Term 3 : Unit 2
Coordinate Geometry
Name : _____________ (
)
Class : ________ Date : ________
2.1 Midpoint of the Line Joining Two Points
2.2 Areas of Triangles and Quadrilaterals
2.3 Parallel and Non-Parallel Lines
2.4 Perpendicular Lines
2.5 Circles
Coordinate Geometry
2.1 Midpoint of the Line Joining Two Points
Objectives
In this lesson, you will learn how to find the midpoint of a line segment and
apply it to solve problems.
Coordinate Geometry
y
A line AB joins points
(x1, y1) and (x2, y2).
B ( x 2 , y 2)
M ( x, y )
A ( x 1 , y 1)
M (x, y) is the
of AB. of
Take midpoint
the x-coordinate
D and the y-coordinate of
Construct
E. a right
angled triangle ABC.
E ( x2 , y)
C ( x 2 , y 1)
D ( x , y 1)
x
x1 Ex2is y1  y12  y2 
 xpoint
D is
M isx1the
 and
2


2
, y1  
 
2
,  x1 ,
 2

2


Construct the midpoints D
and E of the line segments AC
and BC. Take the mean of the
coordinates at the endpoints.
Coordinate Geometry
Example 1
P, Q, R and S are coordinates of a parallelogram and M is the midpoint of
PR. Find the coordinates of M and S and show that PQRS is a rhombus.
y
Q ( 9 , 6)
R ( – 2 , 4)
 4   2   4   4 
M 
,
  1,0 
2
2


M is also the midpoint of QS.
M M ( 1 , 0)
OO
P ( 4 , – 4)
S ( a– , 7 b ,) – 6 )
x
9a 6b
M 
,
  1,0 
2 
 2
a  7, b  6 S =  7, 6 
PQ 
 9  4    6   4  
QR 
9   2   6  4  125
2
2
2
 125
2
PQ  QR  the parallelogram PQRS is a rhombus
Coordinate Geometry
Example 2
3 points have coordinates A(–1, 6), B(3, 2) and C(–5, –4). Given that D and
E are the midpoints of AB and AC respectively, calculate the midpoint and
length of DE.
y
A ( – 1 , 6)
M ( – 1 , 2
E ( – 3 , E1 )
D ( 1 , 4)
1
2
)
B ( 3 , 2)
M
x
C ( – 5 , – 4)
  1  3 6  2 
D
,
  1, 4 
2 
 2
  1   5  6   4  
E 
,
   3,1
2
2


Let M be the midpoint of DE.
 1   3 4  1 
1
M 
,


1,
2


2
2 
 2
DE 
 3  1  1  4
2
2
5
Coordinate Geometry
Example 3
If A(2, 0), B(p, –2), C(–1, 1) and D(3, r) are the vertices of a parallelogram
ABCD, calculate the values of p and r.
Let M be the midpoint of AC.
y
 2   1 0  1  1 1
M 
,
  2, 2
2
2 

D(3, 3
r ))
C ( – 1 , 1)
M
M (
1
2
,
1
2
)
A ( 2 , 0)
BB(( p
– ,2 ,– –2)2)
x
M is also the midpoint of BD.
 p  3  2   r  1 1
M 
,
  2, 2
2 
 2
p  2, r  3
Coordinate Geometry
2.2 Areas of Triangles and Quadrilaterals
Objectives
In this lesson, you will learn how to find the areas of rectilinear figures given
their vertices.
Coordinate Geometry
Area of Triangles
y
ABC is a triangle.
We will find its
area.
CC (( 44 ,, 33 ))
x
Construct points D and
E so that ADEC is a
trapezium.
AA (( –– 22 ,, –– 11 ))
D ( – 2 , – 3)
BB (( 22 ,, –– 33 ))
E ( 4 , – 3)
Area of ABC  Area of trapezium ADEC  Area of ADB  Area of BEC
1
1
1
  2  6  6   4  2   2  6
2
2
2
 24  4  6
 14 square units
Coordinate Geometry
y
B ( x 2, y 2 )
ABC is a triangle. The
vertices are arranged in
Construct points D, E and
an anticlockwise
F on the x-axis as shown.
direction. We will find
its area.
C ( x 3, y 3 )
A ( x 1, y 1 )
x
D ( x 3, 0 )
E ( x 2, 0 )
F ( x 1, 0 )
Area of ABC 
Area of ABEF  Area of BCDE  Area of ACDF
 BE  AF   EF  12  CD  BE   DE  12  CD  AF   DF
 12  y2  y1  x1  x2   12  y3  y2  x2  x3   12  y3  y1  x1  x3 

1
2

1
2

1
2
 x1 y2  x1 y1  x2 y2  x2 y1  x2 y3  x2 y2  x3 y2  x3 y3  x1 y3  x1 y1  x3 y1  x3 y3 
 x1 y2  x2 y3  x3 y1  x2 y1  x3 y2  x1 y3 
Coordinate Geometry
y
B ( x 2, y 2 )
x1
x2
x3
x1
y1
y2
y3
y1
C ( x 3, y 3 )
A ( x 1, y 1 )
Definition
x
x1 y2  x2 y3  x3 y1  x2 y1  x3 y2  x1 y3
1
Area of ABC   x1 y2  x2 y3  x3 y1  x2 y1  x3 y2  x1 y3 
2
From the
previous slide,
we know that
1 x1

2 y1
x2
x3
x1
y2
y3
y1
Coordinate Geometry
Example 4
Find the area of a triangle with vertices
A(–2, –1), B(2, –3) and C(4, 3).
y
C ( 4 , 3)
x
The vertices A, B and C follow an
anticlockwise direction.
1 2 2
Area of ABC 
2 1 3
A ( – 2 , – 1)
4
2
3
1
B ( 2 , – 3)

1
 6  6  4  2  12  6 
2
 14 square units
Coordinate Geometry
Area of Quadrilaterals
y
Find the area of a quadrilateral with vertices
A(x1, y1 ), B(x2, y2 ), C(x3, y3 ) and D(x4, y4 ),
following an anticlockwise direction.
D ( x4 , y4 )
C ( x3 , y3 )
A ( x1 , y1 )
x
Area of ABCD  ABC  ACD
1 x1 x2 x3 x1 1 x1 x3 x4 x1


2 y1 y2 y3 y1 2 y1 y3 y4 y1
1
Area of ABCD   x1 y2  x2 y3  x3 y1  x2 y1  x3 y2  x1 y3 
Split the quadrilateral
21
into two triangles.
  x1 y3  x3 y4  x4 y1  x3 y1  x4 y3  x1 y4 
2
1
  x1 y2  x2 y3  x3 y4  x4 y1  x2 y1  x3 y2  x4 y3  x1 y4 
2
B ( x2 , y2 )
Coordinate Geometry
y
D ( x4 , y4 )
C ( x3 , y3 )
A ( x1 , y1 )
x
The area of a quadrilateral with vertices
A(x1, y1 ), B(x2, y2 ), C(x3, y3 ) and D(x4, y4 ),
following an anticlockwise direction.
B ( x2 , y2 )

1
 x1 y2  x2 y3  x3 y4  x4 y1  x2 y1  x3 y2  x4 y3  x1 y4 
2
1 x1

2 y1
x2
x3
x4
x1
y2
y3
y4
y1
The method for finding the
area of quadrilaterals is very
similar to that of triangles.
Coordinate Geometry
Example 5
y
Q ( – 4 , 3 )
Find the area of a quadrilateral with vertices
P(1, 4 ), Q(–4, 3), R(1, –2) and S(4, 0),
following an anticlockwise direction.
P ( 1 , 4 )
x
S ( 4 , 0 )
R ( 1 , – 2 )
1 1
Area 
2 4
4
1
4
1
3
2
0
4
1
  3  8  0  16   16   3   8   0 
2
 24 square units
Coordinate Geometry
3.3 Parallel and Non-Parallel Lines
Objectives
In this lesson, you will learn how to apply the conditions for the gradients of
parallel lines to solve problems.
Coordinate Geometry
yy
yy == mm 1 1xx ++ cc 1 1
 1 1
Consider the straight line with equation
y = m1 x + c1 that makes an angle of θ1
with the positive x-axis.
yy == mm 2 2xx ++ cc 2 2
Translate the line parallel to the x-axis.
 2 2
xx
OO
The new line has equation
y = m2 x + c2 and makes an angle of θ2
with the positive x-axis.
The lines are parallel to each other.
The lines make the same angle with the x-axis.
θ1 = θ2
The lines have the same gradient.
m1 = m2
Coordinate Geometry
Example 6(a)
The diagram shows a parallelogram ABCD with
A and C on the x-axis and y-axis respectively.
The equation of AB is x + y = 2 and the equation
of BC is 2y = x + 10.
y
C
2 y = x + 10
B
x + y = 2
D
A
O
At A, y  0.
x
(a) Find the coordinates of A, B and C.
At B, x  y  2
2 y  x  10
Since x  y  2, x  2.
Solving x  2, y  4.
A is  2, 0 
B is  2, 4  .
At C , x  0.
Since 2y  x  10, y  5.
C is  0,5  .
Coordinate Geometry
Example 6(b)
The diagram shows a parallelogram ABCD with
A and C on the x-axis and y-axis respectively.
The equation of AB is x + y = 2 and the equation
of BC is 2y = x + 10.
y
C (0,5)
2 y = x + 10
B
x + y = 5
x + y = 2
2y = x – 2
D
A
O
(2,0)
x
(b) Find the equations of AD and CD.
AD is parallel to BC (2y = x + 10).
CD is parallel to AB (x + y = 2 ).
Gradient of AD = gradient of BC = 0.5 Gradient of CD = gradient of AB = –1
y0 1
Equation of AD is

x2 2
2y  x  2
Since A is (2,
0)
Equation of AD is
Since C is (0,
5)
y 5
 1
x0
x y 5
Coordinate Geometry
Example 7
Find the equation of the line which passes
through the point (–2, 3) and is parallel to the
line 2x + 3y – 3 = 0.
2x  3 y  3  0
3 y  2 x  3
y   23 x  1
m in y = mx
Rearrange
in +
thec
is the
form
y =gradient
mx + c.
of the line.
The gradient of the line is  23 .
The line through  2,3 with gradient  23 is
y
( – 2, 3)
2x + 3y – 5 = 0
2x + 3y – 3 = 0
O
y 3
2

x   2 
3
3  y  3  2  x  2 
3 y  9  2 x  4
The equation of the line is 2 x  3 y  5  0.
x
Coordinate Geometry
2.4 Perpendicular Lines
Objectives
In this lesson, you will learn how to apply the conditions for the gradients of
perpendicular lines to solve problems.
Coordinate Geometry
Consider the straight line with
equation y = m1 x + c1 that makes an
angle of θ1 with the positive x-axis.
y
y = m 1x + c 1
Rotate the line clockwise through 90°.
A
y = m 2x + c 2
1
O
B

2
D
x
C
AD
AC
m1 
 tan  1 
BD
AB
AC  AB 
m1  m 2 

  1
AB  AC 
The new line has equation y = m2 x + c2
and makes an angle of θ2 with the
negative x-axis.
m2  
AD
AB
  tan  2  
DC
AC
1
m2  
m1
Applies to any two
perpendicular lines.
Coordinate Geometry
Example 8
y
B ( 4 , 15 )
Two points have coordinates A(–2, 3) and B(4, 15).
Find the equation of the perpendicular bisector of AB.
Hence calculate the coordinates of the point P on the
line 3y = x + 1 if P is equidistant from A and B.
M ( 1, 9)
2 y = – x + 19
A( – 2, 3)
3y = x + 1
O
P ( 11 , 4 )
x
 2  4 3  15 
The midpoint of AB is M . 
,
  1,9 
2 
 2
15  3
Gradient of AB 
2
4   2 
Gradient of perpendicular bisector =  12
Equation of perpendicular bisector is
y  9   12  x  1
2 y   x  19
Find P.
2 y   x  19
3y  x 1
5 y  20
12  x  1
P  11, 4 
Solve
Adding
Substitute
simultaneou
forthe
y
s equations
equations
y4
x  11
Coordinate Geometry
Example 9
y
The points A and B have coordinates (5, 2) and (3, 6)
respectively. P and Q are points on the x-axis and
y-axis and both P and Q are equidistant from A and B.
B( 3, 6)
2y = x + 4
M ( 4, 4)
(a) Find the equation of the perpendicular bisector of AB.
Q
( 0, 2)
A( 5, 2)
P
 35 6 2 
The midpoint of AB is M . 
,
   4, 4 
2 
 2
62
(b) Find the coordinates of P and Q.
Gradient of AB 
 2
35
At P y  0, 2 y  x  4
Gradient of perpendicular bisector = 12
x  4
P   4, 0 
Equation of perpendicular bisector is
( – 4, 0)
O
x
y  4  12  x  4 
2y  x  4
At Q x  0, 2 y  x  4
y2
Q 
 0, 2 
Curves and Circles
2.5 Circles
Objectives
In this lesson, you will learn to
• recognise the equation of a circle,
• find the centre and radius of a circle,
• find the intersection of a circle and a straight line.
Curves and Circles
y
Consider the point C(a, b) and a
point P(x, y).
P (x,y)
The distance between C and P is
x
O
The equation of the circle is
 x  a   y  b
2
PC 
C (a,b)
2
The locus of P as the line rotates
around C is a circle of radius r.
 x  a   y  b
2
2
 r2
Curves and Circles
y
The circle with centre C(a, b) and
radius r has equation
P (x,y)
 x  a   y  b
2
r
2
 r2
x 2  2ax  a 2  y 2  2by  b2  r 2
C (a,b)
x 2  y 2  2ax  2by  a 2  b2  r 2  0
The general form of the equation is
x
x 2  y 2  2 gx  2 fy  c  0
O
g  a, f  b c  a  b  r
2
2
2
centre  C   g ,  f 
radius 
f 2  g2  c
Curves and Circles
Example 10
Find the equation of the circle whose centre
is C(–3, 4) and which touches the x–axis.
y
The radius of the circle is 4 units.
C ( – 3, 4)
The equation of the circle is:
 x   3    y  4 
2
2
4
4
2
x
O
x 2  6 x  9  y 2  8 y  16  16
x2  y 2  6x  8 y  9  0
The radius is the ycoordinate of C.
Curves and Circles
Example 11
Find the coordinates of the points of
intersection of the line 2y + x = 12 with the
circle x2 + y2 – 6x – 4y – 12 = 0.
y
x
2
2
+ y – 6 x – 4 y – 12 = 0
5
2 y  x  12
x  12  2 y
Substitute for x
into the circle
equation.
2 y + x = 12
x  y  6 x  4 y  12  0
2
12  2 y 
2
2
5
10
 y 2  6 12  2 y   4 y  12  0
144  48 y  4 y 2  y 2  72  12 y  4 y  12  0
5 y 2  40 y  60  0
5  y  2  y  6   0
y2
or
y6
x 8
or
x0
Using
x = 12
– 2y.
The
points
are
8, 2 
and
 0, 6 
x
Curves and Circles
Example 12
(a) Give the equation of the circle with centre C(–2, 0) and radius r = 3.
The equation is
2
x


2

y

0

3






2
2
x2  4x  4  y 2  9
x2  y 2  4 x  5  0
(b) Find the coordinates of the centre and the radius of the circle
x2 + y2 – 2x – 6y + 1 = 0.
x 2  y 2  2 gx  2 fy  c  0
2 g  2,
g  1
2 f  6, f  3
c 1
centre  C   g ,  f 
 1,3
radius 
3
 3   1  1
2
2