View
1
Download
0
Category
Preview:
Citation preview
1
Lesson １４
Graphs and Equations (I)
１4A • Distance Between Two Points • Division of a Line Segment • Straight Lines
Course I
2
Distance Formula
Distance Formula When two points and are given, the distance between these points is given by
212
212 )()(AB yyxx −+−=
),( 11 yxA ),( 22 yxB
),( 12 yx),( 11 yx
),( 22 yx
3
Division of a Line Segment
Internal Division Point
),( 11 yxA ),( 22 yxBTwo points and are given.
nmxxxx :)(:)( 21 =−−
nmmxnxx++
= 21
nmmynyy++
= 21
Therefore
External Division Point
nmxxxx :)(:)( 21 =−−
Therefore nmmxnxx
−+−
= 21
nmmynyy
−+−
= 21
B`
4
Example [ Example 14.1] Find the point C which is located on the -axis and has equal distances from points A(-1, 3) and B(2, 4).
Ans. Let the coordinates of point C be . Then Square both sides Therefore,
)0,(x
x
2222 )40()2()30()1( −+−=−++ xx
16)2(9)1( 22 +−=++ xx
35
=x
x
y
O
(-1,3)
(2,4)
x
That was too easy!
5
Straight Line
Various Forms
General form of a straight line 0=++ cbyax
(1) Point-slope form From point and slope ),( 11 yx m
)( 11 xxmyy −=−
From intercept and slope −y m
nmxy +=
(2) Slope-intercept form 1
1
xxyym
−−
=
),0( n
0−−
=xnym
x
y
O
m
1
),( 11 yx
x
y
O
n
m
1
6
Straight Line - Cont.
(3) Line through two points and ),( 11 yx ),( 22 yx
)( 112
121 xx
xxyyyy −
−−
=− (when ) 21 xx ≠
Slope 12
12
xxyym
−−
=
( This represents a vertical line. ) 1xx =
∗
(when ) 21 xx =
x
y
O
),( 11 yx
),( 22 yx
12 xx −
12 yy −
Vertical line
1yy =Horizontal line 1xx =
x
y
O1x
1y
∗
7
Example (Intercept-Intercept Form) [ Example 14.2] Find the expression of the straight line which passes two points (4, 0) and (0, -3).
Ans. From the expression mentioned above, we have
or 1243 =− yx 1)3(4=
−+yx
Then we have
(4) Intercept-intercept form When a straight line intersects the -axis at and the -axis at , it is expressed as
xa y b
1=+by
ax
x
y
Oa
b
)4(40030 −
−−−
=− xyx
y
O (4, 0) (0, -3)
8
Exercise
Ans.
[ Exercise 14.1] Three points A(-1, 1), B and C are on the straight line in this order. Find the value of and the equation of in the
following steps. (1) Assume the line as and find the conditions that points A, B
and C are on the line. (2) Find which satisfy these conditions.
(3) Select appropriate values.
),3( a )7,3( +al la
nmxy +=
Pause the video and solve the problem by yourself.
A B
C
9
Exercise
Ans. lnm +−=1 nma += 3 nam ++= )3(7
(1) Because the line passes points A, B, and C, we have (i), (ii) , (iii)
33 >+a
4=a47
43
+= xy
(3) Considering the order, we have
Therefore, the latter satisfies the request of the problem. Finally, we obtain and
.
14 += ma ma )4(6 +=a 0)34)(2( =−+ mm
)1,7,2(),,( −−−=nam )47,4,
43(
, . (2) From (ii)-(i) and (iii)-(i), we have and .
Therefore, we obtain and . By eliminating , we have
[ Exercise 14.1] Three points A(-1, 1), B and C are on the straight line in this order. Find the value of and the equation of in the following
steps. (1) Assume the line as and find the conditions that points A, B and C are on
the line. (2) Find which satisfy these conditions. (3) Select appropriate values.
),3( a )7,3( +al la
nmxy +=
A B
C
10
Lesson １４
Graphs and Equations (I)
１4B • Parallel Straight Lines • Perpendicular Straight Lines • Reflected Images • Distance to a Straight Line
Course I
11
Parallel Straight Lines
For two straight lines
11 nxmy +=
22 nxmy +=
Parallel Lines Condition
21 mm =
x
y
O
11 nxmy +=
22 nxmy +=
1m
2m
1
1
1n
2n
12
Perpendicular Straight Lines For two straight lines
11 nxmy += 22 nxmy +=
Perpendicular Lines Condition
121 −=mm
Proof: ΔFor ABC: AB2+BC2=AC2 (i)
221 m+
,
For ABD: AB2=BD2+AD2= ΔFor BCD: BC2= Δ 2
1)(1 m−+
Therefore, from (i) 2
122
122 )()(11 mmmm −=−+++
121 −=∴ mm
(ii) (iii)
x
y
O
B
A
C
11 nxmy +=
22 nxmy +=
1m−
2m
13
Reflected Image
Put the image point be Q .
[ Example 14.3] Find the reflected image of point P(2, 3) about the mirror line .
Ans. ),( qq yxLine PQ is perpendicular to the mirror line:
123
2 −=⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−⋅
q
q
xy
The middle point of segment PQ is on the mirror line:
From (i) and (ii), we have .
52 += xy
52
22
23
+⎟⎟⎠
⎞⎜⎜⎝
⎛ +⋅=
+ qq xy
527,
514
=−= qq yx
(i)
(ii)
x
y
O
52 += xy
)3,2(P
),( qq yxQ
14
Distance To the Straight Line [ Example 14.4] Find the shortest distance between the origin and the straight line Ans.
0=++ cbyaxThis line is rewritten as .
bcx
bay −−=
),( 11 yx Point A on the given line is nearest point from the origin.
Point A is on the line : 011 =++ cbyax
Segment OA is perpendicular to the line :
11
1 −=⋅⎟⎠
⎞⎜⎝
⎛−xy
ba
From (i) and (ii), we obtain 22221 ,babcy
baacx
+−
=+−
=
Therefore, the distance is 22
21
21
ba
cyxd
+=+=
(i)
(ii) x
y
O
),( 11 yxA
0=++ cbyax
15
Exercise [ Exercise 14.2] Find the area of the triangle in the following steps. (1) Shift the triangle by 1 leftward and find the coordinates of the new
position of points A, B, and C. (2) Find the length AH using the result of Example 14.4. (3) Find the length AH. (4) Find the area of triangle ABC.
Ans. Pause the video and solve the problem by yourself.
x
y
O )0,1(A
)1,6(B
)4,2(CH
16
Exercise [ Exercise 14.2] Find the area of the triangle in the following steps. (1) Shift the triangle by 1 leftward and find the coordinates of the new position of points A, B, and C. (1) Find the length AH using the result of Example 14.4. (3) Find the length AH. (4) Find the area of triangle ABC.
Ans. (1) The new positions are A(0, 0), B(5, 1) and C(1, 4). (2) The equation for the straight line CB is
(3) From Ex.14.4, length OH is (4) BC= Therefore, the area is
01943)1(15414 =−+∴−−−
=− yxxy
519
43
1922=
+
−=d
5)41()15( 22 =−+−
2195
519
21
=××
x
y
O )0,1(A
)1,6(B
)4,2(CH
x
y
O
H
)1,5(B
)4,1(C
Recommended