Embed Size (px)
Citation preview
Instructors : Dr. Lee Fong LokDr. Leung Chi Hong
Student Name : Chan Mei Shan
Student ID : S98039740
Topic : Coordinate Geometry (F3 Mathematics)
Distance Formula Slope Straight Line Drawing
Review:Distance and Slope
Equation of Straight Lines
Points of Division
Perpendicular and Parallel Lines
Intersection of Two Straight Lines
Equation of Special Lines
Two Point Form
Point-Slope Form
Slope-Intercept Form
Intercept Form
General FormBack =>
Topic: Distance and Slope (Review)
Back=>
1 2 3 4 5 6 7 8-1-2
1
2
3
4
5
-1
-2
-3
-4
-5
x
y P (x1, y1)
Q (x2,y2)
221
221 )()( yyxxPQ
y1 – y2
distance ?distance ?By Pythagoras Theorem,
x1 – x2
y1
y2
x2x1
DistanceDistance
ExampleExample A = (-4, 2) B=(2, -4)
72
6)6(
))4(2()24(
22
22
(x1,y1) (x2,y2)
221
221 )()( yyxxAB
Class WorkClass Work (a) Find AB if A=(4,0) and B=(9,a) (Give the answer in terms of a.)
(b) If AB= , find a.34
2
22
25
)0()49(
a
aAB
(a)
3 3
9
3425
3425
2
2
2
ora
a
a
a(b)
2 min 2 min
1 2 3 4 5 6 7 8-1-2
1
2
3
4
5
-1
-2
-3
-4
-5
x
yP (x1, y1)
Q (x2,y2)
x1 – x2
y1 – y2
slope slope ??
21
21
xx
yy
mslope PQ
SlopeSlope
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y (4, 5)
(1, 1)
3
414
15slope
ExampleExample
(x2,y2)
(x1,y1)
21
21
xx
yyslope
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y B(a2, ab)
A(b2, –ab)
Class WorkClass Work
22
22
2
ba
abba
ababslope
Slope of AB?
3 min
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(-4, 2) (2,2)
06
0
)4(2
22
slopeIf a line//If a line//xx--
axisaxis
slope = 0slope = 0
ExampleExample
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(2,-3)
(2,2)0
522
32
slope
If a line // If a line // yy--axisaxis
slope is slope is undefinedundefined
ExampleExample
zero!
Back=>
Topic: Point of Division
Back=>
(9-y)
(y -2)
(8-x)
(x-1)
A= (1,2), B = (8,9) and AC : CB = 3 : 4Find C(x,y)
4 5 6 7 821-1-2
5
6
7
8
9
2
1
x
y
3
3
4
Point C?
A(1,2)
B(8,9)
C(x,y)
4
3
43
1483
148343
144383
x
xx
xx
3
4
2-y
y-9 and
3
4
1
8
x
x
43
2493
249343
244393
y
yy
yy
B(8,9)
(9-y)
(8-x)
4
C(x,y) D
(y -2)
(x-1)A(1,2)
C(x,y)
3
E
∵ ΔBCD ～ ΔCAE
B(8,9) A(1,2)C(x,y) 43
ObservationObservation
x =3 x 8 +4 x 1
3 + 4 43
1483
x
Calculation
43
2493
yy =
3 x 9 +4 x 2
3 + 4
Section Formula
A (1, 2)
B (4, 8)P (a, b)1
2
221
)1)(2()4)(1(a
421
)2)(2()8)(1(b
What are the coordinates of What are the coordinates of P ?P ?
Ans: P = (2, 4)Ans: P = (2, 4)
ExampleExample
A (a, b)
B (4, 9)P (3, 1)
52
Find the values of a and Find the values of a and bb
ClassWorkClassWork
5 min
2
1
2202125
2)4)(5(3
a
a
a
19b
b245725
b2)9)(5(1
Solution
A (3, -7)
B (5, 3)P (a, b)
11
Find the coordinates of point PFind the coordinates of point P
411
)5)(1()3)(1(a
22
37b
Class WorkClass Work
3 min
A(x1,y1)
B (x2,y2)P (a, b)
11
2
2
21
21
yyb
xxaThen
P is the mid-point of AB
1
21
2
1 2 B (5, -2)
C (-2, -5)
A (3, 4)
P
G
12
24
42
53
b
a
(4, 1)
121
)5)(1()1)(2(q
221
)2)(1()4)(2(p
Let P = (a, b) & G = (p, q)
13
524q
23
253p
ExampleExample
B (x2,y2)
C (x3, y3)
A (x1,y1)
G(x,y)
ObservationObservation
Given : G is the centroid of △ABC
Then:
3
3213
321
yyyy
xxxx
Back=>
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
A (-3, 4)
B (2, 1)
P
Find the ratio of AP : Find the ratio of AP : PB.PB.
ChallengeChallenge
Answer =>
3
2k
k320k1
)3(k)2)(1(0
Let AP : PB = 1 : kLet AP : PB = 1 : k
Solution
Back =>
Topic: Equations of Special Lines
Back=>
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(1, 3) (3, 3)(-3, 3)
(-1, -3) (2, -3)(-5, -3)
x = -3
y = 3
y = -3
Horizontal LinesHorizontal Lines
(2,1)y = 1
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
yVertical LinesVertical Lines
(2, 2)
(2, 0)
(-3, -3)
x = -3 x = 2x = -1
x = -3(-1, -4)
(a, b) LL22
LL11
PP
Ans:Ans:1.1. LL11 : : x = ax = a
L L22 : : y = by = b
2.2. P=P= (0, b)(0, b)
Class WorkClass Work
0
x
y
Find Find
• The equations The equations ofofLL1 1 and Land L2;2;
• The The coordinatescoordinates of point P.of point P.
3 min
(1,1)
(3,3)
(-3,-3)
y = x
(1,-2)
(-2,4)
(-1,2)
y =-2xStraight lines Passing through origin
Straight lines Passing through origin
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(a,b)
xa
by
ObservationObservation
Class WorkClass Work
(6,7)
x
yLL
22
LL
11
(4,-3)
Ans:Ans:
LL11 : :
LL2 2 ::
xy4
3
xy6
7
3 min
Find the equatiFind the equations of Lons of L1 1
and Land L2.2.
Back=>
Topic: Two-Point Form
Back=>
1 2 3 4 5 6 7 8 9-1
1
2
3
4
5
6
7
8
9
-1
x
y
15
38
07y4x5
5x512y44
5
1x
3y
Find the equation Find the equation of L.of L.
1x
3y
B(5, 8)
A(1, 3)
P(x, y)
LL
MMAPAP = = M MABAB
B(5, 8)
A(1, 3)
P(x, y)
1 2 3 4 5 6 7 8 9-1
1
2
3
4
5
6
7
8
9
-1
x
y
L: 5x-L: 5x-4y+7=04y+7=0MMBPBP = = M MABAB
07y4x5
25x532y44
5
5x
8y15
38
5x
8y
Will the result be the Will the result be the
same if We consider same if We consider
MMBP BP instead of instead of MMAP AP ? ?
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(-4, 4)
(2, -3)
LL
(-2, b)
PP
)k,0(PLet
3
2k
04k6)0(7
)3
2,0(P,thus
3
5b
04b6)2(7
equationtheoint)b,2(Put
04y6x7
28x724y66
7
4x
4y
24
)3(4
)4(x
4y
(a) Find the equation (a) Find the equation of L.of L.
(b)(b) Find the Find the value of b.value of b.
(c)(c)Find the coordinates Find the coordinates of P.of P.
L: 7x + 6y + 4 = 0
ExampleExample
23
)1(2
)3(
2
x
y
(a)(a)Find the equation of the Find the equation of the straight line joining (-3, 2) straight line joining (-3, 2) and (2, -1). and (2, -1). (b)(b)Does the point (7, -4) lie on Does the point (7, -4) lie on
the straight line ?the straight line ?(c)(c) State whether the point (3, -State whether the point (3, -
2) lies on the straight line 2) lies on the straight line or not.or not.
.S.H.R
0
12021
1)4(5)7(3.S.H.L
The point (7, -4) lies on the The point (7, -4) lies on the straight line.straight line.
.S.H.R
2
1109
1)2(5)3(3.S.H.L
The point (3, -2) does not lie The point (3, -2) does not lie on the line.on the line.
ExampleExample
L: 3x - 5y + 1 = 0
0153
931055
3
3
2
yx
xyx
y23
)1(2
)3(
2
x
y
Class WorkClass Work
(a)(a)Find the equation of the straight Find the equation of the straight line which passes through (0,0) line which passes through (0,0) and (-4,-6). and (-4,-6). (b)(b)If the point A(a,3) lies on L, find a. If the point A(a,3) lies on L, find a.
Solution.
023
644
6
)4(0
)6(0
)0(
0
yx
xyx
y
x
y(a) (b)
2
0)3(23
023
int)3,(
a
a
yx
oaPut
7 min
Back=>
Topic: Point-Slope Form
Back=>
1 2 3 4 5 6 7 8 9-1
1
2
3
4
5
6
7
8
9
-1
x
y
1x
7y
04yx3
3x37y
3
Point-slope Point-slope FormForm
A(1, 7)
LL
B(x, y)
slope = 3
MMABAB = Slope = Slope
Find the equation of the Find the equation of the line which passes through line which passes through (-1,-5) and has slope -3 :(-1,-5) and has slope -3 :
ExampleExample
Working?
083
335
31
5
yx
xyx
y
SolutionSolution3
)1(
)5(
x
y
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
(a) Find the equation of (a) Find the equation of L.L.
(b)(b) What is the What is the value of b ?value of b ?
Put B(2, b) into the equationPut B(2, b) into the equation
L: x + 3y - 3 = 003y3x
3x6y33
1
3x
2y
3
1
)3(x
2y
3
1b
03b32
ExampleExample
(-3, 2)
LL
B (2, b)3
1slope
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
yFind (a) The equation of L.Find (a) The equation of L.
(b) The coordinates of P(b) The coordinates of P
(c) The coordinates of Q(c) The coordinates of Q LL
P (a, 2)
Q
3
4slope
(-2, 0)
Class WorkClass Work
10 min
Solution.
08y3x4
8x4y33
4
2x
y
3
4
)2(x
0y
(a) 5.3a
08)2)(3(a4
Put P(a, 2) into L
therefore P = (-3.5, 2)
(b)
Let Q = (0, b)
3
8b
08b3)0(4
Q = (0, )3
8
(c)
Back=>
Topic: Slope-Intercept Form
y-intercepty-intercept
x-in
tercep
t
x-inter
cept
LL1 1
(0, 3)
(-2, 0)
LL1 1 cuts the y- cuts the y-axis axis
at point (0,3)at point (0,3)
LL1 1 cuts the x- cuts the x-axis axis
at point (-2,0)at point (-2,0)
InterceptsIntercepts
-3 -2 1 2 3 4 5-1
-2
-1
1
2
3
4
5
6
-3
x
y
-4 0
1 2 3 4 5 6 7 8 9-1
1
2
3
4
5
6
7
8
9
-1
x
y
0x
4y
3
What is the What is the equation of equation of
L ?L ? slopeslope
y-intercepty-intercept
4x3y
x34y
3x
4y
(x, y)
LL
(0, 4)4
slope = 3
Slope-intercept FormSlope-intercept Form
ExampleExample (a)(a)Find the equation of the Find the equation of the straight line with straight line with y-y-intercept –1intercept –1 and and slope –3slope –3 in in the slope-intercept form.the slope-intercept form.
(b)(b)What is theWhat is the slope slope and the and the y-y-interceptintercept of the straight of the straight line line y = 3x - 7y = 3x - 7 ? ?
3
1erceptinty,
3
2slope
y=y=3x3x11
(c)(c) What is theWhat is the slope slope and the and the y-y-interceptintercept of the straight of the straight line line 2x + 3y – 1 = 02x + 3y – 1 = 0 ? ?
7erceptinty
3slope
3
1x
3
2y
1x2y3
01y3x2
Slope-intercept Form
ExampleExample L : kx + 3y – 2k = 0 with slope –2.L : kx + 3y – 2k = 0 with slope –2.
(a) Find the value of k .(a) Find the value of k .
3
k2x
3
ky
k2kxy3
0k2y3kx
6k
23
k
thus,thus,
(b)(b)What is theWhat is the y-intercepty-intercept of L ?of L ?
43
)6)(2(3
k2erceptinty
Slope-intercept Form
7 minSlope-intercept Form
Class WorkClass Work Find the value of k for the Find the value of k for the following straight line, L.following straight line, L.
L : 3x + 4y + k = 0 withL : 3x + 4y + k = 0 with y-y-intercept 5intercept 5..
4
kx
4
3y
kx3y4
0ky4x3
20k4
k5
thus,thus, 20k
0k)5(4)0(3
Alternatively,Alternatively,
Put (0, 5) into the Put (0, 5) into the equation of L.equation of L.
Ans.
Back=>
Topic: Intercept Form
Back=>
1 2 3 4 5 6 7 8-1-2
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
22
33
B(2, B(2, 0)0)
A(0, A(0, 3)3)
6y2x3
x36y22
3
x
3y20
03
0x
3y
LL
y-intercepty-interceptx-interceptx-intercept
13
y
2
x thusthus,,
P(x, y)
MMAPAP = M = MABAB
What is the equation What is the equation of L ?of L ?
1 2 3 4 5-1-2-3-4-5
1
2
3
4
5
-1
-2
-3
-4
-5
x
y
Find the equation of L in Find the equation of L in intercept form.intercept form.
ExampleExample
13
y
4
x
Ans:
Do the point (4, 6) and (12, Do the point (4, 6) and (12, 9) lie on L ?9) lie on L ?
x-interceptx-intercept
-4
y-intercepty-intercept 33
L
.S.H.R
1
213
6
4
4.S.H.L
The point (4, The point (4, 6) lies on L.6) lies on L.
Put (4, 6) Put (4, 6) into the into the equationequation
Put (12, 9) into Put (12, 9) into the equationthe equation
.S.H.R
0
333
9
4
12.S.H.L
(12, 9) does not (12, 9) does not lie on L.lie on L.
(a)(a) Convert Convert 7x + 4y + 28 = 07x + 4y + 28 = 0 into the into the intercept formintercept form..
17
y
4
x
128
y4
28
x7
28y4x7
028y4x7
(b)(b) What are the What are the x-interceptx-intercept and and
y-intercepty-intercept of the straight of the straight line ?line ?
x-intercept = -4x-intercept = -4 and and y-intercept y-intercept = -7= -7
ExampleExample
Find the area of the Find the area of the shaded region.shaded region.
125
110
5
10
2
1052
01052
yx
yx
yx
yx
The area of the shaded The area of the shaded region isregion is
units.sq52
)2)(5(
ExampleExample
x
y
0
L : 2x+ 5y + 10 L : 2x+ 5y + 10 = 0= 0
Intercept form
-2
-5
Class WorkClass Work (a) Find the intercepts of L(a) Find the intercepts of L11..(b) Find the equation of L(b) Find the equation of L22..
(c) Find the area of the shaded region.(c) Find the area of the shaded region.
x
y
0
LL1: 1: 3x + 5y-3x + 5y-15=015=0
LL22
-2-2
10 min
Solution.
13
y
5
x
115
y5
15
x3
15y5x3
015y5x3
x-intercept = 5 x-intercept = 5
and y-intercept = 3 .and y-intercept = 3 .
(a)
06y2x3
6y2x3
13
y
2
x
The equation The equation of Lof L22 is is
(b)
The area of the shaded The area of the shaded region isregion is
(c)
unitssq.5.102
)3)(7(
x
y
0
LL1: 1: 3x + 5y-3x + 5y-15=015=0
LL22
-2-2 5
3
Back=>
Topic: General Form
Back=>
Ax + By + C = 0Ax + By + C = 0
Convert into the general forConvert into the general form.m.
52
y3
5
x
050y15x2
50y15x2
52
y3
5
x
General FormGeneral FormGeneral FormGeneral Form
Class WorkClass Work
Convert into the general forConvert into the general form.m. 2
3
4x
5y2
022y4x3
12x310y42
3
4x
5y2
What are the What are the slopeslope and the and the y-intercepty-intercept of the straight of the straight line line 4x – 3y + 7 = 04x – 3y + 7 = 0 ? ?
3
7x
3
4y
7x4y3
07y3x4
3
7erceptintyand
3
4slope
ExampleExample
Find the equation of L in the Find the equation of L in the general form.general form.
07yx2
7x2y
x
y
0
-7-7slope = -2slope = -2
LL
ExampleExample
Find the Find the x-interceptx-intercept and the and the y-intercepty-intercept of the straight line of the straight line 12x – 7y + 4 = 012x – 7y + 4 = 0..
17/4
y
3/1
x
14
y7
4
x12
4y7x12
04y7x12
7
4erceptintyand
3
1erceptintx
ExampleExample
Back=>
Topic: Parallel Lines and Perpendicular Lines
Back=>
IfIf LL11 // L // L2 2 , ,
thenthenmmL1L1 = m = mL2 L2
What will happen ifTwo lines L1 and L2 Are parallel?A FACT to know...
Conversely, if mmL1L1 = m = mL2 L2
ThenThen LL11 // L // L2 2
ExampleExample
LL
11
LL
22
slope = 2slope = 2)
3
5,3(
)3
7,1(
Determine whether LDetermine whether L11 // L // L22
22
413
)37
(35
m2
Since mSince m11 = m = m22= = 2, 2, then, Lthen, L11 is is parallel toparallel to L L22
Find the equation of LFind the equation of L22
02:
2
2
2
yxL
xyx
y
x
y
0(0, 0)
LL11 : m = 2 : m = 2
LL
22
ExampleExample
mmL2L2 = m = mL1L1 = 2 = 2
20
0
x
y
By point-slope form,
4 min
(a) Find the equation of L(a) Find the equation of L22..
-5
x
yLL11 : slope = : slope = -3-3
0(-5, 0)(-5, 0)
LL
22015yx3
15x3y
35x0y
(b) Does the point (-3, -5) lies on L(b) Does the point (-3, -5) lies on L22 ? ?
L.H.S. L.H.S. ==
= 3(-3) + (-5) = 3(-3) + (-5) + 15+ 15
= 1= 1
R.H.S.R.H.S.Thus, (-3, -5) does not Thus, (-3, -5) does not lie on Llie on L22
Class WorkClass WorkMore...More...
Find the equation of LFind the equation of L22..
x
y
0
LL11 : 4x + 3y + 12 = 0 : 4x + 3y + 12 = 0LL22
-6(-6, 0)
4x34
y
12x4y3
012y3x4
02434
24433
4
6
0:2
yx
xyx
yL
Step 1: Express LExpress L11 into slope intercept into slope intercept
form.form.Step 2: Find the slope of Find the slope of L L22 Step 2: Find the slope of Find the slope of L L22
mL2 = mL1 = 3
4
Step 3: Use point-slope form to find LUse point-slope form to find L22.. ExampleExample
6 min
Steps : Steps : 1.1.Express the given line into Express the given line into slope-intercept form. slope-intercept form.
2.2.Find the slope of L1.Find the slope of L1.
3. 3. Use point-slope form to find Use point-slope form to find the equation of the line. the equation of the line.
Find the equation of the line L1 which Find the equation of the line L1 which is parallel to 3x + 2y – 5 = 0 and passes is parallel to 3x + 2y – 5 = 0 and passes through (4, -1).through (4, -1).
Class WorkClass Work
010y2x3
12x32y223
4x1y
25
x23
y
5x3y2
05y2x3
Solution.
Step 1:
Express the given lineExpress the given lineinto slope-intercept form.into slope-intercept form.
Step 2: Find mFind mLL..
mL = 3
4
Step 3: Use point-slope Use point-slope form to find the form to find the equation equation
IfIf LL11 L⊥ L⊥ 2 2 , ,
thenthenmmL1L1 x m x mL2 L2 =-1=-1
One more FACT...
Conversely, if mmL1L1 x m x mL2 L2 =-1=-1
ThenThen LL11 L⊥ L⊥ 2 2
Find the coordinates of P.Find the coordinates of P.(Hint: Let P = (a,0)(Hint: Let P = (a,0)
5.0a
3a5.3
1)3a
)47
(0(2
thus, P = (-0.5, thus, P = (-0.5, 0)0)
LL
11
x
y
0
LL
22
slope = 2slope = 2
)4
7,3(
P
ExampleExample
∵ ∵ LL11 L⊥ L⊥ 22
∴ ∴ mmL1L1 x m x mL2 L2 =-1=-1
Find the equation of LFind the equation of L22..
52
3
1032
01023
xy
xy
yx
0532
42933
2
2
3:2
yx
xyx
yL
Step 1: Express LExpress L11 into slope intercept into slope intercept
form.form.Step 2: Find the slope of Find the slope of L L22 Step 2: Find the slope of Find the slope of L L22
mL2 = -1÷mL1
=-1÷3
2
2
3
Step 3: Use point-slope form to find LUse point-slope form to find L22.. ExampleExample
x
y
0(-2, -3)
LL11 : : 3x-2y +10 =03x-2y +10 =0
LL22
6 min
Steps : Steps : 1.1.Express the given line into Express the given line into slope-intercept form. slope-intercept form.
2.2.Find the slope of L.Find the slope of L.
3. 3. Use point-slope form to find Use point-slope form to find the equation of the line. the equation of the line.
Find the equation of the line L which is pFind the equation of the line L which is perpendicular to 3x - 2y + 6 = 0 and pases terpendicular to 3x - 2y + 6 = 0 and pases through (-4, 3).hrough (-4, 3).
Class WorkClass Work
01732
82933
2
4
3
yx
xyx
y
32
3
632
0623
xy
xy
yx
Solution.
Step 1:
Express the given lineExpress the given lineinto slope-intercept form.into slope-intercept form.
Step 2: Find mFind mLL..
mL = 3
2
2
31
Step 3: Use point-slope Use point-slope form to find the form to find the equation equation
Back=>
Find the equation of the perpendicular Find the equation of the perpendicular bisector of the line segment joining (3, -5) bisector of the line segment joining (3, -5) and (-7, 9).and (-7, 9).
[ Ans.: 5x - 7y + 24 = 0 ][ Ans.: 5x - 7y + 24 = 0 ]
Steps : Steps : 1.1.Find the coordinates of the midpoint.Find the coordinates of the midpoint.2.2.Find the slope of the line segment. Find the slope of the line segment. 3.3.Find the slope of the perpendicular bisectorFind the slope of the perpendicular bisector4.4.Use point-slope form to find the equation ofUse point-slope form to find the equation of
the line.the line.
ChallengeChallenge
Back=>
Topic: Point of Intersection
Back=>
x
y
0
y = 7
x = 5
PP
What are the coordinates of P ?What are the coordinates of P ?
A. P = (-5, -7)A. P = (-5, -7)
B. P = (-5, 7)B. P = (-5, 7)
C. P = (5, -7)C. P = (5, -7)
D. P = (5, 7)D. P = (5, 7)
E. P = (7, 5)E. P = (7, 5)
You are wrong !You are wrong ! Don’t give up …Don’t give up … Try it again …Try it again …
return
SorrSorry !y !
Correct !
Return
What are the coordinates of P ?What are the coordinates of P ?
A. P = (-5, 7)A. P = (-5, 7)
B. P = (5, 7)B. P = (5, 7)
C. P = (7, 2)C. P = (7, 2)
D. P = (7, 13)D. P = (7, 13)
E. P = (13, 7)E. P = (13, 7)x
y
0
x = 36
PP y = 3x – 8
What are the What are the coordinates of P ?coordinates of P ?
)2(
)1(
05y2x3
016y5x2
025y10x15
032y10x4
3x
057x19
:)4()3(
)4(
)3(x
y
0
2x – 5y + 16 = 0
3x + 2y + 5 = 0
PPP = (-3, 2)P = (-3, 2)
2y
4y2
05y2)3(3
:)2(oint3xSub
ExampleExample
011yx3
6x52
y
Find the coordinates of the point of intersection ofFind the coordinates of the point of intersection of
011yx3
6x52
y
)2(
)1(
:)2(oint)1(Sub
5x
17x5
17
0116x52
x3
:)2(oint5xSub
4y
011y)5(3
The coordinates are The coordinates are (5, 4)(5, 4)
ExampleExample
P = (1, 2)P = (1, 2)
What are the coordinates of P ?What are the coordinates of P ?
y
0
3x – 4y + 5 = 0
PP
2
4
LL
4yx2
14y
2x
L :L :
05y4x3
4yx2
05y4x3
16y4x8)2(
)1(
:)2()1(
2
1
16511
y
x
x
ExampleExample
Back=>
