Upload
lammien
View
300
Download
20
Embed Size (px)
Citation preview
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 1
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)
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 2
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:
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 3
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 4
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
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 5
(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
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 6
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 .
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 7
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
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 8
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 9
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
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 10
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).
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 11
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 .
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 12
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 13
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?
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 14
(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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 15
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 16
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 17
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.
Problem 4: Solution:
The points P, Q, R, and S are respectively (3, 2), (3, – 2), (2, – 3) and ( – 2, 3). The
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 18
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.
Problem 6: Solution:
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 19
Therefore ,3
1a b = – 2 .
3
7 ba
Problem 8: Solution:
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:
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 20
12
12
1
1
xx
yy
xx
yy
7
43
10
13
17
4
1
x
y 471 xy
0297 yx
Problem 9: Solution:
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 21
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 .
Problem 10: Solution:
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 .
Problem 11: Solution:
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 .
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 22
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 .
Problem 12: Solution:
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.
Problem 13: Solution:
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
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 23
Problem 14: Solution:
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 .
Problem 15: Solution:
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 .
Problem 16: Solution:
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 .
Problem 17: Solution:
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 24
Problem 18: Solution:
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.
Problem 19: Solution: (C).
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 25
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).
Problem 23: Solution: (E).
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).
Problem 24: Solution: (E).
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 26
Problem 26: Solution: (C).
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 .
Problem 27: Solution:
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 .
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 27
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 28
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?
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 29
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 30
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.
50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines
http://www.mymathcounts.com/50-AMC-Lectures-Program.php 31
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.