Coordinate Geometry I
Y-axis
Q(x2,y2)
Distance PQ: d  (x2  x1)2  (y2  y2 )2
y2

x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y1
P(x1,y1)
X-axis 
0
x1
x2
Gradient of PQ : m 
y 2  y1
x 2  x1

BASIC Point Formulas
Distance, Midpoint, Gradient
By Mr Porter
YOU NEED TO
KOWN THE
FORMULAE !
Distance between 2 Points P(x1, y1) and Q(x2, y2)
(x1, y1)
(x2, y2)
Example 1: Find the distance between A(1,-3) and B(5,7).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
1) Formula : Dis tance PQ : d  (x2  x1 )2  (y2  y2 )2
2) Show substitution into formula clearly.

AB: d  (x2  x1 )2  (y2  y2 )2
d  (5 1)2  (7  3)2
B(5,7)
10
4
A(1,-3)
Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
d  116



YOU NEED TO
KOWN THE
FORMULA !
Use calculator to
evaluate (5-1)2 + (7--3)2
Simplify the surd.
d  4  29
d  2 29
Theexact distance between A(1, -3) and B(5, 7) is
AB  2 29


(x1, y1)
(x2, y2)
Example 2: Find the distance between P(-3,5) and Q(5,-5).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
1) Formula : Dis tance AB : d  (x2  x1 )2  (y2  y2 )2
2) Show substitution into formula clearly.

P(-3, 5)
AB: d  (x2  x1 )2  (y2  y2 )2
d  (5  3)2  (5  5)2
8
10
d  164


Q(5,-5)
Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1

YOU NEED TO
KOWN THE
FORMULA !
Use calculator to evaluate
(5--3)2 + (-5-5)2
Simplify the surd.
d  4  41
d  2 41
Theexact distance between P(-3, 5) and B(5, -5) is


PQ  2 41
Mid-Point of 2 Points P(x1, y1) and Q(x2, y2)
(x1, y1)
(x2, y2)
Example 1: Find the Mid-Point of A(1,-3) and B(5,7).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
1) Formula: (average point / co-ordinates)
x  x y  y 
Mid  point of PQ :  2 1 , 2 1 
 2
2 
2) Show substitution into formula clearly.

B(5,7)
10
4
A(1,-3)
Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1

YOU NEED TO
KOWN THE
FORMULA !
x  x y  y 
Mid  point of AB   2 1 , 2 1 
 2
2 
5 1 7 3 
 
,

 2
2 
6 4 
  , 
2 2 

 3,2
The mid-pointof A(1, -3) and B(5, 7) is (3, 2).

(x1, y1)
(x2, y2)
Example 2: Find the mid-point of P(-3,5) and Q(5,-5).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
P(-3, 5)
1) Formula :
x  x y  y 
Mid  point of PQ :  2 1 , 2 1 
 2
2 
2) Show substitution into formula clearly.
8

x  x y  y 
Mid  point of AB   2 1 , 2 1 
 2
2 
10
Q(5,-5)
5  3 5  5 
 
,

 2
2 
2 0 
  , 
2 2 

Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1

 1,0
The mid-point 
of P(-3,5) and Q(5,-5) is (1, 0).

YOU NEED TO
KOWN THE
FORMULA !
Gradient, m, of line joining 2 Points P(x1, y1) and Q(x2, y2)
(x1, y1)
(x2, y2)
Example 1: Find the gradient, m, of the line passing through points A(1,-3) and B(5,7).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
1) Formula : grad, m 
y2  y1
x2  x1
YOU NEED TO KOWN
THE FORMULA !
2) Show substitution into formula clearly.

B(5,7)
grad, m 
y2  y1
x2  x1
7  (3)
5 1
10
m
4
5
Always try to write the gradient, m,
m
as a simplified IMPROPER fraction
2
grad, m 
10

A(1,-3)
4


Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
m  tan
The 
gradient of the line passing through A(1,-3)
and B(5,7) is :
5

Gradient : m 
2
(x1, y1)
(x2, y2)
Example 2: Find the gradient, m, of the line passing through points P(-3,5) and Q(5,-5).
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or
realistic for the problem.
P(-3, 5)
1) Formula : grad, m 


Q(5,-5)
Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
m  tan

YOU NEED TO KOWN
THE FORMULA !
2) Show substitution into formula clearly.
8
10
y2  y1
x2  x1


y2  y1
x2  x1
5  5
grad, m 
5  (3)
10
m
8
5 Always try to write the gradient, m,
m
as a simplified IMPROPER fraction
4
grad, m 
The 
gradient of the line passing through P(-3,5)
and Q(5,-5) is :
5

Gradient : m 
4

Exercise
(x , y )
(x , y )
1) Given the two points A(-3,7) and B(5,1) find the
a) distance AB
b) point C, the mid-point of interval AB
c) gradient, mAB, of the line joining AB
1
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or

realistic for the problem.
2
1
2
a) Dis tan ce : d 

x 2  x1   y 2  y1 
2
1 (3)
 32
2
2
 5 1

B(5,1)

x 2  x1 y 2  y1 
b) Mid  point C  
,
 2

2 
Distance PQ: d  (x2  x1)2  (y2  y2 )2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
5  (3) 1 7 
C  
,
 2

2 
2 8 
  , 
2 2 

 1, 4
m  tan
 The mid-point C is (1,4)


y 2  y1
x 2  x1
2
m
Simplify all surds!
4 2

The distance AB is d = 4 2

A(-3,7)
c) Gradient, m 

1 7
5  (3)
6
8
3
m
4
m


3
The gradient of line AB is m 
4


YOU NEED
TO
KOWN THE
FORMULAE !
Exercise
(x , y )
(x , y )
2) Given the two points P(-5,-3) and Q(6,4) find the
a) distance PQ
b) point R, the mid-point of interval PQ
c) gradient, mPQ, of the line joining PQ
1
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or

realistic for the problem.
2
1
2
a) Dis tan ce : d 

x 2  x1   y 2  y1 
2
2

The distance PQ is d = 170


x 2  x1 y 2  y1 
b) Mid  point R  
,
 2

2 
P(-5,-3)
Distance PQ: d  (x2  x1)  (y2  y2 )
2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
m  tan

6  (5) 4  (3) 
R  
,
 2

2 
2
c) Gradient, m 
6  (5)  4  (3)
 170
Q(6,4)
2
1 1 
  , 
2 2 

1 1 
  , 
2 2 
1 1 
 The mid-point R is  , 
2 2 


2
y 2  y1
x 2  x1
m
4  (3)
6  (5)
m
7
11


7
The gradient of line PQ is m 
11


YOU NEED TO
KOWN THE
FORMULAE !
Exercise
(x , y )
(x , y )
3) Given the two points L(8,-3) and M(-1,5) find the
a) distance LM
b) point K, the mid-point of interval LM
c) gradient, mLM, of the line joining LM
1
Helpful Hints:
1) It is usually a good idea to make a little
sketch.
2) Write down ALL point formulae used in
coordinate geometry.
3) Label points clearly (x1, y1) & (x2,y2)
4) Is your answer what you expect or

realistic for the problem.
1
2
2
a) Dis tan ce : d 

x 2  x1   y 2  y1 
2
2

The distance LM is d = 145


x 2  x1 y 2  y1 
b) Mid  point K  
,
 2

2 
L(8,-3)
Distance PQ: d  (x2  x1)  (y2  y2 )
2
x  x1 y 2  y1 
Mid  point of PQ:  2
,

 2
2 
y y
Gradient of PQ : m  2 1
x 2  x1
m  tan

(1)  8 5  (3) 
K  
,
 2

2 
2
c) Gradient, m 
(1)  8  5  (3)
 145
M(-1,5)
2
7 2 
  , 
2 2 

7 
  ,1
2 
7 
 The mid-point K is  ,1
2 


2
y 2  y1
x 2  x1
m
5  (3)
(1)  8
m
8
9


8
The gradient of line LM is m 
9


YOU NEED TO
KOWN THE
FORMULAE !
FORMULAE TEST
Given the two points P(x1,y1) and Q(x2,y2)., write down the formula for:
a) Distance PQ
b) Mid-point of the interval PQ
c) Gradient of the interval PQ.
Dis tance PQ  (x2  x1 )2  (y2  y1 )2

x  x y  y 
Mid  point PQ   2 1 , 2 1 
 2
2 
Gradient, mPQ 


YOU NEED TO
KOWN THE
FORMULAE !
y2  y1
x2  x1
These will be revised again in HSC application type questions.