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

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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

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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

∗

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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)

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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

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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

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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

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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

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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

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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

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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