analytical geometry - home - maths at sharp
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ChapterChapterChapterChapter Analytical GeometryAnalytical GeometryAnalytical GeometryAnalytical Geometry
Analytical Geometry is the use of formulae to analyse the length, gradient etc. of
certain points and their lines.
Section A - Coordinates
In grade 10 you learnt that every coordinate is represented by ��; �� and that the
same coordinate can be plotted on the Cartesian plane where � is your position
from left to right and � is your position upwards or downwards.
Section B - Distance between two points
Remember the theorem of Pythagoras – it states that the hypotenuse squared is
equal to the sum of the squares of the other two sides.
i.e. �� � �� ��
To find the distance of any point:
• First find the distance of ‘�’ from the two coordinates – e.g. ���; ��� and
��� ; � �. ∴ � � � � ��
• Next find the distance of ‘�’ from the two coordinates:
∴ � � � � ��
• Then Substitute it into the original Pythagoras equation:
∴ �� � �� � ���� �� � ����
So the distance formula for any point is:
�
�
� � ��� � ���� �� � ����
Example: Find the distance from the point A(-2; 3) to point B (3; -2) .
Answer: ∴ � � ��� � ���� �� � ����
∴ � � ��3 � ��2��� ��2 � 3�� ∴ � � ��5�� ��5��
∴ � � √25 25
∴ � � √50 ∴ � � 5√2�����
Activity 1
1. Determine the distance between the two given points that follow:
a) !5; "�# and $ !"% ; "&# b) ' !8; ")# and *��13;�2� c) , !�12;� &
%# and -��5;�1� d) . !�16;� "&# and 0�2; 6�
e) 1 !"& ; 1# and 2�6; 4�
f) 4!�10;� ")# and 5��4; 20� g) 6�6; 1� and 7�1; 6�
h) 8 !") ; "�# and 9 !1 &: ; �9# i) < !2; ":# and = !2;� "
)#
j) >��3; 25� and ? !3;�1 "�#
2. Determine the missing variable’s value(s) if the length and points are given:
a) Distance = 7.3276…units @�1;�2� and A !B; ")#
b) Distance = √�C:% units D�4;�1� and E !� "
% ; F#
c) Distance = √178 units �1; H� and ��4; 16� d) Distance =
√%"I� units !�;�1 "
�# and $��4; 4� e) Distance =
√JI& units ' !�8;�1 "
&# and *�K; 1�
You can put this straight into your calculator
by pressing these buttons:
Section C - Gradient
The gradient works out the slope of the graph – or the ratio of the height to the
length.
The formula for gradient is:
Example – the gradient between two points A!"& ; �2# and B�2; 1� ∴ L � MNOMPQNOQP ∴ L � "O�O���O!PR#
∴ L � "�I
Co-linear points
Three points are collinear when all three points lie on the same line and we check
this by using gradient, i.e. points A, B and C are collinear when L� � L S � L�S.
E.g. Show that the following points are collinear: D ��3;�3� and '��1; 3� and
*�2; 12�.
Answer: LTU � %O�O%�O"O�O%� LUV � "�O%�O�O"� LTV � "�O�O%��O�O%�
� 3 � 3 � 3
∴ The points D, E and F are collinear.
L ��� � �"�� � �"
In other words – for every 12 units you
go up, you go across 7 units.
You put this into your calculator by pressing these buttons:
Activity 2
1. Determine whether the following points are collinear:
a) A��1;�16� B�3;�8� C�7; 0�
b) D�7;�10� E�1;�13� F!�5;�3 "�#
c) G!�1;� �%# H!3;�16 �
%# I!0;�4 �%#
d) J�1; 12� K��2; 10� L!3; 13 "%#
e) M!"& ; �4# N!1; "�# O!5; 10 "&#
2. Given the gradient, determine the missing variables:
a) L ��16 P��3;�6� Q�B; 2� b) L ��1 R��3; 9� S�1; F� c) L � �
% T!� "& ; H# U!�3 "
& ; 2#
d) L �� %� V��;�8� W!1 "
� ; �11#
e) L � 10 X!� "% ; � �
%# Y�0; K�
Parallel and Perpendicular Gradients
When one line is parallel to another line, the gradients are equal:
When one line is perpendicular (at a right angle) to another line, the gradients are
the negative inverse of each other or gradient 1 times gradient 2 is equal to negative
1.
L" � L�
L" ×L� � �1
Activity 3
1. Say whether line AB (given the points of A and B) is parallel or perpendicular
to line DE (given the points D and E). Show all working out.
a) A!"� ; 3 "�# B�12; 15� D�2;�3� E��3; 2�
b) A�2; 0� B!� "� ; �%# D!�5;� "
%# E!2;�2 ":#
c) A!� :� ; �6# B!� "
� ; 1# D�7;�1� E�14;�3� d) A�4;�2� B�1; 8� D!� :
� ; 16# E!�15; 12 "&#
e) A�2; 3� B�1; 6� D��2;�1� E��4; 5�
2. Look at the following equations for these straight lines and state whether they
are parallel, perpendicular or neither.
a) � � 6� � 5 � � 6� � 1
b) � � �% � 10 � � &
) � � 16
c) � � 8� "� � � � "
X � "%
d) � � "� � 4 � � &
X � � 2
e) � � �1� � 16 � � "� � 5
f) � � "& � "
: � � �4� � 2
g) � � 2� 5 � � &X � � 1
h) � � �3� �% � � 3� � 4
i) � � � � 10 � � 2!� "� � 3#
j) � � 9� 9 � � � "J � � 3
Section D – Midpoint
To find the midpoint of two coordinates you have to find the “average” of the two
�-coordinates and find the average of the two �-coordinates.
So the formula to find the midpoint is:
e.g. Find the midpoint of the points A�2; 5� and B��3; 8�
4� !QPYQN� ; MPYMN� #
∴ 4� !�O%� ; :YX� #
∴ 4� !� "� ; 6 "
�#
Activity 4
1. Determine the midpoint for each of the following pairs of coordinates.
a) A��1;�9� B�1; 3� b) C!� "� ; �2# D!�1 "
� ; 7#
c) E�3; 6� F�5;�5� d) G!2 �% ; 1# H�3; 2�
e) I ��2; 3� J�7;�1�
2. Determine the other coordinate if the first coordinate and the midpoint is
given.
a) K��2; 1� 4Z[��6;�6� b) N�8;�3� 4\]�6; 5� c) P�1;�8� 4^_ !": ; "&#
d) R!�1;� &:# 4`a !� "
� ; � ":#
e) T�4;�5� 4bc��5;�2�
4�Q;M� d�" ��2 ;�" ��2 e
Section E – Equations of Straight Lines
The straight line equation has 4 parts to it
� → depends on your �-value i.e. the dependent variable.
→ changes as � changes
L → gradient (Watch out for parallel and perpendicular lines).
Vertical lines have a undefined gradient because fghijklmn
oipg is undefined,
therefore the equation of the line will be � � Hq���B�� Horizontal lines have a gradient of 0 because
oipgfghijklmn is zero, therefore the
equation of a horizontal line is � � Hq���B��. � → determines your �-value. i.e. the independent variable
H → your �-intercept (where your line crosses the �-axis).
To Solve
• Two coordinates
If you are given two coordinates work out the gradient, then substitute it into
the formula along with one of your coordinates and solve for H. Finally, write
down the equation.
e.g. Find the equation of the line DE if D �1;�7� and E��3; 1�.
1. find the gradient:
L � MNOMPQNOQP
∴ L � "O�OI�O%O�"� ∴ L � �2
2. then find c:
� � L� H Substitute L � �2 and �1;�7� �7 � �2�1� H �5 � H
3. Finally write down the equation properly
∴ � � �2� � 5
� � L� H
• Given the gradient parallel or perpendicular to another line:
Determine if your gradient is the same (parallel lines) or the negative inverse
of the other gradient (perpendicular lines).
Then substitute your point to find c.
Finally write down your equation properly.
e.g. Determine the equation of the straight that is parallel to the line
� � 3� 4 and passing through the point ��1; 3�
1. Determine the gradient:
From the parallel line the gradient is 3.
2. Now find the c value:
� � 3� H Substitute the coordinate
3 � 3��1� H 6 � H
3. Write down the formula properly:
� � 3� 6
Activity 5
1. Given two points on the line – determine the equation of the line.
a) A!&: ; �5# and B!� "% ; � "
%#
b) C!1 "� ; 3# and D!�1;� "
�#
c) E��3; 4� and F!�3;� "&#
d) G�1; 2� and H��1; 8� e) I�3; 2� and J�6; 2�
2. Determine the equation if:
a) the line passes through the point K��1; 2�and is parallel to the line
� � 2� 1.
b) the line passes through the point L�3; 8� and is perpendicular to the
line � � �4� � 2.
c) the line passes through the point M!9;� "�# and is perpendicular to the
line � � 6� � :).
d) the line passes through the point N!� ": ; 9# and is parallel to the line
� � � "% � � 4
e) the line passes through the point P�4;�6� and is perpendicular to the
line � � 5� � 4.
Section F – Angle of Inclination
The angle of inclination tells you at what angle the straight line crosses the �-axis.
To work out the angle of inclination first work out your gradient, then substitute it
into the formula and solve for the angle:
Remember that when your gradient is positive you can simply use the angle given
when you work it out, however, if your gradient is NEGATIVE then work out the
angle using a positive value of the gradient (i.e. times your gradient by negative 1)
then subtract the angle from 180°.
Example: Positive Gradient:
Determine the angle of inclination for the line � � 3� � 1
∴ L � 3
∴ 3 � tanw
∴ w � 72°
L � �B�w
Example: Negative Gradient:
Determine the angle of inclination for the line � � � "& � � 6
∴ L � � "& ∴ Positive value � "
&
∴ "& � tanw
∴ w � 14° ∴ angle of inclination � 180° � 14° � 166° Activity 6
1. Determine the angle of inclination for the following lines:
a) � � 8� ") b) � � �6� � 3
c) � � &: � � 3 d) � � �2� 1
e) � � �3� � %: f) � � � "
% � � &%
g) 2� � � 3 h) �4� � � � 6
i) � � %� � � 2 j) � � �1� 2
2. Given the angle of inclination – determine the gradient round off to one
decimal place.
a) w � 60.95° b) w � 99.46° c) w � 75.96° d) w � 45° e) w � 63.43° f) w � 9.46° g) w � 98.13° h) w � 101.31° i) w � 120.96° j) w � 85.24°
Section G – Equation of a Circle
Things to remember about circles before we can begin:
• The diameter is twice the radius i.e. � � 2�q� "�� � �
• The radius is the same throughout the circle.
• The tangent is perpendicular to the radius
• A normal is a line perpendicular to the tangent at the point of contact – the
normal is not the radius but can go through the circle or be outside the circle.
• A secant cuts the circle twice.
• A chord touches the circle twice internally and divides the circle into segments
• A sector is the middle piece between two radii.
• A chord divides a circle’s circumference into different arcs.
• A circumference is the distance around the circle and is given by the formula
2x�.
Circle with centre at the origin i.e. (0; 0)
When the circle has a centre at the origin the formula is:
This formula should remind you of Pythagoras ☺
r is the radius and � and � is the coordinate at a point through the circle.
Circle with any centre
Essentially a circle with any centre is simply a circle with a centre at the origin that
has been shifted up or down and left or right.
The formula for a circle with any centre is:
• r is the radius
• a is the �-coordinate of the centre
• b is the �-coordinate of the centre.
�� � �� ��
�� � �� � B�� �� � F��
Example: Given the centre and the radius
Determine the equation of the circle with centre C��6; 3� and radius = 3.
∴ Substitute all points into the equation and simplify.
∴ �3�� � �� 6�� �� � 3��
∴ 9 � �� 12� 36 �� � 6� 9
∴ �36 � �� 12� �� � 6�
Example: Given the centre and another point on the circle.
Determine the equation of the circle with centre C��2; 4� and a point passing
through the circle P!"� ; �6#.
∴ Substitute the centre coordinates into a and b and the point
coordinates into � and � and solve for ��. Then write down the formula
and simplify again.
∴ �� � !"� 2#� ��6 � 4��
∴ �� � 106 "&
∴ 106 "& � �� 2�� �� � 4��
∴ 106 "& � �� 4� 4 �� � 8� 16
∴ 86 "& � �� 4� �� � 8�
Example: Determine the radius and the centre of the circle with equation
107 � �� 2� �� � 6�
∴ You need to complete the square for both �and �.
∴ 107 � �� 2� !��#� � !��#
� �� � 6� !O)) #� � !� )
)#�
∴ 107 1 9 � ��� 2� 1� ��� � 6� 9� ∴ 117 � �� 1�� �� � 3��
∴ � � √117 K���K��1; 3� ∴ � � 3√13
Activity 7
1. Determine the equation of the circle given the centre and the radius in the
form: �� � �� � B�� �� � F��
a) C�9; 8� � � 12 b) C�3;�1� � � 6
c) C�7;�5� � � 1 d) C��1; 4� � � 3
e) C!� "& ; 4# � � 5
2. Determine the equation of the circle given the centre and another point on
the circle in the form: �� � �� � B�� �� � F��
a) C!1 "� ; 8# P!2;� "
�# b) C�1; 6� P�10; 12� c) C�5; 6� P!�1; �%# d) C��2; 3� P��1;�2� e) C��3;�1� P�2; 2�
3. Determine the radius and the centre coordinates given the equation:
a) 9 � �� � 8� �� 4�
b) 81 � �� � 4� �� 8�
c) �20 � �� 16� �� 18�
d) 128 � �� 14� �� � 2� e) 2 "%
") � �� � 6� �� � 2�
Equation of the tangent to the circle
A tangent is a straight line that is drawn perpendicular to the circle’s radius and
touching the circle at only one point.
To work out the equation of the tangent use the straight line formula:
�B��� → are the coordinates that the tangent touches the circle at
L → is the gradient – you find the gradient by working out the gradient of
the gradient of the radius from the centre of the circle to the point
where the tangent touches the circle. Then find its negative inverse –
this is the gradient of the tangent.
H → is the �-intercept – substitute �, � and L into the straight line
equation and solve for H.
Example: Determine the equation of the tangent that touches the circle with the
equation 48 � �� � 2� �� 12� at the point �10;�4�.
First find the centre of the circle’s coordinates:
∴ 48 � �� � 2� !��#� � !��#
� �� 12� !"�� #� � !"�� #
�
∴ 48 1 36 � ��� � 2� 1� ��� 12� 36� ∴ 85 � �� � 1�� �� 6��
∴ Centre at �1;�6�
Next find the gradient of the radius:
∴ L � O)Y&"O"C ∴ L � �J
∴ the gradient of the tangent is �J ×L� ��1
∴ L� �� J�
Now find the c value by substituting in �, � and L.
∴ �4 � � J� �10� H
∴ 41 � H
Finally, write down the equation properly
∴ � � � J� � 41
� � L� H
Activity 8
1. Determine the equation of the tangent given the centre’s coordinates, C, and
the point, P, where the tangent touches the circle.
a) C�6;�6� P��1; 1� b) C��1;�3� P�4; 6� c) C��7; 4� P��8;�2� d) C�8; 3� P��6; 4� e) C�5; 3� P��5; 4�
2. Given the equation of the circle and the point on the circle determine the
equation of the tangent to the circle at that point.
a) 10 � �� 12� �� � 4� P�1; 3� b) 8 � �� 8� �� � 2� P��7;�3� c) 124 � �� 4� �� � 12� P�8;�2� d) �48 � �� 14� �� 16� P��6; 0� e) 67 � �� 10� �� 6� P�5;�4�
3. Given the radius and the point of the tangent that touches the circle,
determine value(s) of the missing variable for the centre of the circle and give
the equation(s) of the circle.
a) � � 3 T��2; 6� C�B; 4� b) � � 7 T�3; 9� C��3; F� c) � � 12 T!8; "�# C�H; 10� d) � � 5 T��3; 6� C��; 2� e) � � 8 T�4; 1� C�5; K�
4. Given the gradient of the radius to the point of the tangent, the equation of
the circle and the point T, where the tangent touches the circle, determine
the value(s) for the missing variable.
a) L � "I 333 � �� 18� �� 12� T�12; ��
b) L �� "& 42 � �� 10� �� � 2� T�K;�1�
c) L �� """ 93 � �� � 10� �� � 4� T��6; z�
d) L � ���Kz��K� �4 � �� 2� �� 4� T�{;�3� e) L �� "&
"% 157 � �� 16� �� � 24� T�5; |�
Activity 9 – Mixed Exercise
1. Given the sketch below with two circles, c and d and the straight line a, all
passing through the point ��4; 0�. The centre of circle d is the �-intercept of
the straight line a, given by the equation � � "& � 1.
a) Determine the equation of circle d.
b) Given that the radius of circle c is 1 and that the centre passes through
the �-axis, determine the equation of circle c.
c) Determine the other point of intersection of the straight line a, with
both of the circles in the sketch.
d) Determine the equation of the tangent to circle c at the point ��3; 1�. e) Where does the straight line in question (d) intersect with the circle d.
The typical analytical geometry questions include all of the
above sections in a mixture. So below is a mixed exercise
for you to practice with.
2. Given the sketch below with triangle ABC with vertex A at (-1; 4), vertex B at
(-2; -3), and vertex C at (2; 0).
a) Prove using two methods that angle }� is a right angle.
b) Determine the midpoint of A and B, called N.
c) Determine the equation of the line through points C and N.
d) What is the angle of inclination for line AC.
3. Given the sketch below with centre at A and diameter BC.
a) Given that the coordinate of B is (0; -2) and the coordinate of C is
(-2; 4), determine the point A – the centre of the circle.
b) Hence, or otherwise, determine the equation of the circle.
c) Determine the equation of the tangent that touches the circle at C.
d) A normal to the tangent in (c) runs through the point A. Determine the
equation of the normal.
4. Given the sketch below with the centre at C(1; -2) and a triangle with
coordinates A(1; 1) and B!�5;� "�#. D is the �-intercept of the circle.
a) Prove that triangle ABC is an isosceles triangle.
b) Find the length of the radius of the circle.
c) Hence, or otherwise, determine the equation of the circle.
d) Show that the �-axis is the tangent to the circle at D.
e) Determine the �-intercept of line AB.
f) Determine the point of intersection where line BC cuts through the
circle (round off your answer to the nearest whole number).
g) Determine the equation of the tangent at the point where line BC cuts
the circle in the diagram.
5. Given the sketch below with the centre of the circle at C, and diameter AB. D
is a point on the circle. (For all of the following questions round off the final
answers to one decimal place – hint use the memory function for accuracy).
a) Given that the coordinates of A and B are (-4; 2) and (4; 4) respectively,
prove that the coordinates of the centre of the circle, C, are (0; 3).
b) Hence, or otherwise, determine the equation of the circle.
c) Given that the angle �~$ is 30,39°determine the equation of the line
BD.
d) Hence, or otherwise, determine the coordinates of D.
e) Prove that the triangle ACD is an equilateral triangle.
f) Determine the equation of the tangent at the point A.
g) Determine the equation of the line DC and then determine where it
cuts the circle again at E if line DC is produced to cut the circle again at
E.
h) Show that the tangent to circle at D is parallel to the tangent at point E.
Answers to Activities
Activity 1
1. a) C !5; "�# and $ !"% ; "&#
$ST ����� � �"�� ��� � �"��
$ST ��!"%� 5#� !"&� "�#
�
$ST � 4.67 units
b) ' !8; ")# and *��13;�2�
$UV ����� � �"�� ��� � �"�� $UV ����13 � 8�� !�2 � "
)#�
$UV � 21.11 units
c) , !�12;� &%# and -��5;�1�
$�� ����� � �"�� ��� � �"��
$�� ����5 12�� !�1 &%#
�
$�� � √&&�% � 7.01 units
d) . !�16;� "&# and 0�2; 6�
$�� ����� � �"�� ��� � �"��
$�� ���2 16�� !6 "&#
�
$�� � 19.05 units
e) 1 !"& ; 1# and 2�6; 4�
$Z[ ����� � �"�� ��� � �"��
$Z[ ��!6 � "&#
� �4 � 1��
$Z[ � √)I%& � 6.49 units
f) 4!�10;� ")# and 5��4; 20�
$�\ ����� � �"�� ��� � �"��
$�\ ����4 10�� !20 ")#
� $�\ � 21,04 units
g) 6�6; 1� and 7�1; 6� $]^ ����� � �"�� ��� � �"��
$]^ ���1 � 6�� �6 � 1��
$]^ � 5√2
$]^ � 7,07 units
h) 8 !") ; "�# and 9 !1 &: ; �9#
$_` ����� � �"�� ��� � �"��
$_` ��!1 &:� "
)#� !�9 � "
�#�
$_` � 9,64 units
i) < !2; ":# and = !2;� ")#
$ab ����� � �"�� ��� � �"��
$ab ���2 � 2�� !� ") � "
:#�
$ab � ""%C ≈ 0,36667 units
j) >��3; 25� and ? !3;�1 "�#
$c� ����� � �"�� ��� � �"��
$c� ���3 3�� !�1 "�� 25#�
$c� � 27,17 units
2. a) Distance = 7.3276… units @�1;�2� and A !B; ")#
∴ $�� ����� � �"�� ��� � �"��
∴ �7.327�� � �B � 1�� !") 2#�
∴ 53.69372176 � �B � 1�� ")J%)
∴ 48.99927732 � �B � 1��
∴ ±6.9999 � B � 1
∴ B � �7 1 or B � 7 1
∴ B � �6 B � 8
b) Distance = √�C:% units D�4;�1� and E !� "
% ; F#
∴ $�� ����� � �"�� ��� � �"��
∴ !√�C:% #� � !� "%� 4#� �F 1��
∴ �C:J � ")J
J �F 1��
∴ 4 � �F 1��
∴ ±2 � F 1
∴ F � �2 � 1 or F � 2 � 1
∴ F � �3 F � 1
c) Distance = √178 units �1; H� and ��4; 16� ∴ $� ����� � �"�� ��� � �"��
∴ �√178�� � �4 � 1�� �16 � H��
∴ 178 � 9 �16 � H��
∴ 169 � �16 � H��
∴ ±13 � 16 � H
∴ �H � �13 � 16 or �H � 13 � 16
∴ H � 29 H � 3
d) Distance = √%"I� units !�;�1 "
�# and $��4; 4� ∴ $ST ����� � �"�� ��� � �"��
∴ !√%"I� #� � ��4 � ��� !4 1 "�#
�
∴ %"I& � ��4 � ��� "�"
&
∴ 49 � ��4 � ���
∴ ±7 � �4 � �
∴ �� � �7 4 or �� � 7 4
∴ � � 3 � � �11
e) Distance = √JI& units ' !�8;�1 "
&# and *�K; 1� ∴ $UV ����� � �"�� ��� � �"��
∴ !√JI& #� � �K 8�� !1 1 "
&#�
∴ JI") � �K 8�� X"
")
∴ 1 � �K 8��
∴ ±1 � K 8
∴ K � �1 � 8 or K � 1 � 8
∴ K � �9 K � �7
Activity 2
1. a) A��1;�16� B�3;�8� C�7; 0�
L� � MNOMPQNOQP L S � MNOMPQNOQP L� �OXY")%Y" L S � CYXIO%
∴ L� � 2 ∴ L S � 2
∴ Points A, B and C are collinear.
b) D�7;�10� E�1;�13� F!�5;�3 "�#
LTU � MNOMPQNOQP LUV � MNOMPQNOQP
LTU �O"%Y"C"OI LUV �O%PNY"%O:O"
LTU � "� LUV �� "J
"�
∴ They are not collinear.
c) G!�1;� �%# H!3;�16 �
%# I!0;�4 �%#
L�� � MNOMPQNOQP L�� � MNOMPQNOQP
L�� �O")N�YN�%Y" L�� �O&N�Y")N�CO%
L�� ��4 L�� ��4
∴ They are collinear
d) J�1; 12� K��2; 10� L!3; 13 "%#
L�Z � MNOMPQNOQP LZ[ � MNOMPQNOQP
L�Z � "CO"�O�O" LZ[ � "%P�O"C%Y�
L�Z � �% LZ[ � �
%
∴ They are collinear.
e) M!"& ; �4# N!1; "�# O!5; 10 "&#
L�\ � MNOMPQNOQP L\] � MNOMPQNOQP
L�\ � PNY&"OPR L\] � "C
PROPN:O"
L�\ � 6 L\] � %J")
∴ They are not collinear
To enter this into the calculator
press the following buttons:
2. a) L ��16 P��3;�6� Q�B; 2� L^_ � MNOMPQNOQP
�16 � �Y)�Y%
�16B � 48 � 8
�16B � 56
∴ B � �3 "�
b) L ��1 R��3; 9� S�1; F� L`a � MNOMPQNOQP
�1 � �OJ"Y%
�4 � F � 9
∴ F � 5
c) L � �% T!� "
& ; H# U!�3 "& ; 2#
Lbc � MNOMPQNOQP
�% � �O�
O%PRYPR
�% � �O�O%
�6 � 6 � 3H
�12 ��3H ∴ H � 4
d) L �� %� V��;�8� W!1 "
� ; �11#
L�� � MNOMPQNOQP
� %� �O""YX"PNO�
�4 "
� 3� ��6
3� � �1 "�
∴ � � � "�
e) L � 10 X!� "% ; � �
%# Y�0; K� L�� � MNOMPQNOQP
10 � iYN�CYP�
"C% � K �
%
∴ K � X% q�2 �
%
Activity 3
1. a) A!"� ; 3 "�# B�12; 15� D�2;�3� E��3; 2�
L� � MNOMPQNOQP LTU � MNOMPQNOQP
L� � ":O%PN
"�OPN LTU � �Y%O%O�
L� � 1 LTU ��1
∴ L" × L� ��1 ∴ � ⊥ $'
b) A�2; 0� B!� "� ; �%# D!�5;� "
%# E!2;�2 ":#
L� � MNOMPQNOQP LTU � MNOMPQNOQP
L� �N�OCOPNO�
LTU �O�P�YP��Y:
L� �� &": LTU �� &
":
∴ � ∥ $'
c) A!� :� ; �6# B!� "
� ; 1# D�7;�1� E�14;�3� L� � MNOMPQNOQP LTU � MNOMPQNOQP
L� � "Y)OPNY�N LTU �O%Y""&OI
L� � 3 "� � I
� LTU �� �I
∴ L" ×L� ��1 ∴ � ⊥ $'
d) A�4;�2� B�1; 8� D!� :� ; 16# E!�15; 12 "
&#
L� � MNOMPQNOQP LTU � MNOMPQNOQP
L� � XY�"O& LTU � "�PRO")
O":Y�N
L� �� "C% LTU � %
"C
∴ L" × L� ��1 ∴ � ⊥ $'
e) A�2; 3� B�1; 6� D��2;�1� E��4; 5� L� � MNOMPQNOQP LTi � MNOMPQNOQP
L� � )O%"O� LTU � :Y"O&Y�
L� ��3 LTU ��3
∴ � ∥ $'
2. a) � � 6� � 5 � � 6� � 1
∴ ∥
b) � � �% � 10 � � &
) � � 16
∴ ∥
c) � � 8� "� � � � "
X � "%
8 � "X ��1
∴⊥
d) � � "� � 4 � � &
X � � 2
∴ ∥
e) � � �1� � 16 � � "� � 5
neither
f) � � "& � "
: � � �4� � 2
"& × �4 � �1 ∴⊥
g) � � 2� 5 � � &X � � 1
� � "� � � 1
∴ neither
h) � � �3� �% � � 3� � 4
neither
i) � � � � 10 � � 2!� "� � 3#
� � �1� 6
∴ 1 ×�1 � �1
∴⊥
j) � � 9� 9 � � � "J � � 3
9 � "J ��1
∴ ⊥
Activity 4
1. a) A��1;�9� B�1; 3� b) C!� "� ; �2# D!�1 "
� ; 7#
4� � !QPYQN� ; MPYMN� # 4ST � !QPYQN� ; MPYMN� #
4� � !O"Y"� ; OJY%� # 4ST � �OPNO"PN� ; O�YI� �
4� � �0;�3� 4ST � !�1; 2 "�#
c) E�3; 6� F�5;�5� d) G!2 �% ; 1# H�3; 2�
4UV � !QPYQN� ; MPYMN� # 4�� � !QPYQN� ; MPYMN� #
4UV � !%Y:� ; )O:� # 4�� � ��N�Y%� ; "Y�� �
4UV � !4; "�# 4�� � !2 :) ; %�#
e) I ��2; 3� J�7;�1�
4�� � !QPYQN� ; MPYMN� #
4�� � !O�YI� ; %O"� #
4�� � !:� ; 1#
2. a) K��2; 1� 4Z[��6;�6�
��6;�6� � !O�YQ� ; "YM� #
∴ � → �6 � O�YQ� ∴ � →�6 � "YM�
∴ �12 � �2 � ∴ �12 � 1 �
∴ � � �10 ∴ � � �13
b) N�8;�3� 4\]�6; 5�
�6; 5� � !XYQ� ; O%YM� #
∴ 6 � XYQ� ∴ 5 � O%YM�
∴ 12 � 8 � ∴ 10 � �3 �
∴ � � 4 ∴ � � 13
c) P�1;�8� 4^_ !": ; "&#
!": ; "&# � !"YQ� ; OXYM� #
∴ ": � "YQ� ∴ "
& �OXYM�
∴ 2 � 5 5� ∴ 2 � �32 4�
∴ 5� � �3 ∴ 4� � 34
∴ � � � %: ∴ � � 8 "
�
d) R!�1;� &:# 4`a !� "
� ; � ":#
!� "� ; � "
:# � �O"YQ� ; OR�YM� �
∴ � "� �O"YQ� ∴ � "
: �OR�YM�
∴ �1 � �1 � ∴ � �: �� &
: �
∴ � � 0 ∴ � � �:
e) T�4;�5� 4bc��5;�2�
∴ ��5;�2� � !&YQ� ; O:YM� #
∴ �5 � &YQ� ∴ �2 � O:YM�
∴ �10 � 4 � ∴ �4 � �5 �
∴ � � �14 ∴ � � 1
Activity 5
1. a) A!&: ; �5# and B!� "% ; � "
%#
L� � MNOMPQNOQP
L� �OP�Y:
OP�OR�
L� �� IC"I
∴ � "% �� IC
"I !� "%# H
∴ H � � �J"I
∴ � � � IC"I � � �J
"I
b) C!1 "� ; 3# and D!�1;� "
�#
LST � MNOMPQNOQP
LST � OPNO%
O"O"PN
LST � I:
∴ 3 � I: !%�# H
∴ H � J"C
∴ � � I: � J
"C
c) E��3; 4� and F!�3;� "&#
LUV � MNOMPQNOQP
LUV � OPRO&O%Y%
LUV � ���Kz��K�
∴ � � �3
d) G�1; 2� and H��1; 8�
L�� � MNOMPQNOQP
L�� � XO�O"O"
L�� ��3
∴ 2 � �3�1� H
∴ H � 5
∴ � � �3� 5
This means that � � Hq���B�� i.e. that the�-value that appears in
each coordinate i.e. -3 is your constant
because you have a vertical line.
e) I�3; 2� and J�6; 2�
L�� � MNOMPQNOQP
L�� � �O�)O%
L�� � 0
∴ � � 2
2. a) the line passes through the point K��1; 2�and is parallel to the line
� � 2� 1.
∴ L � 2
∴ 2 � 2��1� H
∴ H � 4
∴ � � 2� 4
b) the line passes through the point L�3; 8� and is perpendicular to the
line � � �4� � 2.
∴ L � "&
∴ 8 � "& �3� H
∴ H � 7 "&
∴ � � "& � 7 "
&
c) the line passes through the point M!9;� "�# and is perpendicular to the
line � � 6� � :).
∴ L � � ")
∴ � "� �� "
) �9� H
∴ H � 1
∴ � � � ") � 1
This means that � � Hq���B�� i.e. that the�-value that appears in
each coordinate i.e. 2, is your constant
because you have a horizontal line.
d) the line passes through the point N!� ": ; 9# and is parallel to the line
� � � "% � � 4
∴ L � � "%
∴ 9 � � "% !� "
:# H ∴ H � 8 "&
":
∴ � � � "% � 8 "&
":
e) the line passes through the point P�4;�6� and is perpendicular to the
line � � 5� � 4.
∴ L � � ":
∴ �6 � � ": �4� H
∴ H � �5 ":
∴ � � � ": � � 5 "
:
Activity 6
1. a) � � 8� ") b) � � �6� � 3
∴ L � 8 ∴ L � �6
∴ tanw � 8 ∴ tanw � �6 ∴ w � 82,87° ∴ w � 80,54
∴ B�{�K � 180 � 80,54
∴ B�{�K � 99,46°
c) � � &: � � 3 d) � � �2� 1
∴ L � &: ∴ L � 2
∴ tanw � &: ∴ tanw � 2
∴ w � 38,66° ∴ w � 63,43
∴ B�{�K � 180 � 63,43 � 116,57°
Remember to find an angle you need
to press gradient.
e) � � �3� � %: f) � � � "
% � � &%
∴ L � 3 ∴ L � "%
∴ tanw � 3 ∴ tanw � "%
∴ w � 71,57 ∴ w � 18,43
∴ B�{�K � 180 � 71,57 ∴ B�{�K � 180 � 18,43
∴ B�{�K � 108,43° ∴ B�{�K � 161,57°
g) 2� � � 3 h) �4� � � � 6
∴ � � "� � %
� ∴ � � � "& � )
&
∴ L � "� ∴ L � "
&
∴ tanw � "� ∴ tanw � "
&
∴ w � 26,57° ∴ w � 14,04
∴ B�{�K � 180 � 14,04
∴ B�{�K � 165,96°
i) � � %� � � 2 j) � � �1� 2
∴ L � %� ∴ L � 1
∴ tanw � %� ∴ tanw � 1
∴ w � 56,31° ∴ w � 45
∴ B�{�K � 180 � 45
∴ B�{�K � 135°
2. a) w � 60.95° b) w � 99.46° ∴ tanw � L ∴ tanw � L
∴ L � 1.8 ∴ L � �6
c) w � 75.96° d) w � 45° ∴ tanw � L ∴ tanw � L
∴ L � 4 ∴ L � 1
e) w � 63.43° f) w � 9.46° ∴ tanw � L ∴ tanw � L
∴ L � 2 ∴ L � ") � 0,2
g) w � 98.13° h) w � 101.31° ∴ tanw � L ∴ tanw � L
∴ L � �7 ∴ L � �5
i) w � 120.96° j) w � 85.24° ∴ tanw � L ∴ tanw � L
∴ L � �1 �% ��1,7 ∴ L � 12
Activity 7
1. a) C�9; 8� � � 12 b) C�3;�1� � � 6
�� � �� � B�� �� � F�� �� � �� � B�� �� � F��
∴ �12�� � �� � 9�� �� � 8�� ∴ �6�� � �� � 3�� �� � ��1��
∴ 144 � �� � 9�� �� � 8�� ∴ 36 � �� � 3�� �� 1��
c) C�7;�5� � � 1
�� � �� � B�� �� � F��
∴ �1�� � �� � 7�� �� � ��5���
∴ 1 � �� � 7�� �� 5��
d) C��1; 4� � � 3
�� � �� � B�� �� � F��
∴ �3�� � �� � ��1��� �� � 4��
∴ 9 � �� 1�� �� � 4��
e) C!� "& ; 4# � � 5
�� � �� � B�� �� � F��
∴ �5�� � d� � !� "&#e
� �� � 4��
∴ 25 � !� "&#
� �� � 4��
2. a) C!1 "� ; 8# P!2;� "
�# b) C�1; 6� P�10; 12�
�� � !� � 1 "�#
� �� � 8�� �� � �� � 1�� �� � 6��
�� � !2 � 1 "�#
� !� "� � 8#� �� � �10 � 1�� �12 � 6��
∴ �� � 72 "� ∴ �� � 117
∴ 72 "� � !� � 1 "
�#� �� � 8�� ∴ 117 � �� � 1�� �� � 6��
c) C�5; 6� P!�1; �%# d) C��2; 3� P��1;�2�
�� � �� � 5�� �� � 6�� �� � �� 2�� �� � 3��
�� � ��1 � 5�� !�%� 6#� �� � ��1 2�� ��2 � 3��
∴ �� � 64 &J ∴ �� � 26
∴ 64 &J � �� � 5�� �� � 6�� ∴ 26 � �� 2�� �� � 3��
e) C��3;�1� P�2; 2�
�� � �� 3�� �� 1��
�� � �2 3�� �2 1��
∴ �� � 34
∴ 34 � �� 3�� �� 1��
3. a) 9 � �� � 8� �� 4�
9 � � � 8� !X�#� � !X�#
� �� 4� !&�#� � !&�#
�
∴ 9 16 4 � �� � 8� 16� ��� 4� 4� ∴ 29 � �� � 4�� �� 2��
∴ � � √29 Centre �4;�2�
b) 81 � �� � 4� �� 8�
81 � �� � 4� !&�#� � !&�#
� �� 8� !X�#� � !X�#
�
∴ 81 4 16 � ��� � 4� 4� ��� 8� 16� ∴ 101 � �� � 2�� �� 4��
∴ � � √101 Centre �2;�4�
c) �20 � �� 16� �� 18�
�20 � �� 16� !")X #� �� 18� !"X� #
� � !"X� #�
∴ �20 64 81 � ��� 16� 64� ��� 18� 81� ∴ 125 � �� 8�� �� 9��
∴ � � √125 Centre ��8;�9� ∴ � � 5√5
d) 128 � �� 14� �� � 2� 128 � �� 14� !"&� #
� � !"&� #� �� � 2� !��#
� � !��#�
∴ 128 49 1 � ��� 14� 49� ��� � 2� 1� ∴ 178 � �� 7�� �� � 1��
∴ � � √178 Centre ��7; 1�
e) 2 "%") � �� � 6� �� � 2�
2 "%") � �� � 6� !)�#
� � !)�#� �� � 2� !��#
� � !��#�
∴ 2 "%") 9 1 � ��� � 6� 9� ��� � 2� 1�
∴ 12 "%") � �� � 3�� �� � 1��
∴ � � �12 "%") Centre �3; 1�
∴ � � √�C:&
Activity 8
1. a) C�6;�6� P��1; 1�
Lp�� � MNOMPQNOQP
Lp�� � "Y)O"O) Lp�� ��1 ∴ Lj�mn � 1
∴ � � 1� H Substitute P ��1; 1� ∴ 1 � 1��1� H ∴ H � 2
∴ � � � 2
b) C��1;�3� P�4; 6�
Lp�� � MNOMPQNOQP
Lp�� � )Y%&Y"
Lp�� � J: ∴ Lj�mn �� :
J
∴ � � � :J � H Substitute P �4; 6�
∴ 6 � � :J �4� H
∴ H � 8 �J
∴ � � � :J � 8 �
J
c) C��7; 4� P��8;�2�
Lp�� � MNOMPQNOQP
Lp�� �O�O&OXYI
Lp�� � 6 ∴ Lj�mn �� ")
∴ � � � ") � H Substitute P ��8;�2�
∴ �2 � � ") ��8� H
∴ H � �3 "%
∴ � � � ") � � 3 "
%
d) C�8; 3� P��6; 4�
Lp�� � MNOMPQNOQP
Lp�� � &O%O)OX
Lp�� �� ""& ∴ Lj�mn � 14
∴ � � 14� H Substitute P ��6; 4� ∴ 4 � 14��6� H
∴ H � 88
∴ � � 14� 88
e) C�5; 3� P��5; 4�
Lp�� � MNOMPQNOQP
Lp�� � &O%O:O:
Lp�� �� ""C ∴ Lj�mn � 10
∴ � � 10� H Substitute P ��5; 4� ∴ 4 � 10��5� H ∴ H � 54
∴ � � 10� 54
2. a) 10 � �� 12� �� � 4� P�1; 3�
4 36 10 � �� 6�� �� � 2��
50 � �� 6�� �� � 2��
Centre ��6; 2�
Lp�� � MNOMPQNOQP
Lp�� � �O%O)O"
Lp�� � "I ∴ Lj�mn � �7
� � �7� H Substitute P �1; 3� ∴ 3 � �7�1� H
∴ H � 10
∴ � � 7� 10
b) 8 � �� 8� �� � 2� P��7;�3�
1 16 8 � �� 4�� �� � 1��
25 � �� 4�� �� � 1��
Centre ��4; 1�
Lp�� � MNOMPQNOQP Lp�� �O%O"OIY&
Lp�� � &% ∴ Lj�mn �� %
&
∴ � � � %& � H Substitute P ��7;�3�
∴ �3 � � %& ��7� H
∴ H � �8 "&
∴ � � � %& � � 8 "
&
c) 124 � �� 4� �� � 12� P�8;�2�
36 4 124 � �� 2�� �� � 6��
164 � �� 2�� �� � 6��
Centre ��2; 6�
Lp�� � MNOMPQNOQP
Lp�� �O�O)XY�
Lp�� �� &: ∴ Lj�mn � :
&
∴ � � :& � H Substitute P �8;�2�
∴ �2 � :& �8� H
∴ H � �12
∴ � � :& � � 12
d) �48 � �� 14� �� 16� P��6; 0�
64 49 � 48 � �� 7�� �� 8��
65 � �� 7�� �� 8��
Centre ��7;�8�
Lp�� � MNOMPQNOQP
Lp�� � CYXO)YI Lp�� � 8 ∴ Lj�mn �� "
X
∴ � � � "X � H Substitute P ��6; 0�
∴ 0 � � "X ��6� H
∴ H � � %&
∴ � � � "X � � %
&
e) 67 � �� 10� �� 6� P�5;�4�
9 25 67 � �� 5�� �� 3��
101 � �� 5�� �� 3��
Centre ��5;�3�
Lp�� � MNOMPQNOQP
Lp�� �O&Y%:Y:
∴ Lp�� �� ""C ∴ Lj�mn � 10
∴ � � 10� H Substitute P �5;�4� ∴ �4 � 10�5� H
∴ H � �54
∴ � � 10� � 54
3. a) � � 3 T��2; 6� C�B; 4� * use distance formula for this question *
$bS ����� � �"�� ��� � �"�� �3�� � �B 2�� �4 � 6��
9 � �B 2�� 4
5 � �B 2��
±√5 � B 2 ±√5 � 2 � B
∴ B � √5 � 2 or B � �√5 � 2
∴ B � 0,24 B � �4,24
b) � � 7 T�3; 9� C��3; F�
$bS ����� � �"�� ��� � �"�� �7�� � ��3 � 3�� �F � 9��
49 � 36 �F � 9��
13 � �F � 9��
±√13 � F � 9
9 ±√13 � F
∴ F � 12,61 or F � 5,39
c) � � 12 T!8; "�# C�H; 10�
$bS ����� � �"�� ��� � �"�� �12�� � �H � 8�� !10 � "
�#�
144 � �H � 8�� 90 "&
53 %& � �H � 8��
±�53 %& � H � 8
8 ±�53 %& � H
∴ H � 15,33 or H � 0,67
d) � � 5 T��3; 6� C��; 2�
$bS ����� � �"�� ��� � �"�� �5�� � �� 3�� �2 � 6��
25 � �� 3�� 16
9 � �� 3��
±3 � � 3
∴ � � �6 or � � 0
e) � � 8 T�4; 1� C�5; K�
$bS ����� � �"�� ��� � �"�� �8�� � �5 � 4�� �K � 1��
64 � 1 �K � 1��
63 � �K � 1��
±√63 � K � 1 ∴ K � 8,94 or K � �6,94
4. a) L � "I 333 � �� 18� �� 12� T�12; ��
36 81 333 � �� 9�� �� 6��
450 � �� 9�� �� 6��
Centre ��9;�6�
Lp�� � MNOMPQNOQP
∴ "I � �Y)"�YJ
∴ �"I � � 6
∴ 3 � 6 � �
∴ � � �3
b) L �� "& 42 � �� 10� �� � 2� T�K;�1�
1 25 42 � �� 5�� �� � 1��
68 � �� 5�� �� � 1��
Centre ��5; 1�
Lp�� � MNOMPQNOQP
∴ � "& �O"O"iY:
∴ 1K 5 � 8
∴ K � 3
c) L �� """ 93 � �� � 10� �� � 4� T��6; z�
4 25 93 � �� � 5�� �� � 2��
122 � �� � 5�� �� � 2��
Centre �5; 2�
Lp�� � MNOMPQNOQP
∴ � """ � �O�O)O:
∴ � """ � �O�
O""
∴ 1 � z � 2 ∴ z � 3
d) L � ���Kz��K� �4 � �� 2� �� 4� T�{;�3�
4 1 � 4 � �� 1�� �� 2��
1 � �� 1�� �� 2��
Centre ��1;�2�
Lp�� � MNOMPQNOQP
∴ ?C �O%Y�nY"
∴ { 1 � 0
∴ { � �1
e) L �� "&"% 157 � �� 16� �� � 24� T�5; |�
144 64 157 � �� 8�� �� � 12��
365 � �� 8�� �� � 12��
Centre ��8; 12�
Lp�� � MNOMPQNOQP
∴ � "&"% � kO"�:YX
∴ � "&"% � kO"�"%
∴ �14 � | � 12
∴ | � �2
Activity 9
1. a) �-intercept
∴ � � "& �0� 1
∴ � � 1 (0; 1) → Centre of the circle
�� � �� � B�� �� � F��
�� � �� � 0�� �� � 1�� Subs (-4; 0)
�� � ��4�� �0 � 1��
∴ �� � 17
∴ � → 17 � �� �� � 1��
b) radius = 1 �-coordinate of centre → 0
�� � �� � B�� �� � F��
�1�� � �� � B�� �� � 0��
1 � ��4 � B�� �0��
∴ ±1 � �4 � B
∴ �B ��1 4 or �B � 1 4
∴ B � �3 ∴ B � �5
From the graph we can see that the centre must be at (-3; 0).
∴ 1 � �� 3�� ��
c) � � "& � 1 d → 17 � �� �� � 1��
c → 1 � �� 3�� ��
∴ d → 17 � �� !"& � 1 � 1#�
17 � �� !"& �#�
17 � �� "") ��
272 � 16�� ��
272 � 17��
16 � ��
∴ � � 4 or � � �4
∴ � � "& �4� 1
∴ � � 2 ∴ �4; 2�
∴c → 1 � �� 3�� !"& � 1#�
1 � �� 6� 9 "") �� "
� � 1
16 � 16�� 96� 144 �� 8� 16
0 � 17�� 104� 144
∴ 0 � �17� 36��� 4� ∴ � � �2 �
"I or � � �4
∴ � � "& !�2 �
"I# 1
∴ � � X"I ∴ !�2 �
"I ; X"I#
d) Lp�� � MNOMPQNOQP Centre ��3; 0� ∴ Lp�� � CO"O%Y%
∴ Lp�� � ���Kz��K� Tangent Point ��3; 1�
∴ Lj�mn � 0
∴ � � 1
e) � � 1
∴ 17 � �� �� � 1��
∴ 17 � �� �1 � 1��
∴ 17 � ��
∴ � � ±√17
∴ � � �4,12 or � � 4,12
∴ ��4,12; 1� �4,12; 1�
Remember that the inverse of
undefined (fghijklmn
C ) is 0, and
that the inverse of 0
(C
fghijklmn) is undefined.
2. a) Method 1 → gradients
L�S � MNOMPQNOQP L S � MNOMPQNOQP
L�S � &OCO"O� L S � CY%�Y�
L�S � � &% L S � %
&
∴ L" ×L� ��1
∴ }�is a right angle.
Method 2 → distance and Pythagoras
∴ $� ����� � �"�� ��� � �"��
∴ $� ����1 2�� �4 3��
∴ $� � 5√2
∴ $ S ����� � �"�� ��� � �"��
∴ $ S ����2 � 2�� ��3 � 0��
∴ $ S � 5
∴ $�� ����� � �"�� ��� � �"��
∴ $�S ����1 � 2�� �4 � 0�� ∴ $�S � 5
∴ �� � � � �
∴ !5√2�#� � �5�� �5��
∴ 50 � 25 25 � 50
∴ }�is a right angle because the hypotenuse squared is equal to the
sum of the squares of the other two sides.
b) 4\ !QPYQN� ; MPYMN� # A(-1; 4)
∴ 4\ !O"O�� ; &O%� # B (-2; -3)
∴ 4\ !� %� ; "�#
c) LS\ � MNOMPQNOQP C (2; 0)
∴ LS\ � COPN�Y�N
∴ LS\ �� "I
∴ � � � "I � H Subs (2; 0)
∴ 0 � � "I �2� H
∴ H � �I
∴ � � � "I � �
I
d) L�S �� &% (From question A)
∴ tanw � &%
∴ w � 53,13° ∴ angle = 180° � 53,13° � 126,87°
3. a) B (0; -2) 4 S !QPYQN� ; MPYMN� #
C (-2; 4) 4 S��1; 1�
b) ∴ �� � �� 1�� �� � 1��
∴ �� � ��2 1�� �4 � 1��
∴ �� � 10
∴ 10 � �� 1�� �� � 1��
c) C(-2; 4) A(-1; 1)
L�S � MNOMPQNOQP
∴ L�S � "O&O"Y�
∴ L�S ��3 ∴ Lj�mn � "%
∴ � � "% � H Subs (-2; 4)
∴ 4 � "% ��2� H
∴ H � 4 �% ∴ � � "
% � 4 �%
d) A Normal is perpendicular to a tangent.
∴ L\ � �3
∴ � � �3� H Subs (-1; 1)
∴ 1 � �3��1� H ∴ H � �2
∴ � � �3� � 2
4. a) isosceles triangle → 2 sides are equal.
∴ $� ����� � �"�� ��� � �"��
∴ $� ���1 5�� !1 "�#
�
∴ $� � 6,18
∴ $ S ����� � �"�� ��� � �"��
∴ $ S ����5 � 1�� !� "� 2#�
∴ $ S � 6,18
∴ $�S ����� � �"�� ��� � �"��
∴ $�S ���1 � 1�� �1 2��
∴ $�S � 3
∴ 2 sides are equal and therefore ∆ABC is an isosceles triangle.
b) Line AC → � � 1
∴ D coordinate (1; 0)
∴ $TS ����� � �"�� ��� � �"��
∴ $TS ���1 � 1�� �0 2�� ∴ $TS � 2
Another Method:
The circle touches the �-axis
therefore the radius equals the
distance from the centre to the �-
axis which is the positive version of
the y-coordinate of the centre.
∴ r = 2
c) �� � �� � B�� �� � F��
�2�� � �� � 1�� �� 2��
4 � �� � 1�� �� 2��
d) Lp�� � MNOMPQNOQP
∴ Lp�� �O�OC"O"
∴ Lp�� � ���Kz��K� ∴ Lj�mn � 0
∴ � � 0
AND because D is the �-intercept of the circle it lies on the
circumference as well as the �-axis
∴ the �-axis is the tangent to the circle.
e) A (1; 1) B !�5;� "�#
L� � MNOMPQNOQP
∴ L� � OPNO"O:O"
∴ L� � "&
∴ � � "& � H (1; 1)
∴ 1 � "& �1� H
∴ H � %&
∴ � � "& � %
&
∴ �-intercept make � � 0
∴ 0 � "& � %
&
∴ � %& � "
& �
∴ � � �3 ∴ ��3; 0�
f) Line BC B !�5;� "�#
L S � MNOMPQNOQP C �1;�2� ∴ L S � O
PNY�O:O"
∴ L S �� "&
∴ � � � "& � H
∴ �2 � � "& �1� H
∴ H � �1 %& ∴ � � � "
& � � 1 %&
∴ 4 � �� � 1�� !� "& � � 1 %
& 2#�
∴ 4 � �� � 2� 1 !� "& � "
&#�
∴ 3 � �� � 2� "") �� � "
X � "")
∴ 48 � 16�� � 32� �� � 2� 1
∴ 0 � 17�� � 34� � 47
∴ � � O�±√�NO&����
∴ � � %&±��%&�NO&�"I��O&I���"I�
∴ � � 2,94 ≈ 3 or � � �0,94 ≈ �1
g) !�1;�1 "�# C �1;�2�
∴ Lp�� � MNOMPQNOQP
∴ Lp�� �O�Y"PN"Y"
∴ Lp�� �� "& ∴ Lj�mn � 4
∴ � � 4� H
∴ �1 "� � 4��1� H
∴ H � 3 "�
∴ � � 4� 3 "�
5. a) A ��4; 2� B �4; 4�
∴ 4� � !QPYQN� ; MPYMN� #
∴ 4� � !O&Y&� ; �Y&� #
∴ 4� � �0; 3�
b) �� � �� � B�� �� � F��
∴ �� � �� � 0�� �� � 3�� A ��4; 2� ∴ �� � ��4�� �2 � 3�� ∴ �� � 17
∴ 17 � �� �� � 3��
c) ∴ L� � MNOMPQNOQP (-4; 2)
∴ L� � �O&O&O& (4; 4)
∴ L� � "&
∴ B�{�K� → tanw � "&
∴ w � 14,04°
∴ 30,39° � 14,04° � 16,35°
∴ angle for BD � 180 � 16,35
� 163,65
∴ tanw � L ∴ L � �0,3
∴ � � �0,3� H Substitute B (4; 4)
∴ 4 � �0,3�4� H
∴ H � 5 ":
∴ � � �0,3� 5 ":
d) ∴ 17 � �� �� � 3��
∴ 17 � �� !�0,3� 5 ":� 3#�
∴ 17 � �� !�0,3� 2 ":#
�
∴ 17 � �� J"CC �� � 1 X
�: � 4 �"�:
∴ 1700 � 100�� 9�� � 132� 484
∴ 0 � 109�� � 132� � 1216
∴ 0 � �109� 304��� � 4� ∴ � � � %C&
"CJ or � � 4
∴ � � �2 X)"CJ ��2,8
∴ � � �0,3��2,8� 5 ":
∴ � � 6
∴ D ��2,8; 6�
e) Equilateral means all three sides are equal. A (-4; 2)
C (0; 3)
∴ $�S ����� � �"�� ��� � �"�� D (-2,8; 6)
∴ $�S ����4 � 0�� �2 � 3��
∴ $�S �√17
∴ $�S � 4,1
∴ $ST ����� � �"�� ��� � �"��
∴ $ST ���0 2,8�� �3 � 6�� ∴ $ST � 4,1
∴ $�T ����� � �"�� ��� � �"��
∴ $�T ����4 2,8�� �2 � 6�� ∴ $�T � 4,2
∴ approximately all the same ∴ ∆ADC is an equilateral triangle.
f) ∴ Lp�� � MNOMPQNOQP
∴ Lp�� � �O%O&OC
∴ Lp�� � "& ∴ Lj�mn ��4
∴ � � �4� H
∴ 2 � �4��4� H ∴ H � �14 ∴ � � �4� � 14
g) ∴ LTS � MNOMPQNOQP
∴ LTS � %O)CY�,X
∴ LTS �� ":"&
∴ � � � ":"& � H
∴ 3 � � ":"& �0� H
∴ H � 3 ∴ � � � ":"& � 3
∴ 17 � �� �� � 3�� ∴ � � � ":"& � 3
∴ 17 � �� !� ":"& � 3 � 3#�
∴ 17 � �� !� ":"& �#
�
∴ 17 � �� ��:"J) ��
∴ 3332 � 196�� 225��
∴ 3332 � 421��
∴ %%%�&�" � ��
∴ � � 2,8 OR � � �2,8
∴ E → � � � ":"& �2,8� 3
� � 0 ∴ E → �2,8; 0�
h) Lp�� �� ":"&
∴ Lj�mn � "&": for both.