CO-ORDINATE GEOMETRY
TRIANGLES, QUADRILATERALS AND CIRCLES
Task 1
A(-3;3)
C(5;3)
1.
A triangle has been drawn
with a median from A to D .
(A median is a line from a corner
D
to the midpoint of the opposite side.)
B(-5;-3)
2.
a)
Calculate the co-ordinates of point D .
b)
Calculate the lengths of each of the sides the triangle.
c)
Calculate the gradient of AD .
d)
Calculate the gradient of BC .
e)
Find the equation of AD .
A triangle is given in the diagram
E(0;4)
and a median is drawn from E to H .
F(-5;3)
H
G(1;-1)
a)
Find the co-ordinates of H .
b)
Find the lengths of the sides of the triangle – what type of triangle is it?
c)
Find the gradient of EH .
d)
Find the equation of EH .
e)
Find the gradient of FG . What can be said about the lines EH and FG ? Is this always
true for a median?
N
3.
In the diagram MK // NL .
M
First find the co-ordinates of K , L M and N
if each grid square is 1 unit long.
Calculate the lengths of each of the three sides
of the two triangles that are drawn.
What can you find out about the ratios
of the various sides?
4.
J(0;0)
P(2;8)
In the diagram you are given a triangle.
a)
Find the equation of the median from P .
b)
Find the equation of the median from Q .
c)
Find the equation of the median from R .
d)
Find the intersection point of the median
from P and the median from Q .
e)
Find the intersection point of the median
from R and the median from Q .
f)
How does this prove that the medians are
concurrent?
L
K
R(8;-2)
Q(-4;-6)
5.
Draw a triangle yourself on the Cartesian plane and, in a similar manner to that used in question 4,
show that the altitudes of your triangle are concurrent.
6.
Now do the same for the perpendicular bisectors of your triangle.
7.
Note that in questions 4, 5 and 6 you have not proved that the medians, altitudes and perpendicular
bisectors of a triangle are concurrent – you have proved it only for the case that was drawn. You
can do it for the general case by using three points ( a; b) , (c; d ) and (e; f ) . The calculations are
horrid, but you may want to really test yourself – and there is some satisfaction in eventually
getting out the proof. But this proof is not for the faint-hearted. The proofs for the concurrency of
altitudes and perpendicular bisectors are equally troublesome.
Task 2
D(6;5)
1.
2.
You are given a diagram of a
parallelogram ABCD . A
parallelogram has its opposite
sides parallel.
Investigate this parallelogram by
finding all the lengths and gradients
that you can.
What conclusions can you draw
about this parallelogram and its
diagonals, besides the fact that
opposite sides are parallel?
A(0;3)
E
C(9;2)
B(3;0)
F(5;0)
The shape alongside is a kite.
G(8;-1)
Investigate this kite by finding
all the lengths and gradients
that you can.
K
What conclusions can you draw
about this kite and its diagonals?
J(4;-3)
H(8;-6)
3.
Draw a square and then a rectangle that is not a square, of your own on the grid provided and
perform a similar investigation.
4.
What type of quadrilateral is
shown alongside?
M(-4;0)
What can you find out about
this shape?
O
Q(-7;-2)
N(-1;-2)
P(-4;-4)
5.
C(7;5)
Here is a diagram of a parallelogram.
Calculate the sum of the squares of the sides.
B(1;3)
Calculate the sum of the squares of the diagonals.
What do you notice?
D(6;0)
A(0;-2)
Task 3
1.
A circle has been drawn alongside. A chord GH
has been drawn and the centre of this chord, K
O(5;4)
has been linked to the centre of the circle.
G(0;4)
Find the gradients of GH and KO .
K
What conclusion can you draw?
H(5;-1)
2.
A circle has been drawn with a pair of
intersecting chords.
M(4;2)
a)
Find the co-ordinates of R .
b)
Calculate PR  RM and QR  RN .
What do you notice? Will this always
be true? Try a few more cases.
Q(-4;-2)
R
P(-2;-4)
N(2;-4)
Task 1
(0;0)
1a)
b)
c)
1
2a)
3.
(2;1)
b)
3
d)
2
Sides are 6;3; 45 and 8;4; 80
4a)
x2
c)
( 2;0)
d)
Task 3
1.
mGH  1 mKO  1
d)
b)
e)
BC  136
AB  40 ; AC  8 ;
3
y  x
e)
5
EF  26 ; EG  26 ; FG  52
3
2
y  x4
 ; perpendicular, no
e)
2
3
and the ratio between sides is always 3 : 4 3
1
2
y  x2
y  x
c)
3
3
( 2;0)
2a)
(1;3)
b) PR  RM  QR  RN  10