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.