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Schaum's outlines: precalculus

Safier, Fred

Chapter 8

Analytic Geometry - all with Video Answers

Educators

Chapter Questions

05:18

Problem 1

Prove the distance formula.
In Fig. 8-6, $P_1$ and $P_2$ are shown. Introduce $Q\left(x_2, y_1\right)$ as shown. Then the distance between $P_1$ and $Q$ is the difference in their $x$-coordinates, $\left|x_2-x_1\right|$; similarly, the distance between $Q$ and $P_2$ is the difference in their $y$-coordinates, $\mid y_2-y_1 \mathrm{~L}$. In the right triangle $P_1 P_2 Q$, apply the Pythagorean theorem: $d^2=\left|x_2-x_1\right|^2+\mid y_2-y_1 \hat{F}=$ $\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$, since $a l^2=a^2$ by the properties of absolute values. Hence, taking the square root and noting that $d$, the distance, is always positive, $d\left(P_1, P_2\right)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$.
(FIGURE CAN'T COPY)

Tom Greenwood
Tom Greenwood
Numerade Educator
00:45

Problem 2

Find the distance $d\left(P_1, P_2\right)$ given
(a) $P_1(-5,-4), P_2(-8,0)$;
(b) $P_1(2 \sqrt{2}, 2 \sqrt{2}), P_2(0,5 \sqrt{2}) ;$
(c) $P_1\left(x, x^2\right), P_2\left(x+h,(x+h)^2\right)$
(a) Substitute $x_1=-5, y_1=-4, x_2=-8, y_2=0$ into the distance formula:
$$
\begin{aligned}
d & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{[(-8)-(-5)]^2+[0-(-4)]^2} \\
& =\sqrt{9+16}=\sqrt{25}=5
\end{aligned}
$$
(b) Substitute $x_1=2 \sqrt{2}, y_1=2 \sqrt{2}, x_2=0, y_2=5 \sqrt{2}$ into the distance formula:
$$
\begin{aligned}
d & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{(0-2 \sqrt{2})^2+(5 \sqrt{2}-2 \sqrt{2})^2} \\
& =\sqrt{(-2 \sqrt{2})^2+(3 \sqrt{2})^2} \\
& =\sqrt{8+18}=\sqrt{26}
\end{aligned}
$$
(c) Substitute $x_1=x, y_1=x^2, x_2=x+h, y_2=(x+h)^2$ into the distance formula and simplify.
$$
\begin{aligned}
d & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{(x+h-x)^2+\left[(x+h)^2-x^2\right]^2} \\
& =\sqrt{h^2+\left(2 x h+h^2\right)^2} \\
& =\sqrt{h^2+4 x^2 h^2+4 x h^3+h^4}
\end{aligned}
$$

Joel Siegel
Joel Siegel
Numerade Educator

Problem 3

Analyze intercepts and symmetry, then sketch the graph:
(a) $y=12-4 x$;
(b) $y=x^2+3$;
(c) $y^2+x=5$;
(d) $2 y=x^3$.

(a) Set $x=0$, then $y=12-4 \cdot 0=12$. Hence 12 is the $y$-intercept.
Set $y=0$, then $0=12-4 x$, thus $x=3$. Hence 3 is the $x$-intercept.
Substitute $-x$ for $x: y=12-4(-x) ; y=12+4 x$. Since the equation is changed, the graph (see Fig. 8-7) does not have $y$-axis symmetry.
Substitute $-y$ for $y:-y=12-4 x ; y=-12+4 x$. Since the equation is changed, the graph does not have $x$-axis symmetry.
Substitute $-x$ for $x$ and $-y$ for $y:-y=12-4(-x) ; y=-12-4 x$. 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.
$$
\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline y & 16 & 12 & 8 & 4 & 0 & -4 & -8 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)
(b) 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.
$$
\begin{array}{|c||c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline y & 3 & 4 & 7 & 12 & 19 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)
(c) Set $x=0$, then $y^2+0=5$; thus $y= \pm \sqrt{5}$. Hence $\pm \sqrt{5}$ 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.
$$
\begin{array}{|c||c|c|c|c|c|}
\hline x & 5 & 4 & 1 & -4 & -11 \\
\hline y & 0 & 1 & 2 & 3 & 4 \\
\hline
\end{array}
$$(FIGURE CAN'T COPY)
(d) Set $x=0$, then $2 y=0^3$; thus $y=0$. Hence 0 is the $y$-intercept.
Set $y=0$, then $2 \cdot 0=x^3$; thus $x=0$. Hence 0 is the $x$-intercept.
Substitute $-x$ for $x: 2 y=(-x)^3 ; 2 y=-x^3$. Since the equation is changed, the graph (Fig. 8-10) does not have $y$-axis symmetry.
Substitute $-y$ for $y: 2(-y)=x^3 ; 2 y=-x^3$. Since the equation is changed, the graph does not have $x$-axis symmetry.
Substitute $-x$ for $x$ and $-y$ for $y:-2 y=(-x)^3 ; 2 y=x^3$. 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.

Check back soon!

Problem 4

Analyze intercepts and symmetry, then sketch the graph:
(a) $y=|x|-4$;
(b) $4 x^2+y^2=36$;
(c) $|x|+|y|=3$;
(d) $x^2 y=12$.
(a) Proceeding as in the previous problem, the $x$-intercepts are \pm 4 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.
$$
\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline y & -4 & -3 & -2 & -1 & 0 & 1 & 2 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)
(b) The $x$-intercepts are \pm 3 and the $y$-intercepts are \pm 6 .
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.
$$
\begin{array}{|c||c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 \\
\hline y & 6 & \sqrt{32}=5.6 & \sqrt{20}=4.4 & 0 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)
(c) The $x$-intercepts are \pm 3 and the $y$-intercepts are \pm 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.
$$
\begin{array}{|l||l|l|l|l|}
\hline x & 0 & 1 & 2 & 3 \\
\hline y & 3 & 2 & 1 & 0 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)
(d) 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.
$$
\begin{array}{|c||c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline y & \text { undefined } & 12 & 3 & 4 / 3 & 3 / 4 \\
\hline
\end{array}
$$
(FIGURE CAN'T COPY)

Check back soon!
06:15

Problem 5

Find the center and radius for the circles with the following equations:
(a) $x^2+y^2=9$;
(b) $(x-3)^2+(y+2)^2=25$;
(c) $(x+5)^2+\left(y+\frac{1}{2}\right)^2=21$
(a) 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 $\sqrt{9}=3$.
(b) 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 $\sqrt{25}=5$.
(c) Comparing the given equation with the form $(x-h)^2+(y-k)^2=r^2,-h=5$ and $-k=\frac{1}{2}$; hence the center is at $(h, k)=\left(-5,-\frac{1}{2}\right)$. Since $r^2=21$, the radius is $\sqrt{21}$.

Alison Rodriguez
Alison Rodriguez
Numerade Educator
02:06

Problem 6

Find the equations of the following circles: (a) center at origin, radius 7 ; (b) center at $(2,-3)$, radius $\sqrt{14}$; (c) center at $(-5 \sqrt{2}, 0)$, radius $5 \sqrt{2}$.
(a) Substitute $r=7$ into $x^2+y^2=r^2$. The equation is $x^2+y^2=49$.
(b) Substitute $h=2, k=-3, r=\sqrt{14}$ into $(x-h)^2+(y-k)^2=r^2$.
The equation is $(x-2)^2+[y-(-3)]^2=(\sqrt{14})^2$ or $(x-2)^2+(y+3)^2=14$.
(c) Substitute $h=-5 \sqrt{2}, k=0, r=5 \sqrt{2}$ into $(x-h)^2+(y-k)^2=r^2$.
The equation is $[x-(-5 \sqrt{2})]^2+(y-0)^2=(5 \sqrt{2})^2$ or $(x+5 \sqrt{2})^2+y^2=50$.

Karla Conrey
Karla Conrey
Numerade Educator
01:49

Problem 7

Find the center and radius of the circle with equation $x^2+y^2-4 x-12 y=9$.
Complete the square on $x$ and $y$.
$$
\begin{aligned}
x^2-4 x+y^2-12 y & =9 & & {\left[\frac{1}{2}(-4)\right]^2=4 ;\left[\frac{1}{2}(-12)\right]^2=36 } \\
x^2-4 x+4+y^2-12 y+36 & =4+36+9 & & \text { Add } 4+36 \text { to both sides } \\
(x-2)^2+(y-6)^2 & =49 & &
\end{aligned}
$$

Add $4+36$ to both sides

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 .

Ernest Castorena
Ernest Castorena
Numerade Educator
06:59

Problem 8

Prove the midpoint formula.
In Fig. 8-15, $P_1\left(x_1, y_1\right)$ and $P_2\left(x_2, y_2\right)$ are given. Let $\left(x_1 y\right)$ be the unknown coordinates of the midpoint $M$. Project the points $M, P_1, P_2$ to the $x$-axis as shown.
(FIGURE CAN'T COPY)
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
$$
\begin{aligned}
-2 x & =-x_1-x_2 \\
x & =\frac{x_1+x_2}{2}
\end{aligned}
$$

Similarly, it can be shown by projecting onto the $y$-axis that $y=\frac{y_1+y_2}{2}$.

John Mcalister
John Mcalister
Numerade Educator
01:00

Problem 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 $M$.

AG
Ankit Gupta
Numerade Educator
01:42

Problem 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.
$$
\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(4-0)^2+[(-1)-6]^2}=\sqrt{16+49}=\sqrt{65}
$$

Step 3. Substitute the calculated radius and coordinates of the center into the standard form for the equation of a circle. $r=\sqrt{65},(h, k)=(4,-1)$.
$$
\begin{aligned}
(x+h)^2+(y-k)^2 & =r^2 \\
(x-4)^2+[y-(-1)]^2 & =(\sqrt{65})^2 \\
(x-4)^2+(y+1)^2 & =65
\end{aligned}
$$

Charles Carter
Charles Carter
Numerade Educator
02:53

Problem 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
$$
\begin{aligned}
& d(A, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{[(-1)-1]^2+(2-3)^2}=\sqrt{5}=c \\
& d(B, C)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{[5-(-1)]^2+[(-5)-2]^2}=\sqrt{85}=a \\
& d(A, C)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(5-1)^2+[(-5)-3]^2}=\sqrt{80}=b
\end{aligned}
$$

Step 2. Apply the converse of the Pythagorean theorem.
Since $a^2=(\sqrt{85})^2=85$ and $b^2+c^2=(\sqrt{80})^2+(\sqrt{5})^2=80+5=85$, the relation $a^2=b^2+c^2$ is satisfied; hence the triangle is a right triangle.

Emily Himsel
Emily Himsel
Numerade Educator
03:20

Problem 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 $P A=P B$, then $P$ lies on the perpendicular bisector of $A B$. From the distance formula,
$$
\begin{aligned}
& d(A, P)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{[(-12)-0]^2+[11-(-3)]^2}=\sqrt{340}=P A \\
& d(P, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{[6-(-12)]^2+(15-11)^2}=\sqrt{340}=P B
\end{aligned}
$$

Hence $P A=P B$ and $P$ lies on the perpendicular bisector of $A B$.

James Kiss
James Kiss
Numerade Educator
03:16

Problem 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 $P A=P B$, then $P$ lies on the perpendicular bisector of $A B$. Let $P$ have the unknown coordinates $(x, y)$. Then, from the distance formula, $P A=P B$ if
$$
P A=\sqrt{(x-7)^2+[y-(-8)]^2}=\sqrt{[x-(-2)]^2+(y-5)^2}=P B
$$

Squaring both sides and simplifying yields
$$
\begin{aligned}
(x-7)^2+[y-(-8)]^2 & =[x-(-2)]^2+(y-5)^2 \\
x^2-14 x+49+y^2+16 y+64 & =x^2+4 x+4+y^2-10 y+25 \\
18 x-26 y & =84 \\
9 x-13 y & =42
\end{aligned}
$$

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

Sam Limsuwannarot
Sam Limsuwannarot
Numerade Educator
03:17

Problem 14

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

Benjamin Chaback
Benjamin Chaback
Numerade Educator
00:46

Problem 15

Find the distance between the following pairs of points: (a) $(0,-7)$ and $(7,0)$; (b) $(-3 \sqrt{3},-3)$ and $(3 \sqrt{3}, 3)$,

Amy Jiang
Amy Jiang
Numerade Educator
02:07

Problem 16

Find the length and the midpoint of the line segment with the given endpoints:
(a) $A(1,8), B(-3,4)$; (b) $A(3,-7), B(0,8)$; (c) $A(1, \sqrt{2}), B(-1,5 \sqrt{2})$

Nick Johnson
Nick Johnson
Numerade Educator
00:51

Problem 17

Analyze the following for symmetry. Do not sketch graphs:
(a) $x y^2=4$;
(b) $x^3 y=4$;
(c) $|x y|=4$;
(d) $x^2+x y=4$ :
(e) $x^2+y+y^2=4 ;$ (f) $x^2+x y+y^2=4$

James Kiss
James Kiss
Numerade Educator
08:54

Problem 18

Analyze symmetry and intercepts, then sketch graphs of the following:
(a) $3 x+4 y+12=0$
(b) $y^2=10+x$
(c) $y^2-x^2=9$
(d) $|y|-|x|=3$
(b) Fig. 8-17: $x$-intercept $-10, y$-intercepts $\pm \sqrt{10}$, $x$-axis symmetry
(c) Fig. 8-18: no $x$-intercept, $y$-intercept \pm 3 , $x$-axis, $y$-axis, origin symmetry
(d) Fig. 8-19: no $x$-intercept, $y$-intercepts \pm 3 , $x$-axis, $y$-axis, origin symmetry
(FIGURE CAN'T COPY)

Gavin Liang
Gavin Liang
Numerade Educator
06:29

Problem 19

8.19. Analyze symmetry and intercepts, then sketch graphs of the following:
(a) $x+y=0$;
(b) $y+|x|=4$;
(c) $x^2=4|y|$;
(d) $|y|+x^2=4$;
(e) $|x|=4 y^2$;
(f) $-x y^2=4$

Gavin Liang
Gavin Liang
Numerade Educator
01:19

Problem 20

Find the equations of the following circles: (a) center $(5,-2)$, radius $\sqrt[4]{10}$; (b) center $\left(\frac{1}{2},-\frac{5}{2}\right)$, diameter 3 ;
(c) center $(3,8)$, passing through the origin; (d) center $(-3,-4)$, tangent to the $y$-axis.

Nick Johnson
Nick Johnson
Numerade Educator
01:03

Problem 21

Find the equations of the following circles: (a) center $(5,2),(3,-1)$ is a point on the circle;
(b) $(5,-5)$ and $(-3,-9)$ are end points of a diameter.

Heather Zimmers
Heather Zimmers
Numerade Educator
06:58

Problem 22

For the following equations, determine whether they represent circles, and if so, find the center and radius:
(a) $x^2+y^2+8 x+2 y=5$;
(b) $x^2+y^2-4 x-8 y+20=0$;
(c) $2 x^2+2 y^2-6 x+14 y=3$;
(d) $x^2+y^2+12 x+20 y+200=0$

Debasish Das
Debasish Das
Numerade Educator
01:09

Problem 23

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

Carson Merrill
Carson Merrill
Numerade Educator
02:10

Problem 24

Show that the triangle with vertices $(4, \sqrt{3}),(5,0)$, and $(6, \sqrt{3})$ is equilateral.

Vikash Ranjan
Vikash Ranjan
Numerade Educator
01:09

Problem 25

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

Carson Merrill
Carson Merrill
Numerade Educator
02:04

Problem 26

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

Erika Bustos
Erika Bustos
Numerade Educator
01:58

Problem 27

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

M S
M S
Numerade Educator
03:16

Problem 28

(a) Find the equation of the perpendicular bisector of the line segment with endpoints $(-2,-5)$ and (7, -1$)$.
(b) Show that the equation of the perpendicular bisector of the line segment with endpoints $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ can be written $\frac{x-\bar{x}}{y_2-y_1}+\frac{y-\bar{y}}{x_2-x_1}=0$, where $(\bar{x}, \bar{y})$ are the coordinates of the midpoint of the segment.

Sam Limsuwannarot
Sam Limsuwannarot
Numerade Educator