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A.1. Attempt ANY FIVE of the following :(i) Slope of the line (m) = – 2
y intercept of the line (c) = 3 ½By slope intercept form,The equation of the line is y = mx + c
y = (– 2)x + 3 y = – 2x + 3 ½ The equation of the given line is y = – 2x + 3
(ii) tan = 3 [Given]
But, tan 60 = 3 ½ tan = tan 60
= 60 ½
(iii) Equation of a line parallel to Y-axis and passing through the 1point (3, 4) is x = 3.
(iv) + = 90º [Given] ½ = (90 – )
cosec = 2 [Given] sec = cosec (90 – ) [ sec = cosec (90 – )] sec a = cosec
sec = 2 ½
(v) y + 3 =12 (x – 5)
Comparing with the equation of a line in slope point form,y – y1 = m (x – x1) ½
m =12
Slope of the line y + 3 =12 (x – 5) is
12 ½
Time : 2 Hours Model Answer Paper Max. Marks : 40
MTMT - GEOMETRY - SEMI PRELIM - I : PAPER - 3
2017 ___ ___ 1100
PAPER - 32 / MT
(vi) + = 90º [Given]
tan =34
[Given]
cot = tan [ cot = tan (90 – )] ½
cot =34
½
A.2. Solve ANY FOUR of the following :(i)
(ii) The terminal arm passes through P –1, 3
x = – 1 and y = 3
r = x y2 2 ½
= 1 3 22
= 1 3
= 4
r = 2 units ½
RO 3.1 cm
(Analytical figure)
O 3.1 cm R
1 mark for circle1 mark for tangent
PAPER - 33 / MT
Let the angle be
sin =yr
=32
cosec =ry =
23
cos =xr
=–12
sec =rx
=– 21
1
tan =yx
=3
–1 = – 3 cot =xy =
–13
(iii) Let A (3, 4) (x1, y1) and m = 5The equation of the line passing through A and having slope 5 ½by slope point form is,
y – y1 = m (x – x1) ½ y – 4 = 5 (x – 3) y – 4 = 5x – 15 ½ 5x – y – 15 + 4 = 0 5x – y – 11 = 0
The equation of the line passing through the points (3, 4) and ½having slope 5 is 5x – y – 11 = 0.
(iv)
1 mark for drawing circle1 mark for drawing tangent
N M•
•L
A
O
3.6 cm N
M•
•L
•A
(Analytical figure)
3.6 c
m
PAPER - 3
(v) L.H.S. = sec2 + cosec2
=1 1
cos sin2 2
1 1sec , coseccos sin
½
=sin + coscos . sin
2 2
2 2 ½
=1
cos . sin2 2 [ sin2 + cos2 = 1] ½= sec2 . cosec2 ½= R.H.S.
sec2 + cosec2 = sec2 . cosec2
(vi) Let, A (2, 3) (x1, y1) B (4, 7) (x2, y2)The line passes through points A and B
The equation of the line by two point form is
x – xx – x
1
1 2=
y – yy – y
1
1 2½
x – 22 – 4 =
y – 33 – 7
x – 2–2 =
y – 3– 4 ½
4 (x – 2) = 2 (y – 3) 4x – 8 = 2y – 6 ½ 2y = 4x – 8 + 6 2y = 4x – 2 y = 2x – 1 [Dividing throughout by 2] ½ y = 2x – 1 is the equation of the line passing through (2, 3)
and (4,7)
A.3. Solve ANY THREE of the following :(i)
(Analytical figure)
7.3 cm B
C
A
D
3.5
cm
3.5 cm
4 / MT
PAPER - 35 / MT
M7.3 cm
C
A
D
B
3.5
cm
3.5 cm
(ii) tan = 1
sincos
= 1
sin = cos .......(i) ½1 + tan2 = sec2
1 + (1)2 = sec2 1 + 1 = sec2 2 = sec2 sec = 2 [Taking square roots] ½
cos =1
sec
cos =12
sin =12 [From (i)] ½
cosec =1
sin
=1
12
cosec = 2 ½
1 mark for circle1 mark for perpendicular bisector1 mark for tangents
PAPER - 36 / MT
sin cos
sec cosec
=
1 12 22 2
½
=2
22 2
=2
2 2 2
=2
2 2
=24
sin cos
sec cosec
=
12
½
(iii) Let, A (1, 2) (x1, y1)
B 1 , 32 (x2, y2) ½
C (0, k) (x3, y3) Points A, B and C are collinear
Slope of line AB = Slope of line BC ½
y – yx – x
2 1
2 1=
y – yx – x
3 2
3 2½
3 – 21 – 12
= k – 310 –2
½
11–2
=k – 3
–12
½
1 = k – 3 k = 1 + 3 k = 4 ½
The value of k is 4.
PAPER - 37 / MT
(iv) 3 tan2 – 4 3 tan + 3 = 0
3 3 tan – 4tan 32 = 0 ½
3 tan – 4tan 32 = 0
3 tan – 3tan – tan + 32 = 0 ½
3 tan tan – 3 – 1 tan – 3 = 0
tan – 3 3 tan – 1 = 0 ½
tan – 3 = 0 OR 3 tan – 1 = 0 tan = 3 3 tan = 1 ½
But, tan 60 = 3 tan =13
tan = tan 60 But, tan 30 =13
= 60 tan = tan 30 1
= 30
(v) Let, A (– 1, 1), B (– 9, 6), C (– 2, 14), D (6, 9)
Slope of a line =y – yx – x
2 1
2 1½
Slope of side AB =6 – 1
–9 – (–1) ½
=5
–9 1
=5– 8
Slope of line AB =–58
½
Slope of line CD =9 – 146 – (–2) ½
=–5
6 2
Slope of line CD =–58
½
Slope of line AB and slope of line CD are equal. line AB || line CD The line joining (– 1, 1) and (– 9, 6) is parallel to the line ½
joining (– 2, 14) and (6, 9).
PAPER - 38 / MT
A.4. Solve ANY TWO of the following :(i) L.H.S. = (1 + tan )2 + (1 + cot )2
= 1 + 2 tan + tan2 + 1 + 2 cot + cot2 ½= 1 + tan2 + 1 + cot2 + 2 tan + 2 cot ½= sec2 + cosec2 + 2 (tan + cot )
[ 1 + tan2 = sec2 , 1 + cot2 = cosec2 ]
= sec2 + cosec2 +sin cos2cos sin
½
= sec2 + cosec2 +sin + cos2cos × sin
2 2
½
= sec2 + cosec2 + 2 ×1
cos × sin ½
[ sin2 + cos2 = 1]= sec2 + cosec2 + 2 × sec × cosec ½= (sec + cosec )2 1= R.H.S.
(1 + tan )2 + (1 + cot )2 = (sec + cosec )2
(ii)
1 mark for drawing triangle1 mark for drawing angle bisectors1 mark for drawing perpendicular1 mark for incircle
S7 cm
6 cm 6.5 cm
R
T
(Analytical figure)
M×ו•
I
S7 cm
6 cm 6.5 cm
R
T×ו•
I
M
PAPER - 39 / MT
(iii) L.H.S. =cos sin
1 – tan sin – cos
2 3
=
cos sinsin sin – cos1 –cos
2 3
½
=
cos sincos – sin sin – cos
cos
2 3
½
=cos sin
cos – sin sin – cos
3 3
+ ½
=cos sin
cos – sin cos – sin
3 3
– ½
=cos – sincos – sin
3 3
=(cos – sin ) (cos + cos . sin + sin )
(cos – sin )
2 2 1
= cos2 + sin2 + sin . cos = 1 + sin . cos [ sin2 + cos2 = 1] 1= R.H.S.
2 3cos sin+
1 – tan sin – cos
= 1 + sin . cos
A.5. Solve ANY TWO of the following :(i)
T
E
A MH
6.3 cm
4.9 cm
120º
A1A2 A3
A4 A5 A6 A7
× ×
•
•
(Analytical figure)
PAPER - 310 / MT
(ii) Let E be the position of the cloudand let BC represent the surfaceof the lake. ½Let A be the point of observerand let F be the reflection ofthe cloud
EC = CF ½Let EC = CF = x mABCD is a rectangle [By definition]
AB = CD = 60 m [Opposite sides ½of rectangle]
EC = ED + DC [E - D - C] ½ x = ED + 60 ED = (x – 60)m
Also,DF = DC + CF [D - C - F] ½
DF = (60 + x) DF = (x + 60) m
T
E
A MH
6.3 cm
4.9 cm
120º
A1
A2
A3
A4
A5
A6
A7
× ×
•
•
A
B
E
D
C
60º30º
60 m
F
60 m
x
x
½ mark for drawing analytical figure1 mark for AMT½ mark for constructing 7 congruent parts1½ mark for constructing HA5A MA7A1½ mark for constructing EHA TMA
PAPER - 311 / MT
In right angled ADE,
tan 30º =EDAD [By definition]
13 =
x – 60AD
AD = 3 x – 60 m ½In right angled ADF,
tan 60º =DFAD [By definition]
3 =x + 60
3 (x – 60) 1 3 (x – 60) = x + 60 3x – 180 = x + 60 3x – x = 60 + 180 2x = 240 x = 120 1
The height of the cloud above the lake is 120 m.
(iii) A (5, 4), B (– 3, – 2), C (1, – 8)seg AD is the median of seg BC
D is midpoint of seg BC
D x + x y + y
,2 2
1 2 1 2
–3 + 1 –2 + (–8),
2 2
–2 –2 – 8,2 2
–10–1 ,2
1
(– 1, – 5)By two point form,The equation of median AD
x – xx – x
1
1 2=
y – yy – y
1
1 2½
x – 5
5 – (–1) =y – 4
4 – (–5) ½
x – 55 + 1 =
y – 44 + 5
x – 5
6=
y – 49
9 (x – 5) = 6 (y – 4)
PAPER - 312 / MT
9x – 45 = 6y – 24 9x – 6y – 45 + 24 = 0 9x – 6y – 21 = 0 3x – 2y – 7 = 0 [Dividing throughout by 3] 1 The equation of median AD is 3x – 2y – 7 = 0
Slope of line AC =y – yx – x
2 1
2 1
=–8 – 41 – 5
=–12–4 ½
= 3 Slope of parallel lines are equal
Slope of the line parallel to line AC is 3The line passes through B (– 3, – 2)
The equation of the line parallel to line AC passing throughpoint B by the slope point form is
y – y1 = m (x – x1) ½ y – (– 2) = 3 [x – (– 3)] y + 2 = 3 (x + 3) y + 2 = 3x + 9 3x – y + 9 – 2 = 0 3x – y + 7 = 0 1 The equation of the line parallel to AC passing through
point B is 3x – y + 7 = 0