Transcript A2 :Lines and Circles

A2 :Lines and Circles
I
(Lines)
A2.1 Lines Parallel to coordinate axes:
• Horizontal line–parallel to x-axisy is given by
y=b,where(0,b) is y-intercept.
y=0 gives x-axis
(0,b)
• Vertical line-parallel
to y-axis is given by
x=a,where (a,0) is x-intercept.
x=0 gives y-axis.
x=a
y=b
(a,0)
Fig1
x
• Non Vertical Lines :
Take A(x1,y1) ,B(x2,y2)
on the line. Let P(x,y)
be a general point on the
line. From Fig2 ∆ACB
and ∆BQP are similar.
P(x,y)
y
A(x1,y1)
B(x2,y2)
Run
Q(x,y2)
Rise
C(x2,y1)
x
Fig 2
:. CB/AC=QP/BQ= Rise/Run =
VerticalChange/Horizontal Change = const (1)
(1)Characterizes straight lines,and the constant is
called the gradient or slope of the straight line .it
is independent of the choice of A& B.
• If Run≠0 then
Gradient = Rise/Run =Vertical
Change/Horizontal Change
(2)
• If Run=0 (No horizontal Change)
we get a vertical line with infinite slope.
• If Rise=0 (No vertical change) we get a
horizontal line with slope=0.
• A2.2 Various Equations of lines:
(i) With slope=m and passing throughA(x1’y1):
m=(Rise from A to P)/(Run from A to P)=
P(x,y)
(y-y1)/(x-x1)
(y-y1)=m(x-x1)
Rise
Or y= y1+ m(x-x1) (3) A(x1,y1)
(x,y1)
Run
Or y=mx+(y1-mx1)
Fig 3
(ii) With gradient m and y- intercept (0,c) A particular
case of (i)
With x1=0 , y1=c
y=mx+c
:. (3) gives:
y= m x+ c
(4)
(iii) Passing through A(x1,y1), B (x2,y2) (0,c)
y-intercept
Here m=[ y2-y1/ x2-x1])
Rewrite (3) as
0
x-intercept
y-y1 = [(y2-y1)/x2-x1)]=(x-x1) (5)
[Do Example(1.1) and activities (1.1)‫(ـــــــ‬1.4)]
• (iv) Note a straight line is given by a linear
equation
A x+By+C=0
(6)
And conversely .(slope=-A/B) y-intercept(0,-C/B),
x-intercept (-C/A,0)
A2.3 Relations between lines:
If ℓ1 has slope m1 and ℓ2 has slope m2:
• If m1= m2 then ℓ1 //ℓ2: and if ℓ1∩ℓ2=ø then
ℓ1and ℓ2 don’t meet and if ℓ1∩ℓ2≠ ø then ℓ1Ξℓ2
• If m1≠m2 then ℓ1 and ℓ2 intersect in one point
only .
• Perpendicular lines: ℓ1 with slope m, and ℓ2
with slope ℓ2 are perpendicular if and only if
m1 m2= - 1, or one line is horizontal (with
slope = 0) and the other is vertical( with
Infinite slope). [Do activities 1.5 , 1.6 ]
•
Example: For the given pairs of lines find points
of intersection , if any.
a) ℓ1: 3x+2y=3, ℓ2:6x+4y=5
b) ℓ1: 3x+2y=3, ℓ3:6x+4y=6
c) ℓ1:3x+2y=3, ℓ4:2x-3y=-11
Solutions : In a) & b) m1=-3/2,m2=m3=-6/4=m1.
So lines ℓ1 , ℓ2 and ℓ3 are parallel .
But in a) 3x+2y=3, 6x+4y=5
has no solution . So ℓ1 does’nt intersect ℓ2.
Whereas in b) 3x+2y=3 is the same equation as
6x+4y=6 So ℓ1Ξℓ2. In c) m1= -3/2, m4=2/3
:. m1m4=-1 and ℓ1┴ℓ4. Solving the equations we
get x=-1, y=3 . Therefore ℓ1and ℓ4 intersect at
(-1,3).
[Do activities (1.7) , (1.8)].
II
CIRCLES
B(x2,y2)
• A2.4; Distance formula:
d
By Pythagoras’ formula
(x ,y )
A(x ,y )
The distance d between
x
any two points A(x1,y1)
Fig 5
and B(x2,y2) is given by
d2= (x2-x1)2 +(y2-y1)2
(7)
• A circle with centre C(a,b) and radius r is
given by (x-a)2 +(y-b)2 =r2 (8)
[Do activities 2.3 , 2.4 ]
y
2
1
1
1
• Midpoint Rule: The midpoint of the segment
between A(x1,y1) and B(x2,y2) has coordinates
(1/2(x1+x2), 1/2(y1+y2)).
• To find the equation of the circle determined by 3
noncollinear points A,B,C:
(i)Find midpoints of AB, BC.
(ii) Find equations of the lines perpendicular to each
segment through its mid point
(iii) The point of intersection of these lines gives the
centre of the circle.
(iv) The distance between the centre and either point
A,B or C gives the radius .
[ Do example 2.1 and activity 2.5].
A2.5 General Equation of circle :
Expanding brackets in (8) :
(x-a)2 + (y-b)2 = r2 gives an equation of the
form x2 +y2 +Ax +By +C = 0
(9)
a second degree equation in x and y with no
mixed term xy.
Conversely (9) represents a circle (or an
empty set ). This could be shown by
completing the square .
A2.6 Completing the square :
For x2 +2px we write:
x2 +2px = (x +p)2 –p2 (10)
• Ex: Verify that the equation :
x2 – 4x + y2 + 2y + 4 =0 represents a circle,
find its centre and radius .
Solution : complete the square in x and y :
x2 – 4x= (x-2)2 -4; y2 +2y =(y+1)2 -1
:. x2 -4x +y2 +2y +4 =(x -2)2 -4 +(y +1)2 -1+4=0
(x -2)2 + (y +1)2 =1
Equation of a circle with radius =1 and centre
C(2, -1).
[ Do activity 2.6 ].
A2.7 Intersections of circles and lines .
A given circle intersects a line y
in
ℓ1
either :
ℓ2
(1) Two points (e.g.ℓ1)
ℓ3
x
(2) one point (tangent)e.g.ℓ2
Fig 6
(3) No point (e.g.ℓ3)
• To find the points of intersection:
(i) From equation of line solve for one variable
(ii) Substitute in equation of circle.
(iii) Solve the resulting quadratic equation.
[ Do example 2.3 and activity 2.7].
• Two given circles intersect either in :
(i) Two points (a) or
a
(ii) one point (b) or
b
(iii) no point (c)
c
Fig 7
To find points of intersection solve the two
equations simultaneously to get a linear
equation then substitute in either circle as
above .
III
Trigonometry
A2.8 Sine and cosine :
Let S1 = { (x,y): x2+y2=1} be the unit circle
with centre at O and radius 1. P(x,y)Є S1
y
P(x,y)
Let θ be the angle between
(o,y)
positive x-axis and OP,
θ
(x,o)
measured anticlockwise .
0
Define:
Fig 8
cos θ=x, sin θ=y (11)
Θ is measured in radians or in degrees.
2π radians = 360 degrees (12).
x
From definition it follows that ,for all θ:
cos2 θ +sri2 θ =1
(13)
cos (θ +2n π) =cos θ ,n any integer (14)
sin (θ +2nπ) =sin θ , n any integer (15)
-1 ≤ sin θ≤ 1, -1≤ cos θ ≤ 1
(16)
Also:
y
cos θ<o
Sin θ>o
cos θ>o
Sin θ>o
x
cos θ<o
Sin θ<o
cos θ>o
Sin θ<o
Fig 9
A2.9
y
Reflections:
y
y
1
o
θ
-θ
Q=(y,x)
P(x.y)
Q=(-x,y)
P(x.y)
1
x
θ
X=y
P(x.y)
θ
x
o
x
π-θ
Q=(-x,y)
Fig 10 shows reflections of the points
P(x,y) on the circle about:
A. x –axis to give Q (x,y).
B. y –axis to give Q (- x y).
C. The line y=x to give Q (y x).
½ π-θ
Using these symmetries we can deduce that
for all θ:
cos(-θ)=cosθ(even),sin(-θ)=-sin θ(odd) (17)
cos (π-θ)= -cos θ, sin(π-θ)= sin θ
(18)
cos (π/2-θ)= sinθ, sin (π/2-θ)= cos θ (19)
[Do activity
3.2 ]
A2.10 Certain Values of sinθ and cosθ :
Given in Table :
θ
0
π/6 π/4 π/3 π/2
π
3π/2
2π
cosθ
1 √3/2 √2/2 1/2
0
-1
0
1
sinθ
0
1/2 √2/2 √3/2
1
0
-1
0
A2.11 Rest of Trigonometric functions :
tan θ= sinθ/cosθ, cotθ= cosθ/sinθ= 1/tanθ
sec θ=1/cosθ, cosecθ=1/sinθ
Their values and properties could be
deduced from those of cos θ and sin θ.
A2.12 Calculations with Triangles .
y
A
P(x,y)
y=sinθ
sinθ
1
θ
0
X=COSθ
x
X
B
Fig 11 Shows part of the unit circle with a
point P(x,y) on it and a right angled triangle .
Triangles OPX and OAB are similar . Hence
OB/OA=OX/OP=cosθ, BA/OA=XP/OP= sinθ.
In Δ 0AB :
OA is called hypotenuse (hyp)
OB is called adjacent (adj)
AB is called opposite (opp)
So we have in a right angled triangle :
sinθ = opp/hyp ; cosθ = adj/hyp
tanθ = opp/adj , cotθ = adj/opp
secθ= hyp/adj , cosecθ=hyp/opp .
[Do activity
3.5] .
Parametric equations:
A2.13 parametric equations of lines:
x and y are given as functions of a third
variable t , called the parameter.
a) For a line with slope m and passing thru
(x1,y1)a suitable parameterization is:
x=t+x1
, y=mt+y1.
(20)
Here (x,y)=(x1,y1) when t=o and t gives the
run from (x1,y1) to (x,y)
b) For a line through (x1,y1) and (x2,y2) a
suitable parameterization is:
x=x1+t (x2-x1),y=y1+t(y2-y1) (21)
When t=o (x,y)=(x1,y1)
y
and when t=1 (x,y)=(x2,y2)
t
.
.A
t=0
.B
(x2,y2)
(x1,y1)
(x,y)
0
[Do activity 4.1]
*Eliminating the parameter t results in a linear
equation in x and y
[Do example 4,1 and activity 4,2]
x
A2.4 Parameteic equations of circles
For a circle of radius r and ceutre C (a,b)
a suitable parameterization is given by:
x=a+r cos (kt)
o≤kt≤2π
y=b+r sin (kt)
o≤t≤2π/k
(22)
Restricting range of t gives portions of a circle.
(22) Could be written as
x=a+r cos θ , y=b+r sin θ , where θ=kt
[Do activities 4.4 & 4.5]