

Step 1. The center is the midpoint of the diameter. Find the coordinates of the center from the midpoint formula.

\[ \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)=\left(\frac{0+8}{2},\frac{6+(-8)}{2}\right)=(4,-1) \]

Step 2. The radius is the distance from the center to either of the given endpoints. Find the radius from the distance formula.

\[ \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{(4-0)^{2}+[(-1)-6]^{2}}=\sqrt{16+49}=\sqrt{65} \]

Step 3. Substitute the calculated radius and coordinates of the center into the standard form for the equation of a circle. $ r = , (h, k) = (4, -1) $ .

\[ (x+h)^{2}+(y-k)^{2}=r^{2} \]

\[ (x-4)^{2}+[y-(-1)]^{2}=(\sqrt{65})^{2} \]

\[ (x-4)^{2}+(y+1)^{2}=65 \]

8.11. Show that the triangle with vertices $ A(1,3) $ , $ B(-1,2) $ , $ C(5, -5) $ is a right triangle.

Step 1. First find the lengths of the sides from the distance formula

\[ d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{[(-1)-1]^{2}+(2-3)^{2}}=\sqrt{5}=c \]

\[ d(B,C)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{[5-(-1)]^{2}+[(-5)-2]^{2}}=\sqrt{85}=a \]

\[ d(A,C)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{(5-1)^{2}+[(-5)-3]^{2}}=\sqrt{80}=b \]

Step 2. Apply the converse of the Pythagorean theorem.

Since $ a{2}=(){2}=85 $ and $ b{2}+c{2}=(){2}+(){2}=80+5=85 $ , the relation $ a{2}=b{2}+c^{2} $ is satisfied; hence the triangle is a right triangle.

8.12. Show that $ P(-12,11) $ lies on the perpendicular bisector of the line segment joining $ A(0,-3) $ and $ B(6,15) $ .

The perpendicular bisector of a segment consists of all points that are equidistant from its endpoints. Thus if PA = PB, then P lies on the perpendicular bisector of AB. From the distance formula,

\[ d(A,P)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{[(-12)-0]^{2}+[11-(-3)]^{2}}=\sqrt{340}=PA \]

\[ d(P,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{[6-(-12)]^{2}+(15-11)^{2}}=\sqrt{340}=PB \]

Hence PA = PB and P lies on the perpendicular bisector of AB.

8.13. Find an equation for the perpendicular bisector of the line segment joining $ A(7, -8) $ and $ B(-2, 5) $ .

The perpendicular bisector of a segment consists of all points that are equidistant from its endpoints. Thus if PA = PB, then P lies on the perpendicular bisector of AB. Let P have the unknown coordinates $ (x, y) $ . Then, from the distance formula, PA = PB if

\[ PA=\sqrt{(x-7)^{2}+[y-(-8)]^{2}}=\sqrt{[x-(-2)]^{2}+(y-5)^{2}}=PB \]

Squaring both sides and simplifying yields

\[ (x-7)^{2}+[y-(-8)]^{2}=[x-(-2)]^{2}+(y-5)^{2} \]

\[ x^{2}-14x+49+y^{2}+16y+64=x^{2}+4x+4+y^{2}-10y+25 \]

\[ 18x-26y=84 \]

\[ 9x-13y=42 \]

This is the equation satisfied by all points equidistant from A and B. Hence, it is the equation of the perpendicular bisector of AB.

SUPPLEMENTARY PROBLEMS

8.14. Describe the set of points that satisfy the relations: (a) x = 0; (b) x > 0; (c) xy < 0; (d) y > 1.

Ans. (a) All points on the y-axis; (b) all points to the right of the y-axis;

  1. all points in the second and fourth quadrants; (d) all points above the line y = 1.

8.15. Find the distance between the following pairs of points: (a) $ (0, -7) $ and $ (7, 0) $ ; (b) $ (-3, -3) $ and $ (3, 3) $ , Ans. (a) $ 7 $ ; (b) 12

8.16. Find the length and the midpoint of the line segment with the given endpoints:

\[ \begin{array}{l}(a)A(1,8),B(-3,4);(b)A(3,-7),B(0,8);(c)A(1,\sqrt{2}),B(-1,5\sqrt{2})\end{array} \]

Ans. (a) length $ 4 $ , midpoint $ (-1,6) $ ; (b) length $ 3 $ , midpoint $ (,) $ ; (c) length 6, midpoint $ (0,3) $

8.17. Analyze the following for symmetry. Do not sketch graphs:

  1. $ xy^{2}=4 $ ; (b) $ x^{3}y=4 $ ; (c) $ |xy|=4 $ ; (d) $ x^{2}+xy=4 $ ;

\[ \left(\mathbf{e}\right)x^{2}+y+y^{2}=4;\left(\mathbf{f}\right)x^{2}+xy+y^{2}=4 \]

Ans. (a) x-axis symmetry; (b) origin symmetry; (c) x-axis, y-axis, origin symmetry;

  1. origin symmetry; (e) y-axis symmetry; (f) origin symmetry

8.18. Analyze symmetry and intercepts, then sketch graphs of the following:

  1. $ 3x + 4y + 12 = 0 $ (b) $ y^{2} = 10 + x $

  2. $ y{2}-x{2}=9 $ (d) $ |y|-|x|=3 $

Ans. (a) Fig. 8-16: x-intercept -4, y-intercept -3, no symmetry

Figure 8-16
  1. Fig. 8-17: x-intercept -10, y-intercepts ±√10, x-axis symmetry

Figure 8-17
  1. Fig. 8-18: no x-intercept, y-intercept $ $ , x-axis, y-axis, origin symmetry

Figure 8-18
  1. Fig. 8-19: no x-intercept, y-intercepts $ $ , x-axis, y-axis, origin symmetry

Figure 8-19

8.19. Analyze symmetry and intercepts, then sketch graphs of the following:

  1. $ x + y = 0 $ ; (b) $ y + |x| = 4 $ ; (c) $ x^{2} = 4|y| $ ;

  2. $ |y| = |4 - x^{2}| $ ; (e) $ |x| = 4y^{2} $ ; (f) $ -xy^{2} = 4 $

Ans. (a) Fig. 8-20: x-intercepts 0, y-intercept 0, origin symmetry

Figure 8-20

  1. Fig. 8-21: x-intercepts $ $ , y-intercept 4, y-axis symmetry

  2. Fig. 8-22: x-intercept 0, y-intercept 0, x-axis, y-axis, origin symmetry

  3. Fig. 8-23: x-intercepts ±2, y-intercepts ±4, x-axis, y-axis, origin symmetry

  4. Fig. 8-24: x-intercept 0, y-intercept 0, x-axis, y-axis, origin symmetry

Figure 8-21

Figure 8-22

Figure 8-23

Figure 8-24
  1. Fig. 8-25: no intercepts, x-axis symmetry

Figure 8-25

8.20. Find the equations of the following circles: (a) center (5, -2), radius $ $ ; (b) center $ (, -) $ , diameter 3;

  1. center $ (3,8) $ , passing through the origin; (d) center $ (-3,-4) $ , tangent to the y-axis.

Ans. (a) $ (x - 5)^{2} + (y + 2)^{2} = $ ; (b) $ (x - )^{2} + (y + )^{2} = $ ;

  1. $ (x - 3)^{2} + (y - 8)^{2} = 73 $ ; (d) $ (x + 3)^{2} + (y + 4)^{2} = 9 $

8.21. Find the equations of the following circles: (a) center $ (5,2) $ , $ (3,-1) $ is a point on the circle;

  1. $ (5,-5) $ and $ (-3,-9) $ are end points of a diameter.

Ans. (a) $ (x - 5)^{2} + (y - 2)^{2} = 13 $ ; (b) $ (x - 1)^{2} + (y + 7)^{2} = 20 $

8.22. For the following equations, determine whether they represent circles, and if so, find the center and radius:

\[ x^{2}+y^{2}+8x+2y=5;(\mathbf{b})x^{2}+y^{2}-4x-8y+20=0;(\mathbf{c})2x^{2}+2y^{2}-6x+14y=3; \]

  1. $ x^{2} + y^{2} + 12x + 20y + 200 = 0 $

Ans. (a) circle; center $ (-4, -1) $ , radius $ $ ; (b) this is not a circle; the graph consists only of the point $ (2, 4) $ ; (c) circle, center $ (, -) $ , radius 4; (d) this is not a circle; there are no points on the graph.

8.23. Show that the triangle with vertices $ (-10,7) $ , $ (-6,-2) $ , and $ (3,2) $ is isosceles.

8.24. Show that the triangle with vertices $ (4, ) $ , $ (5, 0) $ , and $ (6, ) $ is equilateral.

8.25. Show that the triangle with vertices $ (6,9) $ , $ (1,1) $ , and $ (9,-4) $ is an isosceles right triangle.

8.26. Show that the quadrilateral with vertices $ (-3,-3) $ , $ (5,-1) $ , $ (7,7) $ , and $ (-1,5) $ is a rhombus.

8.27. Show that the quadrilateral with vertices $ (7,2) $ , $ (10,0) $ , $ (8,-3) $ , and $ (5,-1) $ is a square.

8.28. (a) Find the equation of the perpendicular bisector of the line segment with endpoints (−2, −5) and (7, −1).

  1. Show that the equation of the perpendicular bisector of the line segment with endpoints $ (x_{1},y_{1}) $ and $ (x_{2},y_{2}) $ can be written $ +=0 $ , where $ ({x},{y}) $ are the coordinates of the midpoint of the segment.

Ans. (a) $ 18x + 8y = 21 $