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50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines

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BASIC KNOWLEDGE

1. Distance formula

The distance (d) between two points ),( 111 yxP and ),( 222 yxP can be calculated by the

following formula:

2

12

2

12 )()( yyxxd (1)

Proof:

Given right triangle ABC, from the Pythagorean Theorem, we have

222 BCACAB

Let AB = d,

2

12

2

12

2 )()( yyxxd

Taking the square root on both sides: 2

12

2

12 )()( yyxxd

Another useful form of the distance formula:

2

12

2

12 )()( yyxxd = 2

12

1212 )(1

xx

yyxx

= 12

21 xxk (2)

Where k is the slope of the line containing the two points and 12

12

xx

yyk

.

2. Finding the coordinates of a point P on AB between points A and B.

P is a point on AB between points A and B. If 1PB

AP , then we have the following

formulas to calculate the coordinates of P:

1

21 xxx , and

1

21 yyy (3)



When = 1, we get the midpoint formula:

The coordinates ),( mm yx of the midpoint of the line segment with endpoints ),( 11 yx and

),( 22 yx are2

21 xxxm

, and

2

21 yyym

. (4)

3. Point to line distance formula

(1). The distance from a point (x1, y1) to a line ax + by +c = 0 is one of those problems

that seems easy, but is very time consuming, unless you know the following formula:

22

11

ba

cbyaxd

(5)

Proof:

Method 1:

Drop perpendiculars to the line l from the given point ),( 000 yxP to

meet l at ) ,(b

caxxN

, to the x-axes to meet l at ) ,( 0

0b

caxxQ

,

and to P0Q from N to meet P0Q at ) ,( 0b

caxxM

.

The triangle P0QN in the picture is a right triangle, so the area can be

found in two ways:

NQdNMQP 2

1

2

10 , or

20

2

00

00

2

1)(

2

1xx

b

cax

b

caxdxx

b

caxy

.

Solving for d:



1

)))

2

2

0

00

0

2

0

2

02

2

00

0

2

0

2

0

00

0

b

axx

xxb

caxy

xxxxb

a

xxb

caxy

xxb

cax

b

cax

xxb

caxy

d

22

00 )

ba

b

caxyb

22

00

ba

cbyax

Method 2:

Find the distance from the point ),( 000 yxP to the line 0: cbyaxl .

0: cbyaxl meets x-axes and y-axes at

),0( ,)0 ,(b

cF

a

cE , respectively.

Drop perpendiculars to the line l from the given point

),( 000 yxP to meeting l at ) ,(b

caxxN

, to the x-axes to meet

l at ) ,( 00

b

caxxQ

,

So Rt∆P0QN ~ Rt∆EFO.

EF

QP

OE

NP 00

ab

bac

b

caxy

a

c

d

22

00

22

00

ba

cbyaxd

(2). Other forms of the formula of point to distance

(i). 2

0

2

022

00 yxba

cbyax

(6)

It tells us that the distance from a point (x0, y0) to the line l: ax + by = 0 is less than or

equal to the distance from the point to origin if the line l passes through the origin and the

point is not on the line.



Note that if we square both sides of (6), we get a famous inequality: Cauchy Inequality:

))(()( 2

0

2

0

222

00 yxbabyax .

(ii). 2

0

2

022

00 )()( yyxxba

cbyax

(7)

It tells us that all the line segments connecting a point that is not on the line l: ax + by + c

= 0 to a point that is on the line l, the perpendicular is the shortest.

(3). If line l1 01 cbyax and line l2 02 cbyax are parallel, the distance between

them is d and 22

21

ba

ccd

(8)

4. Angle formed by two lines

The slopes of line l1 and line l2 are k1 and k2, respectively. The angle formed by lines l1

and l2 is and

tan θ = 21

12

1 kk

kk

(9)

When 01 21 kk , = 2

.

The slopes of line l1 and line l2 are k1 and k2, respectively. The angle from lines l1 to l2 is

and

21

12

1tan

kk

kk

(10)

When 01 21 kk , = 2

.

5. Pattern in Reflections

(1). ),( 00 yxP is a point. The image of P under reflections:

(a). In the xaxes ),( 00 yx

(b). In the yaxes ),( 00 yx



(c). In the x = a ),2( 00 yxa

(d). In the y = a )2,( 00 yax

(e). In the y = x ),( 00 xy

(f). In the y = x ),( 00 xy

(g). In the y = x + m ),( 00 mxmy

(h). In the y = x + n ),( 00 xnyn

(i). In the point A(a, b) )2,2( 00 ybxa

(j). In the Ax + By +C = 0

)()(

022:),(

1010

1010

11

xxByyA

Cyy

Bxx

Ayx

The figures for (a) to (f) are showing below.

(a) (b) (c)

(d) (e) (f)

(2). The reflection of the line Ax + By +C = 0 in the point P( a, b ): Ax + By – (2aA +

2bB +C) = 0



Example 1: Find the image of P(– 5,13) under the reflection in line 2x – 3y – 3 = 0.

Solution:

Let the image be Q ),( 11 yx and R be the midpoint of PQ.

13

52

)3(2

3133)5(2

22

PR

So 13

1042 PRPQ .

The slope of the line PQ is 2

3PQk .

By the formula (2), we have

13

1045 )

2

3(1 1

2 xPQ

Therefore 1651 x 111 x .

We also have: 2

3

16

13

5

13 1

1

1

y

x

ykPQ 111 y .

So the image Q is (11, –11).

Example 2: A line passing through (1, 2) is cut by two parallel lines 4x + 3y + 1 = 0 and

4x + 3y + 6 = 0. The length of the segment cut is 2 . Find the equation of the line.

Solution:

Let k be the slope of the line.

Let the line be )1(2 xky . The two intersecting points with two parallel lines be

),( 111 yxP and ),( 222 yxP .

0134

)1(2

yx

xky

43

731

k

kx .



0634

)1(2

yx

xky

43

1232

k

kx .

Since ,1 12

2

21 xxkPP 243

73

43

1231 2

k

k

k

kk .

Simplifying we get: 07487 2 kk

Solve for k: 7

11 k , 72 k .

The line seeking is )1(7

12 xy and )1(72 xy .

Or x + 7y – 15 = 0 and 7x – y – 5 = 0.

Example 3: Show that 222 cyx and 222 cba if 22 )()( byaxcbxay

with 0bxay . x, y, a, b, and c are real numbers.

Solution:

Rewrite the given equation as

cbyax

bxay

22 )()(

This is the distance from (0, 0) to the line 0)()(: bxayyxaxybl .

Points (a, b) and (x, y) are on the line.

Therefore ,22 cba cyx 22 or 222 cba , 222 cyx .

Example 4: Find the smallest value of yxyxu 4222 if x – 2y + 2 = 0.

Solution:

From yxyxu 4222 , we have 22 )2()1(5 yxu .

Examining the point (1, – 2) and the line x – 2y + 2 = 0:

5

7

5

241)2()1( 22

yx



Squaring both sides: 5

49)2()1( 22 yx .

That is 5

495 u

5

24u .

The smallest value of5

24is achieved when y = 2 and x = 1

5

7 .

Example 5: Line 5)12()1( mymxm always contains the point P for any real

value of m. What is the coordinates of the point P?

Solution: (9, – 4).

Write the given equation in the following form: 0)5()12( yxyxm

For any real number of m, the line always goes through the point of intersection of lines

012 yx and 05 yx .

Solving x and y:

012 yx

05 yx x = 9 and y = – 4.

The answer is: (9, – 4).

Example 6: If the y-intercept of the line 1)2()1( 2 mymmxm is 1, what is the

value of the real number m?

Solution: 3.

Since the y-intercept is 1, (0, 1) is on the line. Substitute (0, 1) into the equation, we have:

122 mmm 0322 mm

Solving for m: m = 3 or m = – 1.

If m = – 1, we have m + 1 = 0 and 1)2()1( 2 mymmxm is not a line. So m = 3

is the answer.

Example 7: Find the distance between the point (2, – 3) and the line 3x − 4y −12 = 0.

Solution: 6/5.



By the point to line distance formula: .5

6

5

6

43

12)3(4)2(322

d

Example 8: What is the closest that the line 4x + 5y = 20 comes to the origin?

Solution:

Since 4x + 5y − 20 = 0, we have .41

4120

41

20

54

20)0(5)0(4

22

d

Example 9: Find the distance between the parallel lines 2x − 3y = 12 and 2x – 3 y= 36.

Solution:

Pick any point on one line and plug it into the distance formula for a point and a line. So

let’s pick (6, 0) from the first line. Now

.13

1324

13

24

32

36)0(3)6(2

22

d

Example 10: What is the closest that the line 5

1

7

4 xy comes to a lattice point? (2003

Duke Math Meet).

Solution:

The line, in standard form, is 20x − 35y + 7 = 0 , so the distance to any lattice point (x, y)

is 22 3520

73520

yx.

So the closest point will occur when the numerator is minimized. Since it is an absolute

value, and must be an integer, we need to see if this quantity can be zero, or if not 0, what

is the smallest possible value. Clearly 20x − 35y will always be a multiple of 5, so we

could make it – 5, making the entire numerator 2. This is the smallest possible value for

the numerator, so the shortest distance is .325

652

655

2

3520

222



Example 11: P(x, y) is a point on a circle of radius of 1. If the center of the circle is (– 2,

0), what is the smallest value of 1

2

x

y?

Solution: 4

33.

Let kx

y

1

2. It is clear that k is the slope of the line connecting points P(x, y) and A(2,

1). The smallest value of 1

2

x

y is the smaller one of the two slopes of the tangent lines to

the circle.

The equation of the line AP is 0)2( kykx

By the point to line distance formula, we have:

11

22

2

k

kk

4

33k

Example 12: What is the shortest distance between the circle x2

+ y2

= 25 and the line

3x + 4y = 48?

Solution: 5

23.

First notice that the closest the line gets to the origin is 5

48

43

4)0(4)0(3

22

, so it is

more than 5 units from the origin. Now subtract the radius of the circle, yielding the

distance .5

235

5

48

Example 13: Find tangent of the angle formed by two lines both passing through the

origin. It is known that these two lines trisect the segment in the first quadrant of the line

2x + 3y = 12.

Solution:

Let A(6, 0) and B(0, 4).



Since 2CB

AC,

3

8

21

42

221

6

C

C

y

x

.

Since 2DA

BD,.

3

4

21

4

421

42

D

D

y

x

C (2, 3

8), D(4,

3

4).

3

1 OCk ,

3

4 ODk

13

9

3

4

3

11

3

1

3

4

tan

.

Example 14: Find the smallest possible value of 8422)( 22 xxxxxf .

Solution:

2222 )20()2()10()1()( xxxf

That is, to find the smallest possible sum of the distances from (x, 0) to points (1, 1), (2,

– 2). That is, the distance between (1, 1) and (2, – 2), which is 10 .



PROBLEMS

Problem 1: What is the closest that the line 7

2

5

3 xy comes to a lattice point?

Problem 2: (1966 AMC) Let m be a positive integer and let the lines 13x + 11y = 700

and y = mx – 1 intersect in a point whose coordinates are integers. Then m can be:

(A) 4 only (B) 5 only (C)6 only (D) 7 only

(E) one of the integers 4, 5, 6, 7 and one other positive integer

Problem 3: (1982 AMC) A vertical line divides the triangle with vertices (0, 0), (1,1) and

(9,1) in the xy-plane into two regions of equal area. The equation of the line is x =

(A) 2.5 (B) 3.0 (C) 3.5 (D) 4.0 (E) 4.5

Problem 4: (NC Math Contest 1997) Let P be the point (3,2). Let Q be the reflection of

P about the x-axis, let R be the reflcetion of Q about the line y = – x and let S be the

reflection of R through the origin. Then PQRS is a convex quadrilateral. What is the area

of PQRS?

(a) 14 (b) 15 (c) 16 (d) 17 (e) 18

Problem 5: (NC Math Contest 2001) Find the slope of the line with a positive rational

slope, which passes throug the point(6, 0) and at a distance of 5 from (1, 3). Write the

slope in the form b

a, where a and b are relatively prime. What is the sum of a and b?

a. 24 b. 23 c. 22 d. 21 e. none of these.

Problem 6: Line 1 (l1) x – 2y + 3 = 0 intersects x-axes at A. Line 2 (l2) is obtained by

rotating l1 45 about A. If the rotation is counterclockwise, find the equation for l2.

Problem 7: Find a + b if the line ax + y + 1 = 0 is parallel to the line (a + 1)x + by + 2 =

0 and two lines have the same distance to the origin.

(A) .3

7 (B) 3. (C)

3

7 or 3. (D) any value except 3, .

3

7

Problem 8: The vertices are A(4, 1), B(7, 5), and C(4, 7) of triangle ABC. Find the

equation of the angle bisector of A.



Problem 9: Find the equation of the line passing through p(1, 2) and the length of its

segment cut between two lines 0134 yx and 0634 yx is 2 .

Problem 10: Find the image of P(2, 1) under the reflection in line 022 yx .

Problem 11: Find the image of line 02 yx under the reflection in line

033 yx .

Problem 12: Find the image of the line 043 yx under the reflection in (1, 1).

Problem 13: Find the greatest possible value of 13452)( 22 xxxxxg .

Problem 14: Find the equation of the line passing through point P(5, 4). It is known

that P divides the segment A(x, 0) and B(0, y) of the line between the x and y axis into the

ratio of 1:2.

Problem 15: A line segment is between two lines: 0103:1 yxl and

082:2 yxl . The midpoint of the segment is P(0, 1). Find the equation of the line

containing the segment.

Problem 16: Find the equation of the angle bisector of the angle formed by lines

02 yx and 047 yx .

Problem 17: Find a if the distance between line l1: 2x + 3y – 6 = 0 and line l2: 4x + 6y +

a = 0 is 26

135.

Problem 18: Find the equation of the line passing through point P(2, 1) and having a

distance 2 from the origin.

Problem 19: (2001 AMC 12) Points A = (3, 9), B = (1, 1), C = (5, 3), and D = (a, b) lie in

the first quadrant and are the vertices of quadrilateral ABCD. The quadrilateral formed by

joining the midpoints of AB , BC , CD , and DA is a square. What is the sum of the

coordinates of point D?



(A) 7 (B) 9 (C) 10 (D) 12 (E) 16

Problem 20: (2003 AMC 12 A) A set S of points in the xy-plane is symmetric about the

origin, both coordinate axes, and the line y = x. If (2, 3) is in S, what is the smallest

possible number of points in S?

(A) 1 (B) 2 (C) 4 (D) 8 (E) 16

Problem 21: (2004 AMC 12 A) Let A = (0, 9) and B = (0, 12).

Points 'A and 'B are on the line y = x, and 'AA and 'BB intersect at

C = (2, 8). What is the length of 'AA ?

(A) 2 (B) 22 (C) 3 (D) 22 (E) 23

Problem 22: (2004 AMC 12 B) The point (3, 2) is rotated 90o clockwise around the

origin to point B. Point B is then reflected in the line y = x to point C. What are the

coordinates of C?

(A) (3, 2) (B) (2, 3) (C) (2, 3) (D) (2, 3) (E) (3, 2)

Problem 23: (2000 AMC12) The point P = (1, 2, 3) is reflected in the xy-plane, then its

image Q is rotated by 180 about the x-axis to produce R, and finally, R is translated by 5

units in the positive-y direction to produce S. What are the coordinates of S?

(A) (1, 7, – 3) (B) (–1, 7, –3) (C) (–1, –2, 8) (D) (–1, 3, 3) (E) (1, 3, 3)

Problem 24: (1981 AMC 12) The lines L and K are symmetric to each other with respect

to the line y = x. If the equation of line L is y = ax + b with a ≠ 0 and b ≠ 0, then the

equation of K is y =

(A) bxa

1

(B) bxa

1

(C) a

bx

a

1 (D)

a

bx

a

1 (E)

a

bx

a

1

Problem 25: (2001 NC Math Contest Algebra II) Given A(1,2) , B(5, 1) , and C(2, 2) ,

find the equation of the angle bisector at A.



a. 5x y 7 b. 7x y 2 c. 7y x 2 d. y x 3 e. none of the above

Problem 26: (1980 AMC 12) The equations of L1 and L2 are y = mx

and y = nx, respectively. Suppose L1 makes twice as large an angle

with the horizontal (measured counterclockwise from the positive x-

axis) as does L2, and that L1 has 4 times the slope of L2. If L1 is not

horizontal, then mn is

(A) 2

2 (B)

2

2 (C) 2 (D) – 2

(E) not uniquely determined by the given information

Problem 27: (1990 AIME) A triangle has vertices P = (– 8, 5), Q = (–15, –19) and R =

(1, –7). The equation of the bisector of P can be written in the form ax + 2y + c = 0.

Find a + c.



SOLUTIONS TO PROBLEMS

Problem 1: Solution:

Let m and n are the coordinates of a lattice point. The distance from this lattice point (m,

n) to the given line is

347

10)53(7

34

7

1053

125

9

7

2

5

3

nm

nmnm

d

We know that )53(7 nm is a multiple of 7. We also know that 3 and 5 are relatively

prime. Since we want to get the smallest value for d, or the smallest value for the

numerator, so )53(7 nm = 7 is the best value we could get (m = – 2 and n = – 1).

238

343

347

3min d

Problem 2: Solution: (C).

Substituting y from the second equation into the first gives 13x + 11(mx – 1) = 700, so

that

.1113

793

1113

711 2

mmx

Since x is to be an integer, the denominator 13 + 11m must be a divisor of the numerator,

and its only divisors are 1, 3, 32, 79, 3·79, 3

2·79. Our task now is to find a positive integer

m such that

13 + 11m = d, or ,11

13

dm

Where d is one of these divisors. Since m > 0, we see that d >13, so the only divisore we

need to test are the last three:

(i) if d = 79, d – 13 = 66, and 611

66m

(ii) if d = 3·79 = 237, d – 13 = 224 is not divisible by 11

(iii) if d = 32·79 = 711, d – 13 = 698 is not divisible by 11.

We conclude that m = 6 is the only positive integer yielding a lattice point for the

intersection of the given lines.



Problem 3: Solution:(B).

Method 1 (official solution): In the adjoining figure,

ABC is the given triangle and x = a is the dividing line.

Since area ,4)8)(1(2

1ABC the two regions must

each have area 2. Since the portion of

∆ABC to the left of the vertical line through vertex A has area less than area ,2

1ABF

the line x = a is indeed right of A as shown. Since the equation of line BC is ,9

xy the

vertical line x = a intersects BC at a point E: ).9

,(a

a Thus

area ),9(9

12

12 a

aDEC

or (9 – a )

2 = 36.

Then 9 – a = ± 6, and a = 15 or 3. Since the line x = a must intersect ∆ABC, x = 3.

Method 2 (our solution):

Extend CA to meet y-axes at D. We see that ∆CDA ∆CGE. Therefore, we get:

CG

CD

GE

DB

CGGE

91

9

CGGE

We also see that the area of ∆CDA = the area of ∆CAB +

the area of at ∆DAB = the area of ∆CGE 2 + the area

of at ∆DAB.

22

22

DBADGECGDBCD

2

112

22

19

GECG 4GECG 4

9

CGCG

362 CG 6CG .

DG = DC – CG = 9 – 6 = 3.

Therefore x = 3.


The points P, Q, R, and S are respectively (3, 2), (3, – 2), (2, – 3) and ( – 2, 3). The



rectangle with vertices( – 2, 3), (3, 3), (3, – 3) and ( – 2, – 3 ) has area 30. When the

three corners, with 2

1,

2

5 and 12 are subtracted, the remaining quadrilateral has area15.

Problem 5: Solution: b.

5DE , 34BE , 5

18OC .

zDBE xDBO

3

5)tan(tan yxz .

5

3

6

5

18

tan y . 15

8

tantan1

tantantan

yz

yzx .

Answer: 8 + 15 = 23.


Let the slope of l2 be k and l1 be .

We have 2

1tan and 3

tan1

tan1)45(tan

k .

Since the intersecting point is A(– 3, 0) , the equation of line 2 is then y = 3(x + 3) or

3x – y + 9 = 0.

Problem 7: Solution: (A).

Since two lines are parallel, ab = a + 1 (1)

Since they have the same distance to the origin,

222 )1(

2

1

1

baa

(2)

Solving the system of equations (1) and (2) gives:

2

1

b

a or

.2

3

1

b

a

When a = 1, b = 2, these two lines are overlapped.



Therefore ,3

1a b = – 2 .

3

7 ba


Let p(x, y) be a point on the angle bisector AD.

Method 1:

The distance from p to AC is the same as the distance from p

to AB.

The equation for AB: )4(3

41 xy or 01334 yx .

The equation for AC: )4(3

41 xy or 01643 yx .

22 )3(4

1334

yx

22 43

1643

yx

1334 yx )1643( yx

037 yx or 0297 yx .

It is easy to know that 037 yx is the equation of the exterior angle of A.

The equation is then 0297 yx .

Method 2:

By the angle bisector theorem, 2

1

10

5

68

43

)71()44(

)51()74(

22

22

22

22

AC

AB

DC

BD

So 2

1

DC

BD , then:

3

10

2

11

)4(2

17

1

21

xxDx , and

3

17

2

11

72

15

1

21

yyDy

The equation is:



12

12

1

1

xx

yy

xx

yy

7

43

10

13

17

4

1

x

y 471 xy

0297 yx


Method 1:

Let the equation be )1(2 xky . The points of intersection with two lines are A and B,

respectively.

Solving the system equations of

)1(2

0134

xky

yx

and

0134

0634

yx

yx

The coordinates of A and B are:

)43

85 ,

43

73(

k

k

k

k and )

43

108 ,

43

123(

k

k

k

k.

Since 2)43

5()

43

5( 22

k

k

kAB , 2

43

15 2

k

k.

The above equation has the form of: (7k + 1)(k – 7) = 0 7

1k or 7.

So the equation is 0153 yx or 057 yx .

Method 2:

The distance between two parallel lines is 134

61

22

d .

Since the length of the segment cut by them is 2 , the angle formed the line in question

and two parallel lines is 45.



Let the equation be )1(2 xky .

21

12

1tan

kk

kk

1

3

41

3

4

k

k

Solving for k: ,7

11 k 72 k .

So the equation is 0153 yx or 057 yx .


Let the image be P (x, y).

1)2

1(

2

1

022

22

2

2

x

y

yx

032

082

yx

yx

Solve for x and y:

.5

19

5

2

y

x

)5

19 ,

5

2( 'P .


Method 1:

Solving

033

02

yx

yx gives the point of intersection: )

2

9 ,

2

5(

Let the slope of the line in question be k:

k

k

31

3

131

13

Solve for k: k = – 7 or k = 1(extraneous).

So the equation of the line is )2

5( 7

2

9 xy or 0227 yx .

Method 2:

Since )2,0(0 P is a point on 02 yx , its image is ),( yxP .



032

2

23

130

2

yx

x

y

Solve for x:

1

3

y

x or

2

0

y

x (extraneous).

So 7

32

5

12

9

k

The equation is )2

5( 7

2

9 xy .


Let the equation we want to find be bxy 3

We know that d, the distances from (1, 1) to these two parallel lines are the same.

10

2

13

4132

d .

Therefore10

2

10

2

b

Solving for b: b = 0 or b = – 4 (extraneous).

So the equation is y = 3x.


2222 3)2(2)1()( xxxg

To find the greatest difference PBPA of the distances from P to A (1, 2) and P to B(2,

3). The desired point P must be the point of intersection of the line segment connecting A

and B and x-axes.

2)32()21( 22 AB




Let the equation of the line be: 1 b

y

a

x. The coordinates of A and B are A(a, 0), B(0,

b.).

Since 2

1 PB

AP, considering P(5, 4), by formula (3), we obtain:

2

15a , b = 12.

The equation is 06058 yx .


Let the equation be y = kx +1. The points of intersection of line y = kx + 1 with

0103 yx and 082 yx are )13

14,

13

7(

k

k

kA and )

2

28,

2

7(

k

k

kB ,

respectively.

By the midpoint formula (4), we get 4

1k .

The equation is 14

1 xy or 044 yx .


Let P(x, y) be a point on the angle bisector. The distances from point P to two given lines

are the same. So

25

47

2

2

yxyx.

The equations are: 0326 yx and 073 yx .


We know that the distance between l1: 2x + 3y – 6 = 0 and l2: l2: 2x + 3y + 2

a= 0 is

26

135.

By the formula (8) 22

21

BA

CCd

, we have:

26

135

94

|62

|

a

.

Therefore 2

5|6

2| a

. Solving for a gives a＝ – 7 or – 17.




If the slope of the line does not exist, the equation of the line is x = 2.

If the slope of the equation exists, let it be k.

The equation is then written as y + 1 = k(x – 2), or kx – y – 2k – 1 = 0.

By the point to line distance formula, we have1

|12|

2

k

k＝2.

Solve for k: 4

3k .

The equation of the line is: )2(4

31 xy or 3x – 4y – 10 = 0.

The equations are x – 2 = 0 or 3x – 4y – 10 = 0.


Let the midpoints of sides AB , BC , CD , and DAbe M, N, P, and Q, respectively. Then

M = (2, 5) and N = (3, 2). Since MN has slope – 3, the slope of MQ must be 3

1, and MQ

= MN = 10 . An equation for the line containing MQ is thus ,3

)2(5

xy or

.3

)13(

xy Therefore Q has coordinates of the form (a, ).

3

)13( a Since 10MQ , we

have 10)53

)13(()2( 22

aa 10)

3

2()2( 22

aa

10)2(9

10 2 a 9)2( 2 a 32 a .

We know that Q is in the first quadrant, so a = 5 and Q = (5, 6). Since Q is the midpoint

of AD and A = (3, 9), we have D = (7, 3), and the sum of the coordinates of D is 10.

Problem 20: Solution: (D).

The set S is symmetric about the line y = x and contains (2, 3), so it must also contain (3,

2). Also S is symmetric about the x-axis, so it must contain (2, – 3) and (3, – 2). Finally,

since S is symmetric about the y-axis, it must contain (– 2, 3), (– 3, 2), (– 2, – 3), and (– 3,

– 2). Since the resulting set of 8 points is symmetric about both coordinate axes, it is also

symmetric about the origin.



Problem 21: Solution: (B).

Line AC has slope 2

1 and y-intercept (0, 9), so its equation is

.92

1 xy

Since the coordinates of A satisfy both this equation and y = x, it follows that A = (6, 6).

Similarly, line BC has equation y = 1 – 2x + 12, and B = (4, 4).

Thus A B 22)46()46( 22 .

Problem 22: Solution: (E).

The rotation takes (3, 2) into B = (2, 3), and the

reflection takes B into C = (3, 2).


Reflecting the point (1, 2, 3) in the xy-plane produces (1, 2, –3). A half-turn about the x-

axis yields (1, –2, 3). Finally, the translation gives (1, 3, 3).


If (p, q) is point on line L, then by symmetry (q, p) must be a point on K. Therefore, the

points on K satisfy x = ay + b.

Solving for y yieldsa

b

a

xy .

Problem 25: Solution: (e).

We find that side AC and AB both have length 5, so the angle bisector is also the median,

passing through A(1,2) and the midpoint (3/2,3/2), of side BC. This makes the slope 7

and the equation y 7(x 1) 2 7x 9.




In the adjoining figure, L1 and L2 intersect the line x = 1 at B and A, respectively; C is the

intersection of the line x = 1 with the x-axis. Since OC = 1, AC is the slope of L2 and BC

is the slop of L1. Therefore AC = n.

Since OA is an angle bisector, AB

AC

OB

OC .

This yields n

n

OB 3

1 and OB = 3.

By the Pythagorean theorem ,9)4(1 2 n so 2

2n .


Method 1:

Let the bisector be PT and the coordinates of T be (x, y).

25247 22 PQ , .15129 22 PR

3

5

15

25

PR

PQ

TR

QT

By formula (3), we have x = – 5,2

23y .

The equation for PT: )8(2

115 xy or 11x + 2y + 78 = 0.

Therefore a + c = 11 + 78 = 89.

Method 2:

Let the slope of the angle bisector be k. The slopes of PR and PQ are kPR and kPQ,

respectively.

3

4

81

57

PRk .

7

24

815

519

PQk

By the formula (1), we have:

k

k

k

k

7

241

7

24

3

41

3

4

22k2 + 117k – 22 = 0 ,

2

11 k

11

2 k .



Since the angle bisector is the interior angle bisector, 2

11k .

)8(2

115

xy or 11x + 2y + 78 = 0

Therefore a + c = 11 + 78 = 89.



EXTRA EXERCISES

Exercise 1: In the xy- plane, how many lines whose x-intercept is a positive prime

number and whose y-intercept is a positive integer pass through the point (4, 3)? (1994

AMC 12).

Answer: 2.

Exercise 2: If the line whose x-intercept and y-intercept have the smallest sum passes

through the point P(1, 4), find the slope of the line. It is known that both the x-intercept

and y-intercept are positive.

Answer: 2 .

Exercise 3: Find the distance between the point (1, – 1) and the line x − y + 1 = 0.

Answer: 2

23.

Exercise 4: What is the positive value of m if a line with the slope of 1 passing through

the point (0, m) and is tangent to the circle x2 + y

2 = 2?

Answer: 2.

Exercise 5: What is the positive slope of the line that goes through the point (–2, 0) and

is tangent to the circle x2 + y

2 = 1? Express your answer in the simplest radical form.

Answer: 3

3.

Exercise 6: P(x, y) is a point on a circle of radius of 2. If the center of the circle is (3, –

2), what is the sum of the smallest and the greatest values of 2x – 3y?

Answer: 24.

Exercise 7: P(x, y) is a point on a circle of radius of 3 . If the center of the circle is (2,

0), what is the largest value of x

y?



Answer: 3 .

Exercise 8: Among all lines passing through the point P(3, 5), line l has the greatest

distance from the origin. Find the equation of the line l.

Answer: 3x + 5y – 34 = 0

Exercise 9: The area of the triangle bounded by the x-axis, the y-axis and the line x – 2y

+ 2k = 0 is 1. What is the greatest value of k?

Answer: k = 1.

Problem 11: The area of the triangle bounded by the x-axis, the y-axis and the line l has

the smallest possible value. What is the slope of the line l?

Answer: –1/2.

Exercise 10: Find the equation of the image of the line 033 yx rotating 30°

counterclockwise with respect to point P( – 3, 0).

Answer: 0333 yx .

Exercise 11: Find the equation of the line passing through point P(1, 2) . It is known

that the angle formed by this line and line 3x – y – 1 = 0 is 45°.

(A) 2y – x + 5 = 0 or y + 2x = 0; (B) 2y – x + 5 = 0 or y – 2x + 4 = 0;

(C) 2y + x + 3 = 0 or y + 2x = 0; (D) 2y + x + 3 = 0 or y – 2x + 4 = 0.

Answer: (A) 2y – x + 5 = 0 or y + 2x = 0.

Exercise 12: Find the range for a if the distance from point (4, a) to line 4x – 3y = 1 is

less than or equal to 3.

(A) (0, 10) (B) ( ,3

1

3

31); (C) (0, 10); (D) (– , 0) (10, +)

Answer: (C).

Exercise 13: Find a if the distance from point P(2, 3) to line ax + (a – 1)y+3 = 0 is 3.



Answer: 7

3or – 3.

Exercise 14: Find a if the distance from point (a, –2) to line 0443 yx is 1.

Answer: 3

7 or

3

17

Exercise 15 . Find the equation of the line passing through point P(1, 2) and having an

angle of 45° to line 0352 yx .

Answer: 01773 yx

Exercise 16: Find the equation of the line passing through the intersecting point of lines

0533 yx and 053 yx . The line and line 03 yx have an angle of

30°.

Answer: 2

5y or xy 3 .

Exercise 17: Find the coordinates of the image of P(a, b) under the reflection in line x –

y + 1 = 0.

(A) (b – 1, a + 1); (B) (b + 1, a –1); (C) (a –1, b + 1); (D) (a + 1, b – 1).

Answer: (A).

Exercise 18: Find the equation of the image of lx + 2y – 4 = 0 under the reflection in line

x = 2.

(A) x – 2y + 2 = 0; (B) x + 2y – 6 = 0; (C) x + 2y – 8 = 0; (D) x – 2y = 0.

Answer: D.

Exercise 19: Find a and b if point A(a + 2, b + 2) is the image of the point B(a + 2, b + 2)

under the reflection in line 4x + 3y = 11.

Answer: a = 4 and b = 2.



Exercise 20: Find the equation of the image of 2x – y + 3 = 0under the reflection in line

x + y – 1 = 0.

Answer: x – 2y + 4 = 0.

Exercise 21: Find the equation of the image of 2x – y + 3 = 0under the reflection in point

A(2,3).

Answer: 2x – y – 5 = 0.

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