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1.
ANALYTICAL GEOMETRY:
THE LINE:
The distance between two points:
B(x2 ; y2)
A(x1 ; y1)
The distance between A and B defined by :
AB 2 = (x 2 - x 1) 2 + (y 2 - y 1) 2
AB
=  (x 2 - x 1) 2 + (y 2 - y 1) 2
Examples:
1. Find the distance between PQ if P(-1 ; 2) and Q(3 ; -4).
Solution:
AB =  (x 2 - x 1) 2 + (y 2 - y 1) 2
=  [3 - (-1)] 2 + (-4 - 2) 2
P(-1 ; 2)
=  (4) 2 + (-6) 2
=  16 + 36
Q(3 ; -4)
=  52
= 2  13 
2.
2. Prove that the triangle with vertices K(-1 ; -1) ; M(3 ; -2) and N(-2 ; 3) are isosceles.
Solution:
N(-2 ; 3)
K(-1 ; -1)
M(3 ; -2)
KM =  (x 2 - x 1) 2 + (y 2 - y 1) 2
=  [3 - (-1)] 2 + [-2 - (-1)] 2
MN =  (x 2 - x 1) 2 + (y 2 - y 1) 2
=  (-2 - 3) 2 + [3 - (-2)] 2
=  (3 + 1) 2 + (-2 + 1) 2
=  (-5) 2 + (3 + 2) 2
=  16 + 1
=  25 + 25
=  17 
=  50 / 5  2 
KN =  (x 2 - x 1) 2 + (y 2 - y 1) 2
=  [-2 - (-1)] 2 + [3 - (-1)] 2
=  (-2 + 1) 2 + (3 + 1) 2
=  1 + 16
=  17 
 Triangle KMN is isosceles because KM = KN =  17 
3.
3. The distance between A(1 ; 3) and B(5 ; a) is 5 units. Find a.
Solution:
AB =  (x 2 - x 1) 2 + (y 2 - y 1) 2
5
=  (5 – 1) 2 + (a – 3) 2
5
=  16 + a 2 – 6a + 9
(5) 2 = ( 16 + a 2 – 6a + 9) 2
25 = a 2 – 6a + 25
= a 2 – 6a
0
a = 0 or 6 
-----------------------------------------------------------------------------------------------------------Exercise:
1. Find the distance between the given pairs of points : (simplest surd form)
(1.1)
(2 ; 3) ; (4 ; 5)
(1.2)
(1.2) (6 ; 1) ; (-6 ; 6)
(1.3)
(1.3) (3 ; -7) ; (-1 ; 3)
2. Given the co-ordinates of the vertices of ABC ; in each case, determine :
(i)
(ii)
(iii)
the perimeter of the 
whether the  is equilateral , isosceles or scalene
whether or not the  is a right angled triangle
(2.1) A(1 ; -3) ; B(7 ; 3) ; C(4 ; 6)
(2.2) A(5 ; 1) ; B(1 ; 3) ; C(1 ; -2)
3. Determine the value(s) of x or y when:
(3.1) the distance of (x ; -3) from the origin is 5 units.
(3.2) the distance between (-7 ; y) and (-3 ; 4) is 4  5 units.
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4.
MIDPOINT OF A LINE :
If A = (x1 ; y1) and B = (x2 ; y2) then the co-ordinates of MIDPOINT (M) between A and B are :
M (x ; y) = x1 + x2 ; y1 + y2
2
2
Examples:
1. Find the midpoint between A(-2 ; -4) and B(3 ; 1).
Solution:
M (x ; y) = x1 + x2 ; y1 + y2
2
2
= -2 + 3 ; -4 + 1
2
2
1
3
= ( /2 ; - /2 ) 
2. P(4 ; 2) is the midpoint of QR where Q(3 ; -5) and R(a ; b). Find a and b.
Solution:
M (x ; y) = x1 + x2 ; y1 + y2
2
2
(4 ; 2) = 3 + a ; -5 + b
2
2
4 = 3 + a and
2 = -5 + b
2
2
3+a = 8
and
-5 + b = 4
a = 5
and
b = 9
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5.
GRADIENT AND INCLINATION OF A LINE:
GRADIENT (m) = y
x
=
INCLINATION / : tan  = m
(  0 ; 180)
y2 – y1
x2 – x1
EXAMPLES:
Determine
(a) gradient
(b) inclination of the line joining the points
(i) (-3 ; 5) and (2 ; 0)
(ii) (-2 ; -3) and (3 ;-1)
Solution:
(i) (a) gradient = y2 - y1 = 0 – 5
= -5 = -1 
x2 - x1
2 – (-3)
5
(b) Inclination: tan  = m
 = tan-1 -1
= -45o
= 180o – 45o
 = 135o 
(ii) (a) gradient = y2 - y1 =
x2 - x1
-1 – (-3)
3 – (-2)
=
2
5 
(b) Inclination: tan  = m
 = tan-1 2/5
 = 21.8o 
COLINEAR POINTS:
These are points that lie on the same straight line.
Examples:
A
B
C
D
A, B, C and D are collinear.
(gradients between the points will be the same)
Prove that A(7;5) ; B(1;-2) and C(-5;-9) are collinear.
Solution:
(m)AB = y2 - y1
x2 - x1
= -2 – (5)
1–7
= 7
6
(m)BC = y2 - y1
x2 - x1
= -9 – (-2)
-5 – 1
= 7
6
Gradients are the same; A, B, and C are collinear. 
(m)AC = y2 - y1
x2 - x1
= -9 – (-5)
-5 – 7
= 7
6 
6.
PARALLEL AND PERPENDICULAR LINES:
PARALLEL LINES:
For any two non-vertical parallel lines,
the GRADIENTS ARE EQUAL.
gradient l1 = gradient l2
1 = 2
Example : y = 2x – 7 and y = 2x + 10
m=2
m=2
 lines are parallel
PERPENDICULAR LINES:
Two lines are perpendicular if the
PRODUCTS OF THEIR GRADIENTS
ARE EQUAL TO –1.
m1 = m2
l1 = l2
Example : y = 3x – 10 and y = -1/3x + 15
m1 = 3
m2 = -1/3
m1 X m2 = 3 X –1/3
= -1
lines are perpendicular
EQUATION OF A LINE:
m = y – y1 / x – x1
y – y1 = m(x – x1)
NB!! To determine the equation of a
line, you need a gradient (m) and
any point (on the line)(x ; y)
7.
Examples : Determine the equations of the following lines :
1. with gradient - 1/4 through the point A(-2 ; 3)
Solution :
y – y1 = m(x – x1)
y – 3 = -1/4[x – (-2)]
y – 3 = -1/4 (x + 2)
y = -1/4x – ½ + 3
y = -1/4x + 5/2
2. through the point P(1 ; 2) and Q(-1 ; 3)
Solution:
m = y2 – y1
x2 – x1
m= 3–2
-1 – 1
m = - 1 /2
y – y1 = m(x – x1)
y - 2 = -1/2(x – 1)
y = -1/2x + ½ +2
y = - 1/2 x + 5/2 
3. Parallel to the line y = -2x + 7 through the point A(- 1/2 ; -2)
Solution:
A = (-1/2 ; -2)
m = -2
(parallel lines)
y – y1 = m(x – x1)
y – (-2) = -2[x – (-1/2)]
{m = -2 ; because lines are parallel}
y + 2 = -2(x + ½)
y = -2x –1 – 2
y = -2x – 3 
4. Perpendicular to the line y = -2x + 7 through the point A(- 1/2 ; 2)
Solution:
m = -2
m1 X m2 = -1 {perpendicular lines}
m2 = -1/m1
= -1/-2
m = ½
y – y1 = m(x – x1)
y - 2 = ½[ x – (-1/2)]
y - 2 = ½ (x + ½)
y = ½ x + 9 /4 
8.
INTERSECTION OF TWO LINES :
Example : Find the co-ordinates of the intersection of the lines : y = -2x + 2 and y = 5x – 19.
-2x + 2 = 5x – 19
-7x = 21
x = 3
Solution:
y = -2(3) + 2
y = -6 + 2
y = -4
co-ordinates = (3 ; -4) 
THE CIRCLE:
y
P (x ; y)
x
O
The equation of the circle at the center (0 ; 0) with radius (r) :
Examples : Determine the equation of the circle at center (0 ; 0) :
1. through the point A(2 ; 3)
Solution:
x2 + y2 = r2
(2)2 + (3)2 = r2
4 + 9 = r2
r2 = 13
x2 + y2 = 13 
x 2 + y 2 = r2
9.
2. with radius 4.
Solution:
x2 + y2 = r2
x2 + y2 = (4)2
x2 + y2 = 16 
3. through the point : (  2 ;  3).
Solution:
x2 + y2 = r2
( 2)2 + ( 3)2 = r2
2 + 3 = r2
r2 = 5
2
x + y2 = 5 
EQUATION OF A TANGENT:
(The same as the equation of a line i.e. y – y1 = m(x – x1)
INTERSECTION OF THE CIRCLE AND THE LINE : (Solve simultaneously )
Example:
Determine the co-ordinates where the circle x2 + y2 = 10 and the line 2y – x = 5 intersect.
Solution :
x2 + y2 = 10 …..(1)
2y – x = 5 …...(2)
From (2) : 2y – x = 5
x = 2y – 5 …..(3)
Substitute (3) into (1) :
(2y – 5)2 + y2 = 10
2
4y – 20y + 25 + y2 = 10
5y2 – 20y + 15 = 0
(5y – 5)(y – 3) = 0
y = 1 or y = 3 
Substitute y = 1 and y = 3 into (3) :
x = 2(1) – 5 or x = 2(3) - 5
x=2–5
x=6–5
x = -3
or
x=1 
 Co-ordinates are (-3 ; 1) and (1 ; 3)
10.
THE CIRCLE WITH ANY CENTRE (a ; b) AND RADIUS (r) :
The equation of the circle with centre (a ; b) and radius (r) is :
(x – a)2 + (y – b)2 = r2
Examples:
1. Determine the co-ordinates of the centre and the radius of each circle :
(1.1) (x – 1)2 + (y + 5)2 = 12
(1.2) x2 + 2x + y2 - 6y = 6
(1.3) 3x2 + 12x + 3y2 + 6y = 2
Solutions:
1. (1.1) centre = (1 ; - 5) ; radius =  12 = 2 3 
(1.2) x2 + 2x + ….. + y2 - 6y + …..
x2 + 2x + 1 + y2 - 6y + 9
 (x + 1)2 + (y - 3)2
 centre = (- 1 ; 3) ; radius
= 6 + ….. + …..
= 6 + 1 + 9
= 16
=  16 = 4 
(1.3)  3 :
x2 + 4x + y2 + 2y = 2/3
x2 + 4x + ….. + y2 + 2y + …..
x2 + 4x + 4 + y2 + 2y + 1
 (x + 2)2 + (y + 1)2
 centre = (- 2 ; - 1) ; radius =  17/3
= 2/3 + ….. + …..
= 2 /3 + 4 + 1
= 17/3

2. Determine the equation of the circle with center (- 2 ; 3) and passing through the point (0 ; 7).
Solution:
Centre (a ; b) = (-2 ; 3)
 (x + 2)2 + (y – 3)2 = r2 ….. (1)
Substitute point (0 ; 7) in (1) :
 (0 + 2)2 + (7 – 3)2 = r2
(2)2 + (4)2 = r2
r2 = 20
 equation (x + 2)2 + (y – 3)2 = 20 