Download [167] Find the value of y which makes the triangle ABC A right

4 5 1

 1
21
1
42
2
m 2 (BC) 

 1
24 2
52
3
m 3 (CA ) 

 1
1 4  3
From ( 1 ) , ( 2 ) and ( 3 )
m 1 ( AB ) 
.......... ( 1 )
.......... ( 2 )
.......... ( 3 )
 A , B and C are collinear
[167] Find the value of y which makes the triangle
ABC A right-angled triangle at C , where A ( 2 , 3
),
B ( 5 , 7 ) and C ( 1 , y ) .
Solution:
y3 y3
y7 y7

, m 2 (CB ) 

1 2
1
1 5
4
  ABC is a right  angled at C
m 1 (CA ) 
 m 1 (CA )  m 2 (CB )  1
y3 y7

 1  y  3 y  7   4
1
4
 y 2  10 y  21  4  0  y 2  10 y  25  0

 y  5   0
2
y  5  0
y  0  5  5
[168] ABC is a triangle , the coordinates of its vertices
Are A ( - 4 , 1 ) , B ( 6 , 3 ) and C ( 4 , 5 ) if AD is
A median, M and N  AD where AM =MN = ND ,
Then find the value of x and y where M ( x , 2 ) and
N(2,y)
Solution:
1
 AD is a median  D is the mid  point of BC
 64 35
D  
,
   5,4 
2 
 2
x  ( 4)2  2  12  x  42  1 unit length
2
2
 MN  x  2   2  y  unit length
2
2
2
 ND  5  2   y  4   9  y  4  unit length
2
2
AD  5  ( 4)  4  1  9 2  3 2
 AM 
 81  9  90 unit length
 AD  3  AM  3  ND

x  42  1
2
90 9  x  4   1
 
9
9
 x2  8 x  7  0
 x 2  8 x  16  1  10  0
x  0  1  1
or
 90  3  9  y  4 
2
90 9  9  y  4 
 
9
9
 y 2  8 y  15  0
y  0  3  3
2
 x  7 x  1  0
 x  0  7  7


 90  9  x  4   1
 90  3 
2

2
 90  9   9  y  4  


 y 2  8 y  16  9  10  0
 y  3 y  5   0
or
y  05  5
[169] ABCD is a parallelogram, A ( - 3 , 0 ) ,
B  x  axis , C ( 7 , 4 ) , D  y  axis find the
Coordinates of D and B
2
Solution:
Let D ( 0 , y ) and B ( x , 0 )
 73 40
The mid - point of AC  
,
   2,2 
2 
 2
 x0 y0  x y
The mid - point of BD  
,
 , 
2  2 2
 2
 ABCD is a parallelog ram
x
x y
 2 , 2    , 
 2
2
2 2
y
  2 y  2 2  4
2
 B (4 , 0 ) and D ( 0 , 4 )
x  2 2  4
[170] Find the equation of the straight line whose slope
1
Is 4 and cuts 5 units of the positive part of the
x-axis.
Solution:
y
1
x5
4
[171] Find the value of k which makes the triangle
XYZ is an isosceles and right-angled triangle at
Y , where X ( 1 , 2 ) , Y ( - 2 , 6 ) and Z ( 2 , k )
Find its area.
Solution:
XY  ( 2  1) 2  (6  2) 2  3 2  4 2  9  16
 25  5 unit length
3
 2  22  k  62  16  k  62
2
2
16  k  6   5  16  k  6   5 2  25
YZ 

 k 2  12 k  36  16  25  0
 k 2  12 k  27  0
 k  3 k  9   0
k  0  3  3
or
m 1 ( XY ) 
k  09  9
k6
k6
,

4
2  ( 2)
4
62

 21  3
  XYZ is a right - angled at Y
m 2 ( YZ ) 
k6
k6 4
 1
 1 

3
3
4
Z ( 2 , 9 )
k  6  3  9

k  6  3
1
 Area   5  5  12.5 squared unit
2
[172] Prove that : The straight line passing through
the Points ( 4 , 2 ) and ( 5 , 6 ) is parallel to the
straight Line passing through the points ( 0 , 5 ) and
(-1,1)
Solution:
62 4
 4
54 1
1 5
4
m2 

4
1 0 1
From (1) and ( 2) : m 1  m 2
m1 
4
........(1)
.........( 2)
 The two straight line are parallel
[173] Prove that : The straight line whose equation is :
3 x – 2 y + 5 = 0 is parallel to the straight line whose
equation is : 6 y – 9 x + 7 = 0
Solution:
 2y  3x
5


2
2 2
3
5
3
y  x 
 m1 
..........(1)
2
2
2
6y 9x 7
6y  9x  7



6
6 6
9
7
9 3
y  x 
 m 2   .........(2)
6
6
6 2
From (1) and ( 2) : m 1  m 2
 The two straight line are parallel
[174] If the straight lines : 2 x + y + 1 = 0 and
x – a y + 1 = 0 are parallel , find the value of a
 2y  3x  5

Solution:
y  2 x  1
 m 1  2
 ay   x  1

1
1
y  x 
a
a
From (1) and ( 2) :
 m1  m 2

 ay  x
1


a
a a
1
 m2 
......( 2)
a
1 2

a
1
5
.......(1)
a 
11  1

2
2
[175] If the straight line : x – y = 5 and the straight line
Passing through the two points ( a , 3 ) and ( 4 , 5 )
Are parallel , find the value of a
Solution:
 y  x  5
y  x  5
 m 1  1 ....(1)
53
2

4a 4a
From (1) and ( 2)
m2 
....( 2)
2
1 2
1
4  a 
2
4a
1
  a  2  4  2
a  2
[176] If the straight line : a x + 3 y – 7 = 0 is parallel to
the Straight line passing through the points ( 2 , 3 )
and
( - 1 , 4 ) , then find the value of a
Solution:
 3y  ax  7
3y  ax 7


3
3
3
a
 m1 
......(1)
3

a
7
y 
x
3
3
34
1
m2 

2  ( 1) 3
.......( 2)
From (1) and ( 2) m 1  m 2

a 1

3
3
 a  1
6
 a  1
[177] Find the equation of the straight line which
Intercepts from the y-axis a positive part of length 3
Units and parallel to the straight line passing
Through the points ( 2 , 3 ) and ( 5 , 1 )
Solution:
For the second straight line : m 2 
31
2

25 3
 The two straight line are parallel  m 1  m 2 
2
3
 The equation of the first straight line is
y
2
x3
3
[178] Find the equation of the straight line passing
Through the origin and parallel to AB where
A = ( 1 , 3 ) and B = ( 5 , 4 )
Solution:

m 2 ( The slope of AB ) 
43 1

51 4

 The straight line parallel to AB
The equation of the straight line is y 
1
x
4
[179] Show that the points A ( - 1 , 1 ) , B ( 0 , 5 ) ,
C ( 4 , 2 ) and D ( 5 , 6 ) are the vertices of a
Parallelogram.
Solution:
7
m 1 (The slope of AB ) 
51
4
  4 .......(1)
0  ( 1) 1
62 4
 4
54 1
21
1
m 3 (The slope of AC ) 

4  ( 1) 5
m 2 (The slope of CD ) 
65 1

50 5
From (1) , ( 2) , ( 3) and (4)
m 4 (The slope of BD) 
........( 2)
........( 3)
...........(4)
 m 1  m 2 and m 3  m 4
 AB // CD and AC // BD
 ABDC is a paralle log ram
[180] Prove that : the points A ( 0 , 2 ) , B ( 4 , 8 ) and
C ( 6 , 11 ) are collinear.
Solution:
82 6 3
 
40 4 2
11  8 3
m 2 (The slope of BC ) 

64 2
From (1) and ( 2)
m 1 (The slope of AB ) 
...........(1)
............( 2)
 m1  m 2
 A and B and C are collinear
[181] Prove that : The point ( 2 , 3 ) lies on the straight
Line passing through the points ( 1 , 1 ) and ( 0 , - 1 )
Solution:
8
31 2
 2
21 1
11  2
m 2 (The slope of (1 , 1) and (0 ,  1) ) 

2
01
1
 m1  m 2
m 1 (The slope of ( 2 , 3) and (1 , 1) ) 
 ( 2 , 3) , (1 , 1) and (0 ,  1) are collinear
[182] If the straight line passing through the two
points
( 1 , 3 ) and ( - 1 , 5 ) is parallel to the straight line
passing through the two points ( 3 , 5 ) and ( x , y ) ,
Find the relation between x and y.
Solution:
53
2

 1
11  2
y5
m 2 (The slope of ( 3 , 5) and ( x , y ) 
x3
 The two straight line are parallel
 m1  m 2
m1 ( The slope of (1 , 3) and (1 , 5) ) 
y5
 1
x3
 y  5  x  3

 y  5  1( x  3)
 y  x  3  5
 y  x  8
[183] If the straight line : 2 x – y + 5 = 0 is parallel to
the Straight line : a x + y – 3 = 0 , then find the
value of a .
Solution:
9
m 1 (The slope of the first straight line )

a
 a
1
 a  2
 coefficient of x  2

2
coefficient of y
1
m2 
 a  2
[184] If the straight line passing through the points
( x , 5 ) and ( 2 , 3 ) is parallel to the straight line
Passing through the points ( 3 , 4 ) and ( 5 , 2 ) , then
Find the value of x .
Solution:
53
2

x2 x2
42
2
m 2 (The slope of ( 3 , 4) and (5 , 2) ) 

 1
35 2
 m1  m 2
m1 (The slope of (x , 5) and (2 , 3) ) 
2
 1
x2
 x  2  2  0

 1( x  2)  2
 x  2  2
[185] Prove that : the straight line passing through the
Points A ( - 1 , 4 ) and C ( - 1 , - 2 ) is perpendicular
To the straight line passing through the points
B ( 1 , 1 ) and D ( - 3 , 1 )
Solution:
10
m 2 (The slope of ( 3 , 4) and (5 , 2) ) 
42
2

 1
35 2
 m1  m 2
2
 1
x2
 x  2  2  0

 1( x  2)  2
 x  2  2
[186] Prove that : the straight lines : 2 x – y = 5 and
x + 2 y = 5 are orthogonal.
Solution:
m1 
2
2
1
 m1  m 2  2 
,
m2 
1
2
1
 1
2
Then the two straight lines are orthogonal
[187] Find the slope of the straight line perpendicular
to The straight line passing through the points
A ( 2 , - 3 ) and B ( 3 , 5 )
Solution:
m 2 (The slope of second straight line ) 
 m1 
5  ( 3 ) 8
 8
32
1
1 1

m2
8
[188] If the straight line passing through the two
points
( 2 , 3 ) and ( 5 , 7 ) is perpendicular to the straight
Line a x – 3 y = 7 , then find the value of a
Solution:
11
73 4

52 3
a a
m 2 (The slope of the second straight line ) 

3 3
 The two straight lines are orthogonal
m 1 (The slope of the first straight line ) 
 m2 

1
4 3
 1  
m1
3
4
a 3

3
4
[189]
a 
3  3  9

4
4


AB is perpendicular to the straight line
5 x – 4 y = 7 where A ( 3 , 4 ) and B ( 5 , y ) , find the
Value of y.
Solution:
5 5

4 4
y4 y4
m 2 (The slope of the second straight line ) 

53
2
 The two straight line are orthogonal
1
y4 4
 m2 


m1
2
5
m 1 (The slope of the first straight line ) 
2  4
8
y  4 
 1.6
5
5
 y  1.6  4  2.4
y  4 
[190] Find the equation of the straight line which
Intercepts from the y-axis a positive part of length 4
Units and is perpendicular to the straight line :
3x–4y+1=0
12
Solution:
m 2 (The slope of the second straight line ) 
 m1 
3 3

4 4
1
3 4
 1  
m2
4
3
The equation of the first equation is y 
4
x4
3
[191] Find the equation of the straight line passing
Through the origin and perpendicular to the
Straight line 2 x – 3 y = 5
Solution:
m 2 (The slope of the second straight line ) 
m1 
2 2

3 3
1
2 3
 1  
m2
3
2
Then the equation of the first equation is y 
3
x
2
[192] Find the equation of the straight line passing
Through the origin point and perpendicular to the
Straight line passing through the points A ( 2 , 5 )
And B ( - 3 , 4 )
Solution:
13
m 2 (The slope of the second straight line ) 
m1 
54
1

2  ( 3 ) 5
1
1
 1   5
m2
5
Then the equation of the first straight line is y  5x
[193] Prove that : The shape whose vertices A ( 1 , 1 ) ,
B ( 4 , - 2 ) , C ( 6 , 0 ) and D ( 3 , 3 ) is a rectangle.
Solution:
1 6 1 0
The mid  po int of AC  
,
  3.5,0.5
2 
 2
 3 4 3 2
The mid  po int of BD  
,
  3.5,0.5
2
2


Then The two diagonals AC and BD bi sects each other
1  ( 2 )
3

 1
1 4
3
0  ( 2) 2
m 2 (The slope of BC) 
 1
64
2
Then m1  m 2  1  1  1
m1 (The slope of AB ) 
Then AB  BC
Then the quadrilate ral ABCD is a rec tan gle
[194] Prove that : the points A ( 1 , 3 ) , B ( 6 , 4 ) ,
C ( 7 , 9 ) and D ( 2 , 8 ) are the vertices of
A rhombus.
Solution:
14
1 7 3 9
The mid  po int of AC  
,
  4,6 
2 
 2
 6 2 4 8
The mid  po int of BD  
,
  4,6 
2 
 2
Then the two diagonals AC and BD bi sec ts each other
93

71
48
m 1 (The slope of BD) 

62
 m 1  m 2  1  1  1
m 1 (The slope of AC ) 
6
1
6
4
 1
4
 AC  BD
Then the quadrilate ral ABCD is a r hom bus
[195] Prove that : ABC is a right-angled triangle at B
Where A ( - 1 , - 1 ) , B ( 2 , 3 ) and C ( 6 , 0 )
Solution:
m 1 (The slope of AB ) 
3  ( 1) 4

2  ( 1) 3
m 2 (The slope of BC) 
30
3

26 4
 m1  m 2 
4 3

 1
3 4
 AB  BC
 ABC is a right  angled triangle at B
[196] If ABC is a triangle in which A ( 5 , - 4 ) ,
B ( 4 , 4 ) and C ( - 2 , 0 ), then:
1) Prove that the triangle ABC is isosceles.
2) Find the area of the triangle ABC.
15
Solution:
2
2
AB  5  4   4  4
 1  64
 65 unit length
BC 
4  (2)2  4  02
 62  42
 36  16  52 unit length
AC 
5  (2)2   4  02
 72  42
 49  16  65 unit length
 AB  AC then  ABC is an isosceles triangle
 4 2 4 0
D (The mid - point of BC )  
,
  1 , 2
2
2


AD 
5  12   4  22
 42  62
 16  36  52 unit length
1
 BC  AD
2
1
1
  52  52   52  26 squared unit
2
2
Area of the triangle 
[197] The circle M passes through the point A ( 2 , 6 ).
If M ( 2 , - 1 ) , find:
1) The circumference of the circle M.
22 





2) Its area, 
7 
Solution:
16
r : ( radius of the circle  AM )

2  22  6  ( 1)2
 7 2  7 unit length.
C ( The circumfere nce of the circle )  2    r
22
 7  44 cm.
7
22
Area of the circle   r 2 
 7 2  154 squared unit.
7
 2
[198] Find the equation of the straight line that passes
Through the point ( 3 , 4 ) and is parallel to the
Straight line 2 x + y + 7 = 0 , then find the X-axis
Intercept and the Y-axis intercept by the
Obtained straight line.
Solution:
17
m 1 ( The slope of the straight line 2 x  y  7  0 )
2
 2
1
m 2 ( The slope of the straight line perpendicu lar to

The straight line 2 x  y  7  0 )  1  2 
1
2
The general form of the straight line y  m x  b
At ( 3 , 4 )
4
1
3b
 b  4  1 .5  2 .5
2
 y  0.5 x  2.5
The intercepte d part of y - axis at x  0
y  0.5  0  2.5
y  2.5
The intercepte d part of x - axis at y  0
0  0.5 x  2.5
0.5 x  2.5

0.5
0 .5
 0.5 x  0  2.5  2.5
x   5
[199] If the points A ( 2 , 3 ) , B ( - 1 , - 1 ) and
C ( 3 , - 4 ) form a triangle ABC , prove that
The triangle ABC is right-angled triangle then
Find its area.
Solution:
2
2
AB  2  (1)  3  ( 1)
 32  42
 25  5 unit length.
18
BC 
3  (1)2   4  (1)2
 42  32
 16  9  25  5 unit length.
3  22   4  32
AC 
 1  72
 50 unit length
 AB   BC  5 2  5 2  50 squared unit.
2
2
AC 2   50   50 squared unit.
2
2
2
 AB   BC  AC 
2
 The triangle ABC is right - angled at B
Area of the triangle 

1
 AB  BC
2
1
 5  5  12.5 squared unit.
2
[200] Graph the points A ( 5 , 6 ) , B ( 4 , 0 ) and
C ( - 1 , 5 ) on the coordinates plane then draw
The triangle ABC. Prove that:
1) The triangle ABC is isosceles
2) Find the equation of the straight line that
Passes through the mid-point of AB
3) Prove that A  The perpendicu lar to BC
Solution:
AB 
5  42  6  02
 1  62
 1  36  37 unit length.
[201] Prove that the points A ( 3 , - 1 ) , B ( - 4 , 6 )
, C ( 2 , - 2 ) which belong to an orthogonal
Cartesian co – ordinates plane lie on the circle
19
whose centre M ( - 1 , 2 ) then find the
circumference of the circle.
Solution
MA   1  3    2  1
2
2
 5 unit length
MB 
 1  4    2  6   5 unit length
MC 
 1  2    2  2   5 unit length
2
2
2
2
 A , B and C are lies on the circle M
Circumference of the circle = 2  r = 2  5  = 10 
[202] Find the value of a in each of the following :
1) If the distance between the two ( a , 7 ) and
( - 2 , 3 ) equals 5
2) If the distance between the two points ( a , 7 )
and ( 3 a – 1 , - 5 ) equals 13
Solution
1) d  5 
 a  2 
2
  7  3
2
 25   a  2   16   a  2   25  16  9
2
2
 a  2   9  3  a  3  2  1
or a  3  2  4
20
 3a  1  a    5  7 
 169   2a  1  144   2a  1
2) d  13 
2
2
2
2
 169  144  25
 2a  1   25  5  2a  5  1  6  a 
6
3
2
4
 2
2
[203] If A ( x , 3 ) , B ( 3 , 2 ) , C ( 5 , 1 ) and if AB = BC
find the value of x
or 2a  5  1  4  a 
Solution
AB   x  3    3  2 
BC   5  3    2  1
2
2
2
2
2
2
5
AB  BC  AB 2  BC2
  x  3  1  5   x  3  5  1  4
2
2
 x  3   4  2
 x  3  2  x  3  2  5 or x  3  2  x  2  3  1
[204] If the points ( 0 , 1 ) , ( a , 3 ) , ( 2 , 5 ) are
collinear
Find the value of a
Solution
21
51
2
20
31 2
m 2 : The slope of the two points ( 0 , 1 ) , ( a , 3 ) =

a0 a
2 2
   a  21  2  1
a 1
m1 : The slope of the two points ( 0 , 1 ) , ( 2 , 5 ) =
[205] If the distance between ( x , 5 ) and the point
( 6 , 1 ) equals 2 5 find the value of x
Solution
 x  6    5  1
2
2

 2 5

2
 20
 x  6   16  20   x  6   20  16  4
2
2
  x  6    4  2
2
 x  6  2  x  2  6  8 or x  2  6  4
[206] In which of the following cases , the point A , B
and C are collinear? Explain your answer.
1) A ( - 1 , 5 ) , B ( 0 , - 3 ) , C ( 2 , 1 )
2) A ( - 2 , 1 ) , B ( 2 , 3 ) , C ( 4 , 4 )
3) A ( 0 , 2 ) , B ( 4 , 8 ) , C ( 6 , 11 )
[207] Identify the type of the triangle whose vertices
are A ( - 2 , 4 ) , B ( 3 , - 1 ) , C ( 4 , 5 ) due to its
sides lengths.
Solution
22
AB 
 4  3    1  5   5 2 unit length
AC 
 2  3 
BC 
 2  3    1  4   5 2 unit length
2
2
2
  4  5   26 unit length
2
2
2
 AB = BC   ABC is isosceles triangle
[208] Prove that triangle whose vertices A ( 5 , - 5 )
, B ( - 1 , 7 ) , C ( 15 , 15 ) is right angled at B ,
then calculate its area.
Solution
AB   5  7    5  1  180 unit length
BC   15  1   15  7   320 unit length
AC   15  5    15  5   500 unit length
2
2
2
2
2
2
2
2
2
 AC2  AB 2  BC2   ABC is right angle at B
1
 180  320 
2
1
=  3 2  10  4 2  120 squared units
2
[209] Prove that ( 5 , 3 ) , ( 6 , - 2 ) , ( 1 , - 1 ) , ( 0 , 4 )
are vertices of a rhombus , then find its area.
Area =
Solution
23
AB 
 5  6    3  2   26 unit length
BC 
 6  1   2  1  26 unit length
CD 
1  0
2
 DA 
2
2
2
2
  1  4   26 unit length
2
 0  5
2
  4  3   26 unit length
2
 AB  BC  CD  DA  ABCD is a rhombus
AC 
 5  1
BD =
 6  0
2
2
  3  1  4 2 unit length
2
  2  4   6 2 unit length
2
1
 Area   4 2  6 2  24 squared unit
2
[210] Prove that the points A ( - 2 , 5 ) , B ( 3 , 3 )
, C ( - 4 , 2 ) are not collinear and if D ( - 9 , 4 )
prove that the figure ABCD is a parallelogram.
Solution
24
AB 
 2  3    5  3   29
BC 
 4  3    3  2   50
AC 
 4  2    5  2   45
2
2
2
2
2
2
 AC  AB  BC  AC  AB
 2  4 5  2 
The midpoint of AC is 
,
   3 ,3.5 
2
2 

 3  9 3  4 
The midpoint of BD is 
,
   3 ,3.5 
2 
 2
Then AC and BD bisect each other.
 ABCD is a parallelogram
[211] Let A ( 5 , - 6 ) , B ( 3 , 7 ) and C ( 1 , - 3 ). Find
the equation of the straight line which passes
through A and the mid – point of BC
Solution
 3  1 3  7 
Let D is the midpo int of BC = 
,
   2, 2 
2
2


6  2 8
 m The slope of AD =

52
3
8
y  m x + b  y =
x+b
3
8
16 22
At  2, 2   2 
2b  b  2

3
3
3
8
22
The equation is : y 
x
3
3


25
[212] Find the equation of the straight line passing
through the point ( 3 , - 5 ) and parallel to the
straight line x + 2 y – 7 = 0
Solution
1
2y 7 x
7 1
1
or 2 y = 7  x 
  y  xm
2
2
2 2
2 2
2
1
1
The equation y =
 b at  3 ,  5   5 
 3  b
2
2
1
 5  1.5  b  b  5  1.5  3.5  y   x  3.5
2
m
[213] Find the equation of the straight line which
intercepts the two axes two positive parts of
lengths 4 and 9 for X and Y – axes respectively.
Solution
The two points are ( 4 , 0 ) and ( 0 , 9 )
90 9

0  4 4
The equation of the straight line is y = m x + b
m( The slope ) =
9
x9
4
[214] If A ( 1 , - 6 ) , B ( 9 , 2 ) find the co-ordinates of
the points which divide AB into four equal parts
in length.
Then the equation is y =
Solution
26
 1  9 6  2 
Let the mid-point of AB is C 
,
 = 5 ,  4
2 
 2
 1  5 6  4 
The mid-point of AC is D 
,
 = 3 ,  5
2 
 2
 9  5 2  4 
The mid-point of BC is E 
,
 =  7 ,  1
2 
 2
[215] Prove that the points A ( 6 , 0 ) , B ( 2 , - 4 ) and
C ( - 4 , 2 ) are vertices of a right angled triangle
at B then find the co – ordinates of the point D
which makes the figure ABCD a rectangle.
Solution
4  0
1
26
4  2
m 2 The slope of BC 
 1
2  4
m1  m 2  1   ABC is right angled at B




m1 The slope of AB 
Let D ( x , y ) then the midpoint of
 6  4 0  2 
AC  
,
   1,1
2 
 2
 2  x 4  y 
The midpoint of BD  
,
   1,1 
2
2


2 x 1

  x2 2 x 0
2
1
4  y 1
and
  4  y  2  y  4  2  6  D ( 0 , 6 )
2
1
27
[216] If the points A ( 3 , 2 ) , B ( 4 , - 3 ) , C ( - 1 , - 2 ) ,
D ( - 2 , 3 ) are vertices of a rhombus find :
a) The co – ordinates of the point of intersection of
its two diagonals
b) The area of the rhombus ABCD.
Solution
AB   3  4    2  3 
2
2
 26 unit length
BC 
 4  1 
CD 
 1  2    2  3   26 unit length
DA 
 2  3 
2
  3  2   26 unit length
2
2
2
2
  3  2   26 unit length
2
 3  1 2  2 
M
,
  1 , 2
2
2


AC 
 3  1 
BD =
 4  2 
2
2
  2  2   4 2 unit length
2
  3  3   6 2 unit length
2
1
 4 2  6 2  24 squared unit
2
[217] If A ( - 1 , - 1 ) , B ( 2 , 3 ) , C ( 6 , 0 ) , D ( 3 , - 4 )
are four points on an orthogonal Cartesian co –
ordinates plane. Prove that AC and BD bisect
each other. What is the name of this figure?
Area =
Solution
28
 1  6  1  0 
The mid-point of AC is 
,
 =  2.5 ,  0.5
2 
 2
 2 3 34
The mid-point of BD is 
,
 =  2.5 ,  0.5 
2 
 2
The figure is a parallelogram
[218] ABCD is a parallelogram where A ( 3 , 4 )
, B ( 2 , - 1 ) , C ( - 4 , - 3 ) , find the co – ordinates
of point D then find the co – ordinates of E such
that the figure ABCE becomes a trapezium in
which AE // BC , AE = 2 BC
Solution
Let the point D ( x , y )
 3  4 4  3   1 1 
The mid-point of AC is 
,
, 
 =
2
2
2
2

 
 x  2 y  1   1 1 
The mid-point of BD is 
,
, 
 =
2
2
2
2

 
x  2 1


 x  2  1  x  1  2   3
2
2
y 1 1
  y 1  1 y  11  2  D (  3 , 2 )
2
2
Let the point E ( s , t )
 s  3 t  4  s  3 3 t  4 2
D (  3 , 2 )  
,

,


2
2
2
1
2
1


s  3  6  s  6  3   9
t  4  4  t  4  4  0  E  9,0 
29
[219] If the straight line L1 passes through the two
points ( 3 , 1 ) and ( 2 , k ) , and the straight line L2
makes with the positive direction of the X – axis
an angle of measure 45 , find the value of K if :
1) L1 // L2
2) L1 = L2
Solution
1)
k 1
 tan 45  1
23
 k  1  1  k   1  1  0
L1 // L 2 
k 1
 tan 45  1
23
k  1  1  1  1  k  1  1  2
2) L1  L 2 
[220] Using the slope to prove that the points A ( - 1 , 3
) , B ( 5 , 1 ) , C ( 6 , 4 ) , D ( 0 , 6 ) are vertices of a
rectangle.
Solution
31
2 1


1  5 6 3
6  4 2 1
The slope of CD  m 2  


0  6 6 3
41 3
The slope of BC  m 3  
 3
65 1
63 3
The slope of DA  m 4  
 3
0  1 1
1
m1  m 2 ,m 3  m 4 ,m1  m 3 
 3  1
3
 ABCD is a rectangle.
The slope of AB  m 1  
30
Model (1)
[1] Complete each of the following:
1) If A ( 1 , 2 ) , B ( 3 , 4 ) , then the coordinates of the
midpoint of AB is ............
2) The equation of the straight line which is parallel to
X – axis and passes through the point ( - 2 , 3 ) is .......
3) If x , y are the measures of two
complementary angles , where
x : y = 1 : 2 then sin x + cos y = .......
4) The distance between the points ( 6 , 0 ) , ( - 4 , 0 )
equals .........
5) If the point ( 0 , a ) belongs to the straight line
3 x – 4 y + 12 = 0 then a = ...........



2
6) If AB // CD and the slope of AB 
3

, then the slope of CD  .........
[2] Choose:
1) If cos 2 x =
1
, then m  x   ......
2
 15 , 30 , 45 , 60 
2) Theslope of the straight line whose equation
2 3 2
 3
2 x – 3 y  5  0 equals ......   ,  , , 
3 2 3
 2
31
3) The length of the line segment which is drawn
between the two points ( 0 , 0 ) , ( 5 , 12 ) equals
........... ( 5 , 7 , 12 , 13 )
1
1

4) tan 45  .......  3 ,
,1, 
2
3

5) In  ABC , if m   B   90 , then
sin A + cos C = ...................
 2 sin A , 2 sin C , 2 sin B , 2 cos A 
2 1
1
6) tan 45 sin 30  .......  , 1 , , 
3 4
2
[3]
ABC is right - angled triangle at B
2 AB = 3 AC , find the trigonometric
ratios of  C
[4] Find the equation of the straight line passes through
the point ( 2 , - 1 ) and parallel to the straight line
2x–y+5=0
[5]
Pr ove that : cos 60  cos 2 30  sin 2 30
[6] ABCD is a parallelogram , its diagonals intersects at
E , if A ( 3 , - 1 ) , B ( 6 , 2 ) , C ( 1 , 6 ) then find :
1) the coordinates of E , D
2) The length of DE
[7]
32
Pr ove that : tan 60  2 tan 30  1  tan 2 30

[8]
Find the slope , intercepted part of Y - axis
of the straight line whose equaation
x y
 1
2 3
Model (2)
[1] Complete each of the following:
1) If the two straight line 2 x + b y + 3 = 0 , 3 x – y + 2 =
0 are perpendicular , then b = ............
2) If sin x = 0.5 , x is an acute angle , then
m   x   ........
3) The distance between the two points ( 5 , 0 ) ,
( 0 , - 12 ) equals ...........
4) sin 60  cos 30  tan 60  .........
5) If the two straight lines k x – 2 y + 3 = 0 , 6 x + 3 y – 5
= 0 are parallel , then k = ............
6) The slope of the perpendicular straight line to the line
which passes through the two points ( 2 , 6 ) , ( - 4 , 1 )
equals .............
[2] Choose:
1) 2 sin 30 cos 30  ..........
 sin 60 , cos 60
, tan 60 , 2 sin 60 
2) The points ( - 3 , 0 ) , ( 0 , 3 ) , ( 3 , 0 ) are the vertices
of a triangle
a) Scalene triangle
b) Equilateral triangle
c) Obtuse – angled triangle
d) Right – angled triangle and isosceles
33
3) The equation of the straight line which passes
through the point ( 2 , - 3 ) , parallel to X – axis is
........... ( x = - 2 , y = - 3 , x = 2 , y = 3 )
4) If the straight line whose equation x + 3 y – 6 = 0 is
perpendicular to the straight line whose equation
a x – 3 y + 7 = 0 , then a = ....... ( 2 , 9 , - 9 , - 2 )
5) If the point ( 0 , 4 ) is the midpoint of the distance
between the two points ( - 1 , - 1 ) , ( x , y ) , then the
point ( x , y ) is ..........


 1 3
(
1
,
9
)
,
(
1
,
9
)
,

,
,
(
1
,
3
)




 2 2


6) In  ABC , if m   B   90 , AB = 3 cm
, BC = 4 cm , then sin A cos C
9 12 16 

= ........  1 ,
,
,

25 25 25 

[3] Find the equation of the straight line which passes
through the point ( 1 , 6 ) and the midpoint of AB ,
where A ( 1 , - 2 ) , B ( 3 , - 4 )
[4] Prove that :
sin3 30  9 cos 3 60  tan 2 45
[5] Prove that the triangle whose vertices A ( 1 , 4 ) ,
B ( - 1 , - 2 ) , C ( 2 , - 3 ) right at B , then find its
area.
[6] In the opposite figure:
34
AD  BC , AB = 13 cm , AC = 15 cm ,
find in the simplest form the value of :
tan   CAD   tan   BAD 
tan   CAD   tan   BAD 
A
15 cm
13 cm
[7] Find the equation of
the straight line which C
D
9
cm
passes through the point
( 3 , 4 ) and perpendicular to the straight line :
5x–2y+7=0
[8]
ABCD is a trapezium in which AD // BC
, m   B   90 if AB = 3 cm , AD = 6 cm
, BC = 10 cm , prove that :
cos   DCB   tan   ACB  
1
2
Model (3)
[1] Complete each of the following :
1) cos2 45 + tan 2 60  sin 30 = .........
2) If A ( 2 , - 1 ) , B ( 5 , 3 ) , then AB = ........
3) If L1 : k x – 2 y + 4 = 0 , L2 : x + 3 y – 7 = 0 and
L1  L2 , then k = ......
4) sin 30 cos 60  cos 30 sin 60  .........
5) The equation of the straight line which passes
through the point ( - 2 , 7 ) , parallel to Y – axis is
35
B
..............
6)  ABC is right - angled at A , if tan B = 1
, then tan C sin C cos C = ..........
[2] Choose:
1) The equation of the straight line whose slope is 1 ,
passes through the origin point is ........
(x=1,y=1,y=x,y=-x)


2) If LM  EO , E ( - 1 , 2 ) , O ( 0 , 0 )
1 1


, then the slope of LM  ......  2 ,  , , 2 
2 2


3) If tan 3 x = 3 , where ( 3 x ) is an acute

, then m (  x ) = .........  10 , 20 , 30 , 60 
4) If the origin point is a centre of a circle of diameter
length 6 unit length , then the point which belongs
to the circle is ........
(6 , 0) , (0 , -6) , ( 8 , 1) , (1 , 5)


5) In  ABC , if  C is right , then
sin B + cos B ....... 1   , > , < ,  
1
6) If sin x = , x is an acute , then
2

1
3
1 
sin 2 x = .......  1 , ,
,

4
2
3

[3] Prove that the triangle whose vertices the points :
36
Y ( 2 , 4 ) , X ( 0 , 6.8 ) , Z ( - 5 , - 1 ) is right –
angled triangle at Y
[4]
ABC is a triangle in which AB = AC = 10 cm
, BC = 12 cm , AD is a perpendicular to BC
intersects it at D , prove that :
1) sin B + cos C = 1.4
2) sin 2 C + cos 2 C = 1
[5] Without using calculator , find the value of :
cos 2 60  cos 2 30  tan 2 45
sin 60 tan 60  sin 30
[6] Represent graphically the points A ( 2 , 3 )
, B ( -1 , -1) , C ( 3 , - 4 ) , D ( 6 , 0 ) , in the
coordinate plane , then prove that they
are vertices of a square , then find its area.
[7]
Find the value of x , where 0 < x < 90
, if sin x sin 45 cos 45 tan 60
 tan 2 45  cos 2 60
1
[8] A straight line, its slope is 2 , intersects a positive
part of Y – axis of length two units , find :
1) The equation of this straight line
2) Its intersection point with the Y – axis.
37
Model (4)
[1] Complete each of the following:
1) The slope of the straight line which is parallel to the
straight line which passes through the two points
( 3 , 1 ) , ( 5 , - 1 ) equals ...............
2) The equation of the straight line which passes
through the origin point and perpendicular to the
straight line y = 2 x is ...................
3) The value of the expression:
sin 60 cos 30  cos 60 sin 30  ..........
4) If tan 3 x = 1 , where 3 x is an acute
angle , then x = ........
5) The slope of the perpendicular straight line to
straight line 3 x + 4 y – 9 = 0 is ...........
x
3
x
6) If cos 
, where is an
3
2
3
acute angle , then x = ........
[2] Choose:
1 1


1) sin 2 60  cos 2 60  ..........  0 , , , 1 
4 2


2) If the origin point is a centre of a circle of radius 3
unit length , then the point .......... belongs to it.
( 1 , 2 ) , ( -2 , 5) , ( 3 , 1) , ( 2 , 1)


3) The slope of the straight line which is parallel to the
X – axis is .......... ( - 1 , 0 , 1 , undefined )
4) If the slope of the straight line a x – y + 3 = 0 equals 1
38
1
 1

, then a  ............   , -1 , , 1 
3
 3

5) The perpendicular distance between the two straight
lines y – 3 = 0 , y + 2 = 0 equals ........ ( 1 , 2 , 3 , 5 )
6) If sin 30  cos  , where  is an acute angle
, then m      .......  60 , 45 , 10 , 30 
[3]
AB is a diameter of circle M if B ( 8 , 11 )
, M ( 5 , 7 ) , then find :
1) the coordinate of A
2) the length of the radius of the circle
3) the equation of the perpendicular straight
line to AB from the point B
[4] Find the value of x , if :
sin x = sin 60 cos 30  cos 60 sin 30
, where 0  x  90
[5]
 ABC is right at C in which AB = 10 cm
, BC = 8 cm. find the value of :
sin A cos B + cos A sin B
[6] Find the equation of the straight line which passes
39
through the two points ( 2 , 3 ) , ( - 3 , 2 )
[7] Without using calculator , find the numerical value
of the expression :
cos 60 sin 30  sin60 cos 30
[8] If the points A ( 1 , 0 ) , B ( - 1 , 4 ) , C ( 7 , 8 ) ,
D ( 9 , 4 ) in the coordinate plane , prove that : ABCD
is a rectangle , find the length of its diagonal.
Model (5)
[1] Complete each of the following:
1) If sin ( y + 7 ) = 0.5 then y = ........
2) The equation of the straight line which passes
through the point ( 3 , - 2 ) and parallel to X – axis is
..............
3) The distance between the point ( 4 , 3 ) , the origin
point in the coordinate plane equals .............
4) If m1 , m 2 are the slopes of two perpendicular
straight lines , then m1  m 2  ..........
5) 2 sin 30 cos 30  sin ........
6) If the straight line y = x sin 30  c passes
through the point ( 4 , 6 ) , then c = ..........
[2] Choose:
1) If 
2 k
,
are the slopes of two parallel
3 2
3 1
 4

straight lines , then k = .......   ,  , , 3 
4 3
 3

40
2) If AB is a diameter of a circle , where
A ( 3 , - 5 ) , B ( 5 , 1 ) , then the centre
of the circle is .........
 (4 , -2) , (4 , 2) , (2 , 2) , ( 8 , -2) 
3) sin 60  cos 30  tan 60  .........

3 ,3 3 , 3 ,2 3

4) If the distance between the two points ( a , 0 ) , ( 0 , 1 )
is 1 length unit , then a = ......... ( - 1 , 0 , 1 , 1 )
5) The straight line which passes through the two points
( 1 , y ) , ( 3 , 4 ) , its slope is tan 45 , then y = ...........
(1,-1,2,4)
6) If  XYZ right at Z , XY = 25 cm , YZ = 24 cm
 31 17

, then sin X + sin Y = ........ 
,
, 2 , 1
 25 25

[3] If the two equations of two straight lines L1 , L2
respectively 2 x – 3 y + a = 0 , 3 x + b y – 6 = 0. Then
find:
1) the value of b which makes L1 // L2
2) the value of b which makes L1  L2
3) If the point ( 1 , 3 ) lies on L1
, then find the value of a.
[4] In cause of wind, the upper part of a tree was
broken made an angle of measure 60 with the
41
Ground. If the point of contact of the top of the tree
with the ground was at distance 4 m from its bottom
, find the length of the tree to the nearest metre.
[5]
Pr ove that : sin 60  2 sin 30 cos 30
[6] ABCD is a parallelogram , A ( x , 2 ) , B ( 3 , 8 )
, C ( 9 , 10 ) and D ( 7 , 4 ) , find x
[7] Prove that the triangle whose vertices A ( 1 , - 2 )
, B ( - 4 , 2 ) , C ( 1 , 6 ) is an isosceles triangle.
[8] In the opposite figure:
m  C   40 , AC = 12 , find to the
A
nearest one decimal place the length
of AB and the length of BC to the
C
40
nearest cm.
Cairo 2012
[1] Complete:
1) If sin x =
3
, where x is the measure
2
of an acute angle , then x = .........
2) If A ( 5 , 3 ) , B ( 1 , 7 ) , then the coordinates
of the mid - point of AB is .......
3) The equation of the straight line whose slope 5 and
intersects 7 units from the positive part of the
y – axis is ...............
42
B
4) If ABCD is a rhombus , where A ( 6 , 4 ) , B ( 4 , - 2 ) ,
then the perimeter of the rhombus ABCD = .........
5) 2 cos 60  ........
6) In the opposite figure :
ABC is a right angled triangle at B
AB = 3 cm , BC = 4 cm , then cos C = ..........
[2] Choose:
1) If cos x = sin 45 , where x is the measure
C
3 cm
A
4 cm
B
of an acute angle , then x = ....... ( 15 , 30 , 45 , 60 )
2) The distance between the point ( 3 , - 5 )
and the x - axis equals ........ unit of length.
( - 5 , 3 , 5 , 34 )
3) If the straight line L is perpendicular to
the straight line whose equation y  2 x = 7
, then the slope of L equals .......
(3,2,
1 1
,
)
3
2

4) The slope of AB which passes through
1
1 

A ( 1 , 5 ) , B ( 2 , 3 ) is .....  2 , ,  2 ,

2
2 

5) The equation of the straight line which is parallel to
the line 3 x + 9 y – 6 = 0 and passes through ( 0 , 5 )
is ..............
6) If 2 sin x = tan 60 where x is an acute angle
, then m   x   ..........
 60 , 45 , 30 , 40 
43
[3]
Without using calculator , find the value
of x which satisfies : tan x = 4 cos 60 sin 30
where x is the measure of an acute angle
[4] If the straight line whose equation a x + 2 y – 3 = 0
is parallel to the straight line passing through the
points ( 2 , 3 ) and ( 1 , 5 ) which lie on the same
plane , then find the value of a
[5]
If AB is a diameter in circle M where
M ( 5 , 2 ) and A ( 1 , 3 ) , then find the
equation of the tangent to the circle at A
[6]The time in which the particle covers in distance of
3.5 metres from the beginning of the movement.
Giza 2012
[1] Complete:
1) cos 45  ............
2) The distance between the point ( - 3 , 4 ) and the point
of origin equals ...............
3) If L1 // L 2 and the slope of L1  1
, then the slope of L 2  ...........
4) 3 tan 60  .........
5) If ABCD is a square , where A ( 7 , - 2 ) , B ( - 3 , 1 ) ,
then the area of the square ABCD = .............
[2] Choose:
44
1) If sin x =
3
, wherexisanacuteangle
2
,thenx  ........  30 , 45 , 60 , 90 
2) If A ( 2 , 4 ) , B ( 6 , 0 ) , then the
coordinates of the mid-point
of AB  ........
3) If sin x = cos 60 , where x is an
acute angle , then m   x   .......
 60 , 45 , 30 , 15 
4) The straight line of equation :
2 y = 3 x + 1 intersects a part
of the y - axis equals ........... units.
1 
1
,
2
,
3
,


2 
2
5) The points A ( - 2 , 4 ) , B ( 1 , - 1 ) , C ( 4 , 5 )
represents : ( three collinear points , equilateral
triangle , isosceles triangle , a scalene
triangle )
6) If m   B   90 , AB = 5 cm , BC = 12 cm
, AC = 13 cm , then sin C = ......................
5 12 13 
 5
,
,
,


 13 12 13 5 
[3]
45
AB is a diameter in circle M where A ( 1 , 5 )
, B ( 2 , 6 ) find the coordinates of B
[4]
ABC is a right - angled in B , 2 AB = 3 AC
, find the main trigonometrical ratios of the angle C.
[5]
If sin X = sin 60 cos 30  sin 30 cos 60
without using calculator , find m   X
where X is an acute angle.
[6] Prove that : the points A ( - 3 , 0 ) , B ( 3 , 4 ) and
C ( 1 , - 6 ) are the vertices of an isosceles triangle of
vertex A.
[7]
Find the equation of the straight line drawn
passing to the point ( 1 , 3 ) and perpendicular
to st. line of equation y =
1
x  2.
4
[8] If the ratio between two measures of supplementary
angles as a ratio 3 : 5. Find the value of each one by
circular measure.
Alexandria 2012
[1] Choose:
46
1)  ABC is right - angled at B , A ( 2 , 5 )
, B ( - 2 , - 3 ) , then the slope of BC equals
.......................  2 ,  0.5 , 0.5 , 2 
1
2) If sin ( x + 5 ) = , then x = .......  5 , 10 , 25 , 30 
2
3) The slope of straight line which is perpendicular
 2  2 3 3 
to straight line 2 x + 3 y = 1 is ..........  ,
, ,

3
3
2 2 

4) The distance between the two straight lines y – 3 = 0 ,
y + 2 = 0 is ........... ( 1 , 2 , 3 , 5 )
5) The slope of straight line which parallel to the x – axis
is ........... ( 1 , - 1 , 0 , unknown )
[2] Complete:
1) Cos 60  ...........
2) The straight line 2 x + 3 y = 6 intercepts from the
y – axis a part of length ..................
3) If L1  L 2 , the slope of L1  0 , then
the slope of L 2  ............
4) If cos A = 0.6217 , where A is an acute
angle , then m   A   ...... ....... \ .....\ \
[3]
XYZ is a right angled triangle at Y in which
XY = 6 cm , XZ = 10 cm , find the value of :
1) tan X = tan Z
2) sin  X  Z  30 
[4] Prove that the points A ( 4 , 3 ) , B ( 1 , 1 ) and
47
C ( - 5 , - 3 ) are located on the collinear.
[5] Find the equation of the straight line passing
through the points ( 1 , 2 ) and parallel to the
straight line passing through the two points
A(-3,2),B(-4,5)
[6] ABCD is a parallelogram in which E is the
intersection point of its diagonals , A ( 3 , - 1 ) ,
B ( 6 , 2 ) , C ( 1 , 7 ) find :
1) Coordinates of each of E , D
2) Length of DE
Cairo 2013
[1] Choose:
1
1 

1) tan 45  ........  3 , , 1 ,

2
3

1
2) tan x  , then m ( x ) = ...... ( 90 , 60 , 45 , 30 )
2
3) The length of the line segment drawn from the point
( 0 , 0 ) to the point ( - 4 , 3 ) equals ........
3 , 4 , 7 , 5 units of length.


4) The slope of the straight line which is
parallel to X - axis equals ......
 ( 4 , 2 ) , ( 1 , - 3 ) , ( 2 , - 6 ) , ( 8 , 4 )
48
6) The equation of the straight line passing
through the origin point and makes an
angle of measure 45 with the positive
direction of X - axis is ........
(X=1,Y=1,Y=X,Y=-X)
[2] Find the numerical value of:
2 sin 45 cos45  4 sin 30 cos60
[3] Prove that the triangle ABC whose vertices are
A ( 5 , 1 ) , B ( 2 , 5 ) , C ( - 1 , 1 ) is an isosceles
Triangle , then find its perimeter.
[4] Prove that: cos 60  cos2 45 .
[5] If C is the mid – point of AB where A ( - 3 , y ) ,
B ( 3 , 12 ) and C ( x , 7 ) , find the value of x and y.
Qalubia 2012
[1] Complete:
1) If A ( 3 , 5 ) , B ( 1 , - 1 ) , C is midpoint
of AB , then C ( ........ , ....... )
2) tan 30  ............
3) If tan  x  15   1 , then m   x   ........
4) If sin x = 2 sin 45 cos 60 , then
m   x   ..........
5) If A ( 5 , 5 ) , B ( 3 , 4 ) , C (  4 , 3 )
are vertices of  ABC , then the length
of its perimeter = ...........
49
[2] Choose:
1) The slope of a straight line which makes an angle
of measure 45 with the positive direction of
1


X - axis = .........  1 ,
, 3 , otherwise 
2



2) 4 cos 30 tan 45  ......... 12 , 3 , 6 , 2 3
3) 4 cos 30 tan 60  ......... 12 , 3 , 4 , 6 

4) The straight line whose equation 4 y = 5 x + 12 cuts
from y – axis a part of length = ............... units.
( 3 , 12 , 5 , otherwise )
5) If A ( 2 , 0 ) , B ( 2 , 5 ) , C ( 1 , 4 ) , D ( 1 , 9 ) are four
points in the same co – ordinate plane , then .........
( AB = CD , AB < CD , AB > CD
, A , B , C and D are collinear )
Solution:
Model (1)
[1] Complete each of the following:
 1 3 2 4
1) 
,
   2, 3 
2 
 2
2) y = m x + b where m = 0
then y = 3
3) x + 2 x = 90  x = 30 , y = 60
1 1
 sin x + cos y = sin 30 +cos 60 = + =1
2 2
50
4)
 6  4   0  0
2
2
 10 unit length
12
5) 3 a  4  0  12  0  3 a  12  a 
 4
3
2
6)
3
[2] Choose:
1) 2 x = 60  x = 60  2 = 30
2 2
2)
=
3 3
3)  5  0    12  0   13
4) 1
5) sin A + cos C = sin A + sin A = 2 sin A
1 1
6) tan 45 sin 30  1  
2 2
2
2
[3]
sin C =
AB
3

AC
2
A
2
1
, cos C =
2
C
, tan C =
3
 3
1
[4]
ym x+b,m=
2
2y 2 x+b
1
51
3
1
B
At  2 ,  1  1  2  2   b  b  1  4  5
y  2 x5
[5]
L.H.S.  cos 60 
1
2
2
2


3
1 1
, R.H.S.= cos 2 30  sin 2 30  

   
 2   2 2
L.H.S.  R.H.S.
[6]
 3  1 1  6 
Let D ( x , y ) then E  
,
   2, 2.5 
2 
 2
x6 2
 x6 y 2  2 5
,

,

 

 

2   1 2
2
1
 2
y2 5
x  6  4  x  4  6  2 ,
 y25
2
2
 y  5  2  3  D  2 , 5 
[7]
L.H.S.= tan 60  3
2

1
 1  
2
R.H.S.  2 tan 30   1  tan 30   2 
 1  
 
3   3  
1 2
 2
  3  L.H.S.  R.H.S.
3 3
52
[8]
x y
3x+2y
 1
1 3 x + 2 y  6
2 3
6
6 3
3
3
 2 y  6 3 x  y =  x  3 x  m 
,b  3
2 2
2
2
Model (2)
[1] Complete each of the following:
2 3
1) m 1  m 2  1 

 1
b 1
6
1


b6
b 1
2) m  X   30
3)
 5  0   0  12  13
2
2
3
3

 30
2
2
 k 6
5) m 1  m 2 

 3 k  12
2
3
12
k
 4
3
1
6  1 6
6) m1 
 1 

m2
24 5
4)
[2] Choose:
1
3
3
1) 2  
=
 sin 60
2 2
2
53
2)  AB    3  0    0  3   18
2
2
2
 BC    0  3    3  0   18
2
2
2
 CA    3  3    0  0   36
  CA    AB    BC  , AB = BC
2
2
2
2
2
2
The triangle is right  angled triangle and isosceles
3) y = - 3
1  a
4) m1  m 2  1 

 1
3 3
a
 1  a  9
9
1  x
1  y
5)
0x1,
4
2
2
A
 1  y  8  y  8  1  9   x , y   1 , 9 
5
3
4 4 16
6)  
5 5 25
C
4 B
[3]
 1  3 2  4 
M
,
   2 ,  3
2 
 2
6 3
m  The slope  
 9  y  m x + c
1 2
At  1 , 6   6  9  1  c  c  6  9  15
y  9 x + 15
[4]
3
1 1
L.H .S  sin 3 30    
 2 8
54
3
9
1
1
R.H.S  9 cos 3 60  tan 2 45  9     12   1 
8
8
 2
 L .H .S  R.H.S
[5]
 AB    1  1   4  2   40
 AC    1  2    4  3   50
 BC    1  2    2  3   10
  CA    AB    BC 
2
2
2
2
2
2
2
2
2
2
2
2
The triangle is right  angled triangle at B
1
1
 Area =  40  10   400  10 squared units
2
2
[6] In the opposite figure:
AD  152  92  12 cm
BD  132  122  5 cm
A
tan   CAD   tan   BAD 
tan   CAD   tan   BAD  15 cm
13 cm
9
5

14 195 7
 15 13 


9 5 195 4
2
B

D
C
9
cm
15 13
[7]
5 5
1 2
m1 
 , m2 

2 2
m1
5
2
6 26
At  3 , 4   4 
3c c  4 
5
5 5
55
y
2
26
x
5
5
[8]
ABCD is a trapezium in which AD // BC
, m   B   90 if AB = 3 cm , AD = 6 cm
, BC = 10 cm , prove that :
cos   DCB   tan   ACB  
1
2
Construction:
Draw DF  BC
Solution:
 AD // BF , AB  BC , DF  BC
 ABFD is a rectangle
 BF  AD  6 cm , FC  10  6  4 cm
In  DFC
DC  3 2  4 2  5 cm
FC 4
 cos(DCF) 

DC 5
In  ABC
 tan( ACB) 
6 cm
D
A
3 cm
4 cm F
C
AB 3

BC 10
 cos(DCF)  tan( ACB) 
4 3
5 1
 

5 10 10 2
Model (3)
[1] Complete each of the following :
56
3 cm
6 cm
B
2
 1 
1) cos 45 + tan 60  sin 30 = 
 
 2
2
2
 2  5   1  3
2
2) AB =
2
 
2
3
1
   3
 2
 5 unit length
1
k
1 3

 1 

m2
2
3 1
k=2  3=6
3) m1 
1 1
3
3
4) sin 30 cos 60  cos 30 sin 60   

1
2 2 2
2
5) X =  2
1
1
1
6) 1 


2
2 2
[2] Choose:
1) y = x
1
20 1
= 1
=
m1
1  0 2
60
3) 3 x = 60  x =
= 20
3
2) m 2 =
4) OA =


8  0  1  0  3  r
The point is

2
8 ,1
2

AC BC AC  BC
+
=
>1
AB AB
AB
3
6) x = 30  sin 2 x = sin 60 =
2
5) sin B + cos B =
57
[3]
XY 2   2  0    4  6.8   4  7.84  11.84
2
2
YZ 2   2  5    4  1  49  25  74
2
2
XZ 2   0  5    6.8  1  25  60.84  85.84
 XZ 2  XY 2  YZ 2
 The triangle is right  angled triangle
A
[4]
2
2
AD = 102  62 = 8 cm
8 6
1) sin B + cos C = + = 1.4
10 10
2
2
2) sin C + cos C
2
2
 8  6
=     1
 10   10 
[5]
10 cm
B
10 cm
6 cm D
cos 2 60  cos 2 30  tan 2 45
sin 60 tan 60  sin 30
2
2
2
1  3
1 3
   1
  
 1
 2  2 

4 4
 21  2
3
1
3
1

 3
2 2
2
2
[6]
BC 
 3  1   4  1  5 unit length
AB 
 2  1   3  1  5 unit length
CD 
 6  3    4  0   5 unit length
2
2
2
2
2
2
58
6 cm C
  3  0   5 unit length
31 4
The slope of AB  m 1  

21 3
4  1 3
The slope of BC  m 2  

31
4
4 3
m1  m 2  
 1  AB  BC
3 4
 ABCD is a square , its area = 5  5 = 25 squared unit
AD 
 6  2
2
2
[7]
2
sin x 

sin x  

 x  60
1
1
1 3
2

 3 =1    
2
2
 2 4
3 3
3
3
3


sin
x




2  4
4 2
2
[8]
y
1
x+2 
2
 0 , 2
Model (4)
[1] Complete each of the following:
11
 1
35
1 1
1
2) m 2 

,y=
x
m1
2
2
1)
59
3) sin 60 cos 30  cos 60 sin 30 
3
3 1 1 1

  
2
2 2 2 2
45
4) 3 x  45  x =
=15
3
1
3 4
5) m 2 
 1 

m1
4 3
x
6)  30  x = 3  30 = 90
3
[2] Choose:
2
2


3
1
1


1) sin 2 60  cos 2 60  


  
2

  2 2
2) OA =
 2  0 
2


50

2
3r
 The point is ( -2 , 5)
3) 0
a
4)
1a 1
1
5) 3 + 2 = 5
1
6)  cos   m      60
2
[3]
 x  8 y  11 
Let A ( x , y ) then 
,
= ( 5 , 7 )
2 
 2
60
x8
 5  x  8  10  x  10  8  2
2
y  11
 7  y  11  14  y  14  11  3
2
2
2
then A  2 , 3  , r = MB =  8  5    11  7 
 5 unit length
11  7 4
m1 The slope of MB 

85 3
 The slope of the perpendicular straight  1 3
=
m2 

 line to AB from the point B
 m1
4


3
y  m 2 x  c  At B ( 8 , 11 )  11 
8 c
4
3
 c  11  6  17  y 
x  17
4


[4]
sin x =
3
3 1 1 1

  =  x = 30
2
2 2 2 2
[5]
AC  102  82  6 cm
sin A cos B + cos A sin B
8 8
6 6
=
   1
10 10 10 10
[6]
m
32 1
 ,y=mx+c
2 3 5
61
A
10 cm
B 8 cm C
At  2 , 3  3 =
1
2 13
 2 + c  c = 3  
5
5 5
1
13
y  x
5
5
[7]
1 1
3
3 1
cos 60 sin 30  sin 60 cos 30   


2 2 2
2
2
[8]
 1 7 0 8
The mid  po int of AC  
,
   4 , 4
2 
 2
 1  9 4  4 
The mid  po int of BD  
,
  4 , 4
2
2


2
2
AC   1  7    0  8   10 unit length
BD 
 1  9 
2
  4  4   10 unit length
2
Model (5)
[1] Complete each of the following:
1) y + 7 = 30  y  30  7  23
2) y =  2
3)  3  0    4  0   5 unit length
4)  1
2
2
1
3
3
5) 2 sin 30 cos 30  2  

 sin 60
2 2
2
1
6) 6 = 4  + c  c  6  2  4
2
62
[2] Choose:
2 k
4
= k 
3 2
3
 3  5 5  1 
2) 
,
 =  4 , 2
2
2


1)
3) sin 60  cos 30  tan 60 
3
3

 32 3
2
2
 a  0    0  1  1  a 2  1  1  a  0
y4
5)
 tan 45  1  y  4  2  y  4  2  2
1 3
24 7 31
6) sin X + sin Y = + =
25 25 25
2
4)
2
[3]
2 3
3  3 9
=
b =
=
3 b
2
2
2 3
2 1
1  2
2) m1  m2  1  
 1 

b
2
3 b
b
1
1
1) m1 =m2 
3) 2  1  3  3   a  0  2  9  a  0
7a  0a  7
A
[4]
30
4
sin 60

AC
1
 AC 
1 4
sin 60
4m
5m
60
C
63
B
[5]
L.H.S.= sin 60 
3
2
1
3
3
R.H.S.  2 sin 30 cos 30  2  

 L.H.S.
2 2
2
[6] ABCD is a parallelogram , A ( x , 2 ) , B ( 3 , 8 )
, C ( 9 , 10 ) and D ( 7 , 4 ) , find x
64