kansd

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http://slidepdf.com/reader/full/kansd 1/15

1. (0+6+x)/3=3,x=3 (0+0+y)/3=2,y=6 Answer is B

2. The two intersections: (0,4) an (y, 0) !o, 4 " y/ 2 = 12 =# y= 6

!$o%e is %ositi&e =# y is 'e$ow the xaxis =# y = 6

3. irst, $et*s rewrite 'oth eations in the stanar -or o- the eation o- a $ine:ation o- $ine l : y = x + 4

ation o- $ine w: y = (1/) x 2

ote that the s$o%e o- $ine w, 1/, is the neati&e reci%roca$ o- the s$o%e o- $ine l .There-ore, we can conc$e that $ine w is %er%enic$ar to $ine l .

ext, since $ine k oes not intersect $ine l , $ines k an l st 'e %ara$$e$. !ince $inew is %er%enic$ar to $ine l , it st a$so 'e %er%enic$ar to $ine k. There-ore , $ines

k an w st -or a riht an$e, an its eree easre is ea$ to 0 erees.

The correct answer is 5.

4. To -in the istance -ro the oriin to any %oint in the coorinate %$ane, we tae

the sare root o- the s o- the sares o- the %oint7s x an ycoorinates. !o,

-or exa%$e, the istance -ro the oriin to %oint W is the sare root o- (a2 + b2).This is 'ecase the istance -ro the oriin to any %oint can 'e thoht o- as the

hy%otense o- a riht trian$e with $es whose $enths ha&e the sae &a$es as the

x an ycoorinates o- the %oint itse$-:

8e can se the 9ythaorean Theore to eterine that a2 + b2 = p2, where p is the$enth o- the hy%otense -ro the oriin to %oint W .

8e are a$so to$ in the estion that a2 + b2 = c2 + d 2, there-ore %oint X an %oint

W are eiistant -ro the oriin. An since e2 + f 2 = g 2 + h2, we now that %oint

Y an %oint Z are a$so eiistant -ro the oriin.

- the istance -ro the oriin is the sae -or %oints W an X , an -or %oints Z an

Y , then the $enth o- WY st 'e the sae as the $enth o- XZ . There-ore, the

&a$e o- $enth XZ $enth WY st 'e 0.

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The correct answer is ;.

. The estion ass s to -in the s$o%e o- the $ine that oes throh the oriin an

is eiistant -ro the two %oints P =(1, 11) an Q=(<, <). t7s i&en that the oriinis one %oint on the reeste $ine, so i- we can -in another %oint nown to 'e on

the $ine we can ca$c$ate its s$o%e. ncrei'$y the i%oint o- the $ine seent 'etween P an Q is a$so on the reeste $ine, so a$$ we ha&e to o is ca$c$ate the

i%oint 'etween P an Q (This %roo- is i&en 'e$ow).

>et7s ca$$ R the i%oint o- the $ine seent 'etween P an Q. R7s coorinates

wi$$ ?st 'e the res%ecti&e a&erae o- P 7s an Q7s coorinates. There-ore R7s xcoorinate ea$s 4 , the a&erae o- 1 an <. ts ycoorinate ea$s , the a&erae

o- 11 an <. !o R=(4, ).

ina$$y, the s$o%e -ro the (0, 0) to (4, ) ea$s /4, which ea$s 2.2 in ecia$

-or.

9roo-

To show that the i%oint R is on the $ine throh the oriin that7s eiistant

-ro two %oints P an Q, raw a $ine seent -ro P to Q an ar R at itsi%oint. !ince R is the i%oint then PR = RQ.

ow raw a $ine L that oes throh the oriin an R. ina$$y raw a %er%enic$ar -ro each o- P an Q to the $ine L. The two trian$es so -ore are

conrent, since they ha&e three ea$ an$es an PR ea$s RQ. !ince the

trian$es are conrent their %er%enic$ar istances to the $ine are ea$, so $ine

L is eiistant -ro P an Q.

The correct answer is B.

6.

To -in the s$o%e o- a $ine, it is he$%-$ to ani%$ate the eation into s$o%e

interce%t -or: y = mx + b, where m ea$s the s$o%e o- the $ine (incienta$$y, b re%resents the y

interce%t). A-ter iso$atin y on the $e-t sie o- the eation, the x coe--icient wi$$

te$$ s the s$o%e o- the $ine.

x + 2 y = 1

2 y = x + 1

y = x/2 + 1/2

The coe--icient o- x is 1/2, so the s$o%e o- the $ine is 1/2.

The correct answer is ;

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<. ach sie o- the sare st ha&e a $enth o- 10. - each sie were to 'e 6, <, @,

or ost other n'ers, there co$ on$y 'e -or %ossi'$e sares rawn, 'ecase

each sie, in orer to ha&e inteer coorinates, wo$ ha&e to 'e rawn on the xor yaxis. 8hat aes a $enth o- 10 i--erent is that it co$ 'e the hy%tonese

o- a %ythaorean tri%$e, eanin the &ertices co$ ha&e inteer coorinates

withot $yin on the x or yaxis.

or exa%$e, a sare co$ 'e rawn with the coorinates (0,0), (6,@), (2, 14)

an (@, 6). (t is teios an nnecessary to -ire ot a$$ -or coorinates -oreach sare).

- we $a'$e the sare abcd , with a at the oriin an the $etters re%resentin %oints

in a c$ocwise irection, we can et the n'er o- %ossi'$e sares 'y -irinot the n'er o- nie ways ab can 'e rawn.

a has coorinates (0,0) an b co$ ha&e coorinates:

(10,0)

(@,6)(6,@)

(0,10)

(6,@)(@,6)

(10,0)

(@, 6)

(6, @)(0, 10)

(6, @)

(@, 6)

There are 12 i--erent ways to raw ab, an so there are 12 ways to raw abcd .

The correct answer is .

@. At the %oint where a cr&e interce%ts the xaxis (i.e. the x interce%t), the y &a$e isea$ to 0. - we %$ y = 0 in the eation o- the cr&e, we et 0 = ( x p)( x

q). This %roct wo$ on$y 'e ero when x is ea$ to p or q. The estion is

asin s i- (2, 0) is an xinterce%t, so it is rea$$y asin s i- either p or q is ea$to 2.

(1) !;T: 8e can*t -in the &a$e o- p or q -ro this eation.

(2) !;T: 8e can*t -in the &a$e o- p or q -ro this eation.

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(1) A5 (2) !;T: Toether we ha&e enoh in-oration to see i- either

p or q is ea$ to 2. To so$&e the two si$taneos eations, we can %$ the p

&a$e -ro the -irst eation, p = @/q, into the secon eation, to coe % with2 + @/q = q.

This si%$i-ies to q2 + 2q @ = 0, which can 'e -actore (q + 4)(q 2) = 0, so q =

2, 4.- q = 2, p = 4 an i- q = 4, p =2. ither way either p or q is ea$ to 2.


. >ines are sai to intersect i- they share one or ore %oints. n the ra%h, $ine

seent QR connects %oints (1, 3) an (2, 2). The s$o%e o- a $ine is the chane in

y i&ie 'y the chane in x, or rise/rn. The s$o%e o- $ine seent QR is(3 2)/(1 2) = 1/1 = 1.

(1) !;T: The eation o- $ine S is i&en in y = mx + b -orat, where m

is the s$o%e an b is the yinterce%t. The s$o%e o- $ine S is there-ore 1, the sae asthe s$o%e o- $ine seent QR. >ine S an $ine seent QR are %ara$$e$, so they

wi$$ not intersect n$ess $ine S %asses throh 'oth Q an R, an ths the entireseent. To eterine whether $ine S %asses throh QR, %$ the coorinates o-

Q an R into the eation o- $ine S . - they satis-y the eation, then QR $ies on

$ine S .

9oint Q is (1, 3):

y = x + 4 = 1 + 4 = 3

9oint Q is on $ine S .

9oint R is (2, 2):

y = x + 4 = 2 + 4 = 29oint R is on $ine S .

>ine seent QR $ies on $ine S , so they share any %oints. There-ore, the answeris Cyes,C >ine S intersects $ine seent QR.

(2) !;T: >ine S has the sae s$o%e as $ine seent QR, so they are

%ara$$e$. They iht intersectD -or exa%$e, i- >ine S %asses throh %oints Q an R. Bt they iht ne&er intersectD -or exa%$e, i- >ine S %asses a'o&e or 'e$ow

$ine seent QR.

The correct answer is A.

10. irst, we eterine the s$o%e o- $ine L as -o$$ows:

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- $ine m is %er%enic$ar to $ine L, then its s$o%e is the neati&e reci%roca$ o- $ine

L7s s$o%e. (This is tre -or a$$ %er%enic$ar $ines.) Ths:

There-ore, the s$o%e o- $ine m can 'e ca$c$ate sin the s$o%e o- $ine L as

-o$$ows:

This s$o%e can 'e %$e into the s$o%einterce%t eation o- a $ine to -or the

eation o- $ine m as -o$$ows:

y = ( p – 2) x + b

(where ( p – 2) is the s$o%e an b is the yinterce%t)

This can 'e rewritten as y = px 2 x + b or 2 x + y = px + b as in answer choice A.

An a$ternati&e etho: 9$ in a &a$e -or p. or exa%$e, $et7s say that p = 4.

The s$o%e o- $ine m is the neati&e in&erse o- the s$o%e o- $ine L. Ths, the s$o%e

o- $ine m is 2.

There-ore, the correct eation -or $ine m is the answer choice that yie$s a s$o%eo- 2 when the &a$e 4 is %$e in -or the &aria'$e p.

(A) 2 x + y = px + < yields y = 2 x + <

(B) 2 x + y –px yields y 6 x

(;) x + 2 y = px + < yields y (3/2) x + </2

(5) y < = x ! ( p 2) yields y (1/2) x + <

() 2 x + y = < px yields y = 6 x + <

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En$y answer choice A yie$s a s$o%e o- 2. ;hoice A is there-ore the correct

answer.

11. The istance 'etween any two %oints an in the coorinate %$ane ise-ine 'y the istance -or$a.

5

Ths, the istance 'etween %oint F an %oint G is A + .

!tateent (1) te$$s s that:

Ths A = 6 or A = 1.

sin this in-oration, the istance 'etween %oint F an %oint G is either 11 or4. This is not s--icient to answer the estion.

!tateent (2) a$one te$$s s that A # 2, which is not s--icient to answer theestion.

8hen we co'ine 'oth stateents, we see that A st 'e 6, which eans the

istance 'etween %oint F an %oint G is 11. This is a %rie n'er an we are

a'$e to answer the estion.


12. The -or$a -or the istance 'etween two %oints ( x1, y1) an ( x2, y2) is:

.

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Ene way to nerstan this -or$a is to nerstan that the istance 'etween anytwo %oints on the coorinate %$ane is ea$ to the hy%otense o- a riht trian$e

whose $es are the i--erence o- the x&a$es an the i--erence o- the y&a$es

(see -ire). The i--erence o- the x&a$es o- P an Q is an the i--erence o-the y&a$es is 12. The hy%otense st 'e 13 'ecase these $e &a$es are %art

o- the nown riht trian$e tri%$e: , 12, 13.

8e are to$ that this $enth (13) is ea$ to the heiht o- the ei$atera$ trian$e XYZ . An ei$atera$ trian$e can 'e ct into two 30600 trian$es, where the

heiht o- the ei$atera$ trian$e is ea$ to the $on $e o- each 30600 trian$e.

8e now that the heiht o- XYZ is 13 so the $on $e o- each 30600 trian$e is

ea$ to 13. sin the ratio o- the sies o- a 30600 trian$e (1: : 2), we can

eterine that the $enth o- the short $e o- each 30600 trian$e is ea$ to 13/

. The short $e o- each 30600 trian$e is ea$ to ha$- o- the 'ase o-

ei$atera$ trian$e XYZ . Ths the 'ase o- XYZ = 2(13/ ) = 26/ .

The estion ass -or the area o- XYZ , which is ea$ to 1/2 H 'ase H heiht:

The correct answer is A.

13. To -in the area o- ei$atera$ trian$e "#$ , we nee to -in the $enth o- one

sie. The area o- an ei$atera$ trian$e can 'e -on with ?st one sie since there

is a nown ratio 'etween the sie an the heiht (sin the 30: 60: 0

re$ationshi%). A$ternati&e$y, we can -in the area o- an ei$atera$ trian$e ?stnowin the $enth o- its heiht.

(1) !;T: This oes not i&e s the $enth o- a sie or the heiht o-

the ei$atera$ trian$e since we on7t ha&e the coorinates o- %oint ".

(2) !;T: !ince $ has an xcoorinate o- 6, the heiht o- the ei$atera$

trian$e st 'e 6.

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14. - we %t the eation 3 x + 4 y = @ in the s$o%einterce%t -or ( y = mx + b), we et:

y (3/4) x + 2

This eans that m (the s$o%e) = 3/4 an b (the yinterce%t) = 2.

8e can ra%h this $ine 'y oin to the %oint (0, 2) an oin to the riht 4 anown 3 to the %oint (0 + 4, 2 3) or (4, 1).

- we connect these two %oints, (0, 2) an (4, 1), we see that the $ine %asses

throh arants , an I.


1. To eterine in which arant the %oint ( p, p q) $ies, we nee to now the sino- p an the sin o- p q.

(1) !;T: - ( p, q) $ies in arant I, p is %ositi&e an q is neati&e. p

q st 'e %ositi&e 'ecase a %ositi&e n'er ins a neati&e n'er isa$ways %ositi&e Je.. 2 (3) = K.

(2) !;T: - (q, p) $ies in arant , q is neati&e an p is %ositi&e.

(This is the sae in-oration that was %ro&ie in stateent 1).

The correct answer is 5.

16. 9oint # is on $ine "$ , twothirs o- the way 'etween 9oint " an 9oint $ . To -inthe coorinates o- %oint #, it is he$%-$ to iaine that yo are a %oint tra&e$in

a$on $ine "$ .

8hen yo tra&e$ a$$ the way -ro %oint " to %oint $ , yor xcoorinate chanes 3

nits (-ro x = 0 to x = 3). Twothirs o- the way there, at %oint #, yor xcoorinate wi$$ ha&e chane 2/3 o- this aont, i.e. 2 nits. The xcoorinate o-

# is there-ore x = 0 + 2 = 2.

8hen yo tra&e$ a$$ the way -ro %oint " to %oint $ , yor ycoorinate chanes 6

nits (-ro y = 3 to y = 3). Twothirs o- the way there, at %oint #, yor y

coorinate wi$$ ha&e chane 2/3 o- this aont, i.e. 4 nits. The ycoorinate o- # is there-ore y = 3 + 4 = 1.

Ths, the coorinates o- %oint # are (2,1).


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1<.2 The eation o- a circ$e i&en in the -or inicates that the circ$ehas a rais o- % an that its center is at the oriin (0,0) o- the xycoorinate

syste. There-ore, we now that the circ$e with the eation wi$$

ha&e a rais o- an its center at (0,0).

- a rectan$e is inscri'e in a circ$e, the iaeter o- the circ$e st 'e a iaona$ o-

the rectan$e (i- yo try inscri'in a rectan$e in a circ$e, yo wi$$ see that it is

i%ossi'$e to o so n$ess the iaona$ o- the rectan$e is the iaeter o- the circ$e).!o iaona$ "$ o- rectan$e "#$& is the iaeter o- the circ$e an st ha&e $enth

10 (ree'er, the rais o- the circ$e is ). t a$so cts the rectan$e into two riht

trian$es o- ea$ area. - we -in the area o- one o- these trian$es an $ti%$y it 'y2, we can -in the area o- the who$e rectan$e.

8e co$ ca$c$ate the area o- riht trian$e "#$ i- we ha the 'ase an heiht. 8e

a$reay now that the 'ase o- the trian$e, "$ , has $enth 10. !o we nee to -in theheiht.

The heiht wi$$ 'e the istance -ro the xaxis to &ertex #. 8e nee to -in thecoorinate o- %oint B in orer to -in the heiht. !ince the circ$e intersects trian$e

"#$& at %oint #, the coorinates o- %oint # wi$$ satis-y the eation o- the circ$e

. 9oint # a$so $ies on the $ine , so the coorinates o- %oint # wi$$ satis-y that eation as we$$.

!ince the &a$es o- x an y are the sae in 'oth eations an since , we

can s'stitte (3 x + 1) -or y in the eation an so$&e -or x:

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!o the two %ossi'$e &a$es o- x are 4 an . There-ore, the two %oints where the

circ$e an $ine intersect (%oints # an $ ) ha&e xcoorinates 4 an , res%ecti&e$y.

!ince the xcoorinate o- %oint $ is (it has coorinates (, 0)), the xcoorinate o-

%oint # st 'e 4. 8e can %$ this into the eation an so$&e -or the ycoorinate o- %oint #:

!o the coorinates o- %oint # are (4, 3) an the istance -ro the xaxis to %oint # is

3, ain the heiht o- trian$e "#$ ea$ to 3. 8e can now -in the area o- trian$e

"#$ :

The area o- rectan$e "#$& wi$$ 'e twice the area o- trian$e "#$ . !o i- the area o-

trian$e "#$ is 1, the area o- rectan$e "#$& is (2)(1) = 30.


18. irst, rewrite the $ine as . The eation is now in the

-or where re%resents the s$o%e an ' re%resents the yinterce%t.

Ths, the s$o%e o- this $ine is .

By e-inition, i- $ine is the %er%enic$ar 'isector o- $ine G, the s$o%e o- $ine

is the neati&e in&erse o- the s$o%e o- $ine G. !ince we are to$ that the

$ine is the %er%enic$ar 'isector o- $ine seent L9, $ine seent L9

st ha&e a s$o%e o- (which is the neati&e in&erse o- ).

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ow we now that the s$o%e o- the $ine containin seent L9 is 't we o

not now its yinterce%t. 8e can write the eation o- this $ine as ,

where ' re%resents the nnown yinterce%t.

To so$&e -or ', we can se the i&en in-oration that the coorinates o- %oint L

are (4, 1). !ince %oint L is on the $ine , we can %$ 4 in -or x an 1 in

-or y as -o$$ows:

ow we ha&e a co%$ete eation -or the $ine containin seent L9:

8e a$so ha&e the eation o- the %er%enic$ar 'isector o- this $ine: .

To eterine the %oint M at which these two $ines intersect, we can set these twoeations to ea$ each other as -o$$ows:

Ths, the intersection %oint M has xcoorinate 2. sin this &a$e, we can -in

the y coorinate o- %oint M:

Ths the %er%enic$ar 'isector intersects $ine seent L9 at %oint M, which has

the coorinates (2, 0). !ince %oint M is on the 'isector o- L9, %oint M re%resents

the i%oint on $ine seent L9D this eans that it is eiistant -ro %oint Lan %oint 9.

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8e now %oint L has an xcoorinate o- 4. This is two nits away -ro the x

coorinate o- i%oint M, 2. There-ore the xcoorinate o- %oint 9 st a$so 'e

two nits away -ro 2, which is 0.

8e now %oint L has a ycoorinate o- 1. This is one nit away -ro the y

coorinate o- i%oint M, 0. There-ore, the ycoorinate o- %oint 9 st a$so 'eone nit away -ro 0, which is 1.

The coorinates o- %oint 9 are . The correct answer is 5.

19.

The ost i--ic$t %art o- this estion is conce%ta$iin what yo7re ase to

-in. The 'est way to han$e tricy %ro'$es $ie this is to 'rea the own into

their co%onent %arts, reso$&e each %art, then reconstrct the as a who$e. Nostart o-- with a coorinate %$ane an -or %oints (A, B, ;, an 5) that -or a

sare when ?oine. Beneath the sare, on the xaxis, $ies another %oint, 9. Then

yo are ase to eterine the %ro'a'i$ity that a $ine rano$y rawn throh 9wi$$ not a$so %ass throh sare AB;5. n any %ro'a'i$ity estion, the -irst

thin yo nee to eterine is the n'er o- tota$ %ossi'i$ities. n this case, the

tota$ n'er o- %ossi'i$ities wi$$ 'e the tota$ n'er o- $ines that can 'e rawnthroh 9. The %ro'$e, thoh, is that there are $itera$$y in-inite$y any $ines

that satis-y this criterion. 8e cannot se in-inity as the enoinator o- or

%ro'a'i$ity -raction. !o what to oO

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This is where soe creati&e thinin is necessary. irst, yo nee to reconie

that there st 'e a $iite rane o- %ossi'i$ities -or $ines that %ass throh 'oth

9 an AB;5. - that were not the case, the %ro'a'i$ity wo$ necessari$y 'e 1 or0. 8hat is this raneO 8e$$, rawin ot a iara o- the %ro'$e wi$$ he$%

enoros$y here. Any $ine that %asses throh 'oth 9 an AB;5 has to %ass

throh a trian$e -ore 'y AB9. This trian$e is an isosce$es riht trian$e. 8enow this 'ecase i- we ro% %er%enic$ars -ro %oints A an B, we en %

with two isosce$es riht trian$es, with an$es o- 4 erees on either sie o- 9.

!ince an$es a$on a straiht $ine st ea$ 1@0, we now that an$e A9B steasre 0 erees. !o i- an$e A9B easres 0 erees, we can se that to

-ire ot the %ro%ortion o- $ines that %ass throh 'oth 9 an AB;5. - we raw

a seicirc$e with 9 as its center, we can see that the %ro%ortion o- $ines %assin

throh 9 an AB;5 wi$$ 'e the sae as the %ro%ortion o- the seicirc$eocc%ie 'y sector A9B. !ince a seicirc$e contains 1@0 erees, sector A9B

occ%ies 1/2 o- the seicirc$e. !o 1/2 o- a$$ %ossi'$e $ines throh 9 wi$$ a$so

%ass thoh AB;5, which eans the 1/2 wi$$ ET %ass throh AB;5 an we

ha&e or answer.


20.Becase we are i&en two %oints, we can eterine the eation o- the $ine. irst, we7$$

ca$c$ate the s$o%e 'y sin the -or$a ( y2 y1) / ( x2 x1):

Becase we now the $ine %asses throh (0,) we ha&e or yinterce%t which is .

9ttin these two %ieces o- in-oration into the s$o%einterce%t eation i&es s y = (/<) x + . ow a$$ we ha&e to o is %$ in the xcoorinate o- each o- the answer choices

an see which one i&es s the ycoorinate.

(A) (14, 10) y = /<(14) + = 1D this oes not atch the i&en ycoorinate.

(B) (<, )

y = /<(<) + = 10D this oes not atch the i&en ycoorinate.(;) (12, 4)

y = /<(12) + = 60/< + , which wi$$ not ea$ an inteerD this oes not atch the

i&en ycoorinate.(5) (14, )

y = /<(14) + = D this atches the i&en ycoorinate so we ha&e -on or answer.

() (21, )

y = /<(21) + = 1/< + , which wi$$ not ea$ an inteerD this oes not atch thei&en ycoorinate. (ote that yo o not ha&e to test this answer choice i- yo7&e

a$reay isco&ere that 5 wors.)

J0 ()K

(< 0)

=

<

7/17/2019 kansd


The correct answer is 5

21.

- we %t the eation 3 x + 4 y = @ in the s$o%einterce%t -or ( y = mx + b), we et:

The s$o%e o- a $ine %er%enic$ar to this $ine st 'e

4

3, the neati&e reci%roca$ o-

3

4

Aon the answer choices, on$y i&es an eation with a s$o%e o- 4/3.

The correct answer is

22.

8e are essentia$$y ase to -in the istance 'etween two %oints. The si%$est etho isto setch a coorinate %$ane an raw a riht trian$e sin the two i&en %oints:

8e can now see that one $e o- the trian$e is 3 an the other $e is . Becase it is a

riht trian$e, we can se the 9ythaorean Theore to ca$c$ate the hy%otense, which is

the $ine seent whose $enth we are ase to ca$c$ate.

32 + 2 = c2 34 = c2

c =

y =

3

4 x + 2, which eans that m (the s$o%e) =

3

4

7/17/2019 kansd


ote: a$ways 'e care-$ !oe %eo%$e wi$$ notice the $enths 3 an an atoatica$$y

asse this is a 34 riht trian$e. This one is '() , howe&er, 'ecase the hy%otense

st a$ways 'e the $onest sie an, in this %ro'$e, the leg is nits $on, not thehy%otense.

The correct answer is ;

23.

or this estion it is he$%-$ to ree'er that $ines are %er%enic$ar when their s$o%esare the neati&e reci%roca$s o- each other.

(1) !;T: Becase we now the $ines %ass throh the oriin, we can -ire ot

i- the s$o%es are neati&e reci%roca$s o- each other. The s$o%e o- m is 1 so i- the s$o%e o-

' is 1 (1/1) then we now the $ines are %er%enic$ar an the an$e 'etween the is 0P.Becase we now two %oints -or $ine n, (0, 0) an (a, a), we can ca$c$ate the s$o%e:

Ths the $ines are %er%enic$ar an the an$e 'etween the is 0P.(2) !;T: Leci%roca$s, when $ti%$ie toether, ea$ 1. !o$&in -or one s$o%e

in ters o- the other, we et x = 1/ y. Ths the s$o%es are the neati&e reci%roca$s o- each

other an there-ore the $ines are %er%enic$ar. Ths the an$e 'etween the st 'e 0P.The correct answer is 5

24.

Two $ines are %er%enic$ar i- their s$o%es are o%%osite reci%roca$s. or exa%$e, the $inesi&en 'y the eations y = 3 x + 4 an y = 1/3 x + < st 'e %er%enic$ar 'ecase the

s$o%es (3 an 1/3) are o%%osite reci%roca$s.

The s$o%e o- a $ine can 'e -on sin the -o$$owin eation:

s$o%e = ( y2 y1) Q ( x2 x1)

8e are i&en the coorinate %airs (3, 2) an (1, 2). The s$o%e o- the $ine on which these %oints $ie is there-ore (2 (2)) Q (3 (1)) = 4/4 = 1. !o any $ine that is %er%enic$ar to

this $ine st ha&e a s$o%e o- 1. ow we can chec the choices -or the %air that oes

ET ha&e a s$o%e o- 1.

(A) (@ ) Q ( 4) = 1/1 = 1.(B) (1 (2)) Q (3 4) = 1/(1) = 1.

(;) (6 ) Q (1 (4)) = 3/3 = 1.

(5) ( 2) Q (2 (3)) = 3/.

() (1 2) Q (< 6) = 1/1 = 1.

The on$y %air that oes not ha&e a s$o%e o- 1 is (2, ) an (3, 2).The correct answer is 5

0 (a)

0 (a)

=

a

a

= 1

kansd

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