1973-11-29 — Page 21

ABRARY.

30

育教僑華頁一第張六第 日五初月一十年丑癸醫夏

1974中學會考試題預習專欈

數學科(課程甲)(m2)

第四講西摩松線.

九點圓.

乘水

WABI KID YAALLO 有一公共弦,且立於同旁,若其 頂角相等,則四頂點共圓) 7. 同理,EQ,R亦在此

Bt (12-6)

郭日僑華

四期星

日九廿月一十年三七九一圈公年二十六國民華中

証力點圓的半徑為原

角形叶樟園半經的 一半。

設面積為A

equation of

exwhe form

ax + by

and.c

0

where a, bi

real numbers, and not. both of a and b are zero. The

solution set of a Linear equation ds the set of ordered

A linear equation is any

8.在什△AYHP為科遍.

AH之中點(題設)

上週習題解答

pairs, (x, y), which satisfy the equation.

+豆+芝

B. One Linear and one quadratic.

Example 1. Find the solution sets

of the following systems.

9.PH=PY(斜邊中點。

距頂等)

△

·C (13.43)

共線(西摩松線定理 Simson Line

例:註明:由滴形之外楼圓 10.<BHY=<PHY(對頂角相等) 上一點毛邊作重線,其三重趾

PHY=<PYH(等腰>角号 底角相等)

-G (1.9)

中點座標為D.

三(x+4x)

(36√3)

4x5x:x

it:

(2)

(1)

PRIAL PS LAB.

<PHY=<PYH=<BHdata)

<BYP=<DBY (同-12)

DO

E{X+X3

= = (3123) = b2No3 (7

1974中學會考試題預習專欄 (2)x29

(2)

Solution:

(1) x 3

12

10

(3)

證

•<FRC = 9.4.

<PRI=<PSA =90°

(直角相等

14. <PYHTC BY D= <BH { TZ DBY (等量加等量,其和枯等)

F(x+x 9.+42)

數學科(課程乙)

10--5y

<$A=90°(已知

15.12 <PIMT <BYD=<PYD

MATHEMATICS (4)

>

Puty

(全量事於各部分量之和)

号 into (2)

Solution to exercise

510x

10x + 5

G

t

11

2AP

·AG

3 PRAS 四點共圓(共四邊形之 时尚等於其内對,則其 四點共圓)

< BMX TC DBY-90°

女锐角口為餘角) 从AD中線中

17. <PYD=90°(代入)

果

4. P.A.DC

共圖(題設)

5-PAS=<PCB国的李四邊形之

X1+ X = +X2 = { (1;+x>+xz) a) (p−q)2 = p2 - 2pq + q2

፩ .

外面等玲其内對角)

共圓)

<PRS=∠PAS(對同弧之圓周

再相案)

19.同理,可證亦在此圖

F. (197) 8-18)

7 <PRS=<PCB (1XA)

8 = <PQ C = < PRC - 90° (ZAV).

9

3 PRQC→點共糰(若两角(伏拉氏線)

有一公共底,且於同東

心叶

共線

若不頂角等,則四點共園)

3.夾再為

• ABC 2 H BE

(1+2+3

(19)-4(3)

b)(p-2q)(2p-9

2 [(p+q)2 - 2pq ] ? (p + q)2

5pq.

← 252 + 2(1 − y)2 -5y(1−y)

4y+

97 +2 2)(3y:

5y

5y2

or

The solution set

{-4}

From (1) ↔

From (2) and (3)

心及

国内接四邊形對角互補)

违証?

共綠

!! XPRS +<P RQ=

180°(代入)

註明

2.夾角為3

設

2. a).

{k+ 1)x=1-k

The solution set is

6 k+1 = 2(k+1)

其叶遭成直線)

-2k

斜率為

and

例二:證明-三角形各邊中點, 高之重趾,重心與各頂點聯 結線之中點,此點同在一

角形之叶心)

AH+ BC, BYLAC (i 浅利為の角形

H. 為重線.. 3. ODMAX, DEN BY (→]},

|} = (1 nime-points circle,

△ABC邊之中

ALLBC,

CZIAF

PQR各為HAHB、H

H為重心

BEC D. E, F, X, Y, Z, P. Q.

→過P、XD㸃作一圓(過濕可

作一園)

tam 0 Mi

}} z = _(−—4)~(~~)

7-5

- — -(-2) + Mimi H(+)(2)

4.聯OH (兩點間直線可作?

115 00 XH BOFY HIZ

為一梯形

7.此两垂直平分線過叫之 中點(由梯形一腰之中點 作平行於底之直線,必遇

2.聯FD,FP(過兩點一直線可它膜之中點)

3 FPMYB, FDICA (1) *3+

腰中點聯線平行底邊,且

為底邊之半)

4 <PFD = <AYB = RT<

900

(西角之相當邊平行而異而,則此 两角相等;若一組平圩邊同向

互補)

5.但<pxD=90°(題設)

6.8

tan B

0= 23°38

+帆

3=83°50'

tan x

8.∴OH之中點為九點圓心解

X=60°151元

N(弦之垂直平分線過 設正文邊形六頂點為ABCDEF 圓心

9.26 TE OHE.

NG,H,四點共線

(VARON 2).

習題:

一邊為長,

b=Y (3)

a = b = (5)2

△ABC之高,AD,BE,CP相定,頂點座標為

AH.DED AB, AL, BE, CHA (b,o), B(3, 3), ((-2, 2√3)

試證明之。

P(-3).

Substitute k = 3 inta

equation

· · 2x2 - (3 + 1

←2x2

▶x = 1(repeated)

The solution set of x is.

b) since the equation has equal

roots,

Discriminant

(5)2 -

4(6)(K)

Substitute k = 21⁄2 into the

equation

144

►(12x + 5)2-

+120x + 25

0

Inconsistance and Dependence two`linear equations.

ax+by

dx + ey + f·

u 'are. the system

of linear equations, it is said to be inconsistant.

If g = 9. (i.e. to solution).

it is Badd to be dependent:

If & - - - - (1.e. many

solutions)

Example 2. What is the condition

on the real numbers a and b such that the system of equations x - y = 4;

−2x+2y=h may have solution. in real mumbers?

Solution: Since the ratios of coefficients of x and y in the system of equation

ons ar

and 2, they an all equal

www.The system of equations may

have solution.

it is dependent.

(repeated)

The solution set of x is

{-2}

Exercise 4.

1. Find the solution set of the

following systems.

13. Let

= 2x + 5 = 0

, q be the roots of 3x

a) 2x-

9

pq=

डे

ра

The required equation is

+ 00

i.e. 5x2

2x +3 = 0

Simultaneous Equationa A. Both Linear.

33 + -2 = 25

b) x +

282

- 24.

2. Test, whether the following

system equations have unique solution, no solution or many solutions.

a) 2x 3y = 4

4x-

b) -5x

5y = 2

10x + 2y =3

*) 3x

X

y = 4

=

3 - 13.