A.1. Attempt ANY FIVE of the following :(i) Slope of the line (m) = 0
y intercept of the line (c) = 2 ½By slope intercept form,The equation of the line is y = mx + c
y = (0)x + 2 y = 2 The equation of the given line is y = 2 ½
(ii) sec = 2 [Given]But, sec 60 = 2 ½
sec = sec 60
= 60º ½
(iii) Slope of the line (m) = 2Its y-intercepts (c) = 5 ½
Equation of the line by slope-intercept form,y = mx + c
y = 2x + 5 ½
(iv) 3 sin – 4 cos = 0 3 sin = 4 cos ½
sincos
=
43
tan =43
½
(v) 3 (x + 3) = y – 1 y – 1 = 3 (x + 3) ½
Comparing with the equation of a line in slope point form,y – y1 = m (x – x1)
m = 3 Slope of the line 3 (x + 3) = y – 1 is 3. ½
Time : 2 Hours Model Answer Paper Max. Marks : 40
MTMT - GEOMETRY - SEMI PRELIM - I : PAPER - 5
2017 ___ ___ 1100
PAPER - 52 / MT
(vi) = – 30º [Given]sin = sin (– 30) ½
= – sin 30
=1–2
sin (– 30) =1–2
½
A.2. Solve ANY FOUR of the following :(i)
(ii) The terminal arm passes through P (1, – 1) x = 1 and y = – 1 ½
r = x y2 2
= (1) + (–1)2 2
= 1 1
= 2 r = 2 units ½
MO 2.9 cm
MO 2.9 cm
(Analytical figure)
1 mark for circle1 mark for tangent
PAPER - 53 / MT
Let the angle be
sin =yr
=–12 cosec =
ry =
2–1 = – 2
cos =xr
=12 sec =
rx
=2
1 = 2 1
tan =yx
=–11
= – 1 cot =xy =
1–1 = – 1
(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 5x – y – 15 + 4 = 0 5x – y – 11 = 0 1 The equation of the line passing through the points (3, 4) and
having slope 5 is 5x – y – 11 = 0.
(iv)
N
M•
•L
•A
(Analytical figure)3.
6 cm
1 mark for drawing circle1 mark for drawing tangent
N M•
•L
A
O
3.6 cm
PAPER - 54 / MT
(v) tan A +1
tan A = 2 [Given]
1tan A
tan A
2
= 4 [Squaring both sides] ½
tan2 A + 2 tan A .1 1
tan A tan A 2 = 4 ½
tan2 A + 2 +1
tan A2 = 4 ½
tan2 A +1
tan A2 = 4 – 2 ½
tan2 A + 21
tan A = 2
(vi) Let, A (– 3, 5) (x1, y1) B (4, – 7) (x2, y2)The line passes through points A and B ½
The equation of the line by two point form is
x – xx – x
1
1 2=
y – yy – y
1
1 2½
x – (–3)–3 – 4 =
y – 55 – (– 7)
x 3
– 7 =y – 512
12 (x + 3) = – 7 (y – 5) ½ 12x + 36 = – 7y + 35 7y = – 12x + 35 – 36 7y = – 12x – 1
y =–12 1x –7 7 [Dividing throughout by 7]
y =–12 1x –7 7 is the equation of the line passing through ½
(– 3, 5) and (4, – 7)
PAPER - 55 / MT
A.3. Solve ANY THREE of the following :(i)
(ii) 3 tan = 3 sin
sin3cos
= 3 sin ½
3
cos = 3
cos =33
cos2 =39
cos2 =13 .......(i) ½
P
8.8 cm
2.9
cm
Q
A
B2.9 cm
M
P 8.8 cm
2.9
cm
Q
A
2.9 cmB
(Analytical figure)
1 mark for circle1 mark for perpendicular bisector1 mark for tangents
PAPER - 56 / MT
sin2 + cos2 = 1 sin2 = 1 – cos2 ½
=11 –3 [From (i)]
=3 – 1
3
sin2 =23 ......(ii) ½
sin2 – cos2 =2 1–3 3 [From (i) and (ii)]
=2 – 1
3
sin2 – cos2 =13 1
(iii) Let, A (k, 3), B (2, – 4), C (– k + 1, – 2)
Points A, B and C are collinearSlope of line AB = Slope of line BC ½
– 4 – 32 – k =
–2 – (– 4)(– k 1) – 2 ½
–7
2 – k =
–2 4– k 1 – 2
–7
2 – k =2
– k – 1 ½
– 7 (– k – 1) = 2 (2 – k) ½ 7k + 7 = 4 – 2k 7k + 2k = 4 – 7 9k = – 3
k =– 39 ½
k =–13
The value of k is–13 . ½
PAPER - 57 / MT
(iv) L.H.S. =tan sec 1
sec 1 tan
=tan (sec + 1)
(sec 1) tan
2 2 ½
=tan sec 2sec 1
(sec 1) tan
2 2 ½
=sec sec 2sec
(sec 1) tan
2 2 [ 1 + tan2 = sec2 ] ½
=2sec 2sec(sec 1) tan
2
=2sec (sec + 1)(sec 1) tan
½
=2sectan
= 2sec tan ½
= 2 ×1 sin
cos cos
=1 cos2
cos sin
sin 1tan , seccos cos
=2
sin ½
= 2 cosec = R.H.S.
tan sec + 1+
sec + 1 tan
= 2 cosec
(v) Let, A (x, – 2) (x1, y1) B (8, – 11) (x2, y2) ½
Slope of line AB =– 34 [Given]
Slope of line AB =y – yx – x
2 1
2 1½
– 34 =
–11 – (–2)8 – x
– 34 =
–11 28 – x
½
– 34 =
–98 – x ½
PAPER - 58 / MT
3 (8 – x) = 9 × 4 24 – 3x = 36 3x = 24 – 36 ½ 3x = – 12
x =–123
x = – 4 ½ The value of x is – 4.
A.4. Solve ANY TWO of the following :(i) sec + tan = p ½
1 sin
cos cos
= p
1 sin
cos
= p ½
(1 sin )
cos
2
2 = p2
1 sin1 – sin
2
2 = p2sin cos 1
cos 1 – sin
2 2
2 2 1
1 sin
1 sin 1 – sin
2
= p2 ½
1 sin1 – sin
= p2
1 sin + 1 – sin1 sin – 1 + sin
=
p 1p – 12
2 [By Componendo-Dividendo] ½
2
2 sin =p 1p – 12
2
1
sin =p 1p – 12
2 ½
2
2p – 1p + 1 = sin [By Invertendo] ½
(ii)
7 cm ST
5 cm
U
120º
(Analytical figure)
M×ו•
PAPER - 59 / MT
(iii) sin2 + cos2 = 1 cos2 = 1 – sin2 cos . cos = (1 – sin ) (1 + sin ) ½
cos
1 – sin
=1 sin
cos
½By theorem on equal ratios,
1 + sin – coscos – (1 – sin )
=
cos1 – sin
=
1 sincos
1 sin – coscos – (1 – sin )
=cos
1 – sin
1
Dividing the numerator and denominator of R.H.S. by cos
1 + sin – coscos – 1 + sin
=
coscos
(1 – sin )cos
1
1 + sin – coscos – 1 + sin
=
11 sin–
cos cos
½
sin – cos + 1sin + cos – 1
=
1sec – tan ½
7 cm ST
5 cm
U
120º• ו ×
I
1 mark for triangle1 mark for angle bisectors1 mark for perpendicular1 mark for incircle
PAPER - 510 / MT
A.5. Solve ANY TWO of the following :(i)
(ii)
Let E be the position of the cloudand let BC represent the surface ½of the lake.Let A be the point of observer andlet F be the reflection of the cloud
EC = CFLet EC = CF = x mABCD is a rectangle [By definition] ½
AB = CD = 60 m [Opposite sidesof rectangle]
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
T
E
A MH
6.3 cm
4.9 cm
120º
A1 A2 A3A4 A5
A6 A7
× ×
•
•
(Analytical figure)
PAPER - 511 / MT
EC = ED + DC [E - D - C] x = ED + 60 ED = (x – 60)m ½
Also,DF = DC + CF [D - C - F] ½
DF = (60 + x) DF = (x + 60) m
In right angled ADE,
tan 30º =EDAD [By definition]
13 =
x – 60AD
AD = 3 x – 60 m 1In 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) Let point P be the point of intersection of lines 4x + 3y + 2 = 0and 6x + 5y + 6 = 04x + 3y + 2 = 0
4x + 3y = – 2......(i)Multiplying throughout by 3 we get,
12x + 9y = – 6 .....(ii) ½6x + 5y + 6 = 0
6x + 5y = – 6 .....(iii) ½Multiplying throughout by – 2 we get,– 12x – 10y = 12 ......(iv)Adding (ii) and (iv), 12x + 9y = – 6– 12x – 10y = 12– y = 6
y = – 6 ½Substituting y = – 6 in equation (i),4x + 3 (– 6) = – 2
4x – 18 = – 2
PAPER - 512 / MT
4x = – 2 + 18 4x = 16
x =164
x = 4 P (4, – 6) ½
Let Q be the point of intersection of lines 4x – 3y – 17 = 0 and2x + 3y + 5 = 04x – 3y – 17 = 0
4x – 3y = 17 .....(v) ½2x + 3y + 5 = 0
2x + 3y = – 5 .....(vi) ½Adding equation (v) and (vi),4x – 3y = 172x + 3y = – 5
6x = 12 x = 2 ½
x = 2 in equation (v),4 × 2 – 3y = 17
– 3y = 17 – 8 – 3y = 9 y = – 3 Q (2, – 3) ½
The equation of line PQ by two point from,y – yy – y
1
1 2=
x – xx – x
1
1 2½
y – (–6)–6 – (–3) =
x – 44 – 2
y 6
–6 3 =
x – 42
y 6–3
=x – 4
2 2 (y + 6) = – 3 (x – 4) 2y + 12 = – 3x + 12 ½ 3x + 2y + 12 – 12 = 0 3x + 2y = 0 The required equation of line is 3x + 2y = 0.
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