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