ATI・ I
110, Assign=Ilent 1
'κ
Problem 1: The πïntercept
intersection of the line,
of the line is 4 and the area of the triangle R rmed by the
π¨
axis, and ν-aⅲ s is 20. Find the line equations and y¨ intercepts.
AIlswer.
一10
intercept 10, the other is y =
ν= - ⅜
몹
"+ 10 with the y—
"
with the y¨ intercept - 10. (Ybu wi11 get fUII marks once you
There are two such lines. one is
요
nd
βolution.
one of them.)
There are two possibilities :
Step1: Find y— intercepts.
Let b the po무 itive ν‥
intercept— Then, the triangles haive base ⅲs 4 and height as ¸
●Since
its area is 20, ¸satis且 es
=:
2●
××
4
b.
Thus, we haNe 20 == 2b and hence b == 10. This imphes v¨ inteycepts are either 10 or -10.
Step2: Find the line eq:uations.
In the case of ¸〓 10, the line passes through (4, 0) and (0, 10). :rhis ilnplies the slope γ
γ
ι
of the Iine is
7γ
ι=
10 - 05
-42●
0 —
By the slope— intercept R rΠ 1, we get the line equation
y
In the case of ¸=
〓
-:" + 10.
-10, the line passes through (4, 0) alld (0, - 10). This iIIlplies the slope
τ
η of the line is
—10
η 〓Tτ
-- =05
,˙
=τ
By the slope— intercept R rII1, we get the line equation
y
二
π
κ
〓뭏
.
ProbleΠ1 2. Find the distance from the
from a point to a line means the shortest
:=fι :ξ
A꾸 s㏆ r・
and낳i뇨
1亢
⅜
않
鹿士료 ∶
I흠
=
丸
ν
∬
H=표盟:#
七
o (1, 3) is the intersection point
π= 3 and pasBes through (1, 3).
the line
to 3ν +
3υ +.
旦些모⊥Find
Let’ s
ⅲ
ⅲ
㏂
㏄
ξ
蟲
τ3. (The
of 3y +
π=
3
κ= 3
七
0 -y
the line which is perpendicu1缸
+
χ=
3and p㎱ ses through (1, 3).
== 3:ㅗ
e囍
≡ξ
∫
f=:볐
rewrite the equation 3g+π
perpendicular line is negative reciprocal of・
ⅲ
ⅱ
娩
표
Ξ
∬∴
⅜
;:㎩
;Ξ
:ㄸ
I√
lt (1, 3) and the siope 3, w˚
=쵸
y - 3 = 3(z - 1) ==수
η요
nd th:빎 袁
:庄 근I:::i#표 ∶
u酢
#누 :f∶
y = 3π
.
to s 01ve the system of equation:
rI1,
3
=
By plugging the 요rst equation into the seco==d eqllation, we get
=
3—
Thus, the intersection point is
1I/
”
10
9ㅗ
＼
/I'＼
3—
(", y) =
旦
느
으
호
르Calculate
=3 ==⇒
〓
”
9–
Since y = 3", y =
= 3. ==李 그0∬
"
9"+∬
the distance between (1, 3) and the intersection point.
:if:1ㅓ :√
℉
get the folloⅵ ng
¸
½
= 1— 홅
쭘
By the distance formula between two point, the distance J is
' J = ' (ξ
;-;
i (j;二
;)¸
)2; (二
馮 T吾 =
Proble∏ 1 3. Sketoh the gr라 ph of the fⅱ nction y
〓|"2
-
-intercepts
ㅚand nnd π
and y—
intercepts,
AIlswer.
● L'
::¨
ㄴI-,, 'υ
y
1
=
—
1
Method1 : Sillce y = |χ 2
= xξ
|-
S olutiol1.
0
>-
0
<-
π2 一 1
1) "2 - 1
E)raw the piece— wise function.
Method2: Since the fⅱ nction is gⅳ en as the fbrm ofy = lg(π
draw
Ω(π )
and re묘 ect the gr— ph 0f
(π ) =
ProbleΠ1 4. Let ∫
蟲
(a) Filld the fbrmula fbr g(↔
Ω(π )
below
-axis ⅲ
-axis.
bout π
π
=
(∫
(b, Find the domⅲ n and range of
1)) is
π2 一
1˚
First,
°∫
)(τ ●
)
= = (〓
∫(∫ (-
g(π ) =
.
Ω(π ) 〓(∫ °∫)(") 〓∫(∫ (π )) 〓
First, since
‖where
11
1
∫(χ ) 十1
≒근
# =1-∴
㎡놈+
1
〓 τㅜΣ
π+ 1
)・
・
Ω
π〓-l is not in the doIIlain
㎁so undenned amd hence -l is
of
∫, ∫(-1)
is unde요 ned. Thus,
n0七 in the do〕 mai파
Also, since g(π ) has a q:uotient fUnction, when
π= 一2, Ξ(z)
Thtls, the domⅲ n of Ωis
R ＼{一
1, -2 .
of g・
is unde요 ned●
θ(-1)
(c) Find
∴
iⅡ teIoept
τ
and y— intercept.
π¨
intercept is given when we plug
(工 f
there’
8 no in¨ rcept, e즈 p1ⅲ n why.)
y = 0.'
0 = "十
1
π+ 2
When
χ
=
一1, the eqμ ati● n holds, but since -l is not in the d。 nⅲ
γ n, there’ s no Π¨
inteTcept,
←intercept
is
μve끄 when we plug Ξ=
0.
0 + 11
ν〓τㅜτ= =˙
1,요
Thus, the ν-iη tercept is
(d) Sketch the gra,ph of Ξ
.
Since
∬) = 1 - ㎡
㏊, the
(π
le魚 by 2●
(Note that
#-ph
is
ⅲhen ∫(")
=
y
μven by shi托 ing the graph of —옳up by l aη d
(τ + 2) + 1 = 1 -ㅗ ∫
㎡
㏊
.)
to the