A.1. Attempt ANY FIVE of the following :(i) Slope of the line (m) = – 4

y intercept of the line (c) = – 3 ½By slope intercept form,The equation of the line is y = mx + c

y = (– 4)x + (– 3) y = – 4x – 3 ½ The equation of the given line is y = – 4x – 3

(ii) (1 + tan2 ) (1 – sin ) (1 + sin )= sec2 (1 – sin2 ) [ 1 + tan2 = sec2 ] ½= sec2 × cos2 [sin2 + cos2 = 1

cos2 = 1 – sin2 ]

=1 cos

cos 2

2 ½

= 1

(iii) Equation of a line parallel to X-axis and passing through the 1point (– 2, 5) is y = 5.

(iv) tan = 15 ½ 1 + tan2 = sec2

1 + 152

= sec2

1 + 15 = sec2 sec2 = 16 ½

sec = 4 [Taking square roots]

(v) y – 2 = 5 (x – 2)Comparing with the equation of a line in slope point form,y – y1 = m (x – x1) ½

m = 5 Slope of the line y – 2 = 5 (x – 2) is 5 ½

Time : 2 Hours Model Answer Paper Max. Marks : 40

MTMT - GEOMETRY - SEMI PRELIM - I : PAPER - 2

2017 ___ ___ 1100

PAPER - 22 / MT

(vi) = – 30º [Given] ½sin = sin (– 30)

= – 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 (– 24, – 7) x = – 24 and y = – 7

r = x y2 2 ½

= (– 24) (–7)2 2

= 576 49

= 625 r = 25 units ½

Let the angle be

MO 2.9 cm

MO 2.9 cm

(Analytical figure)

1 mark for circle1 mark for tangent

PAPER - 23 / MT

sin =yr

=– 725

cosec =ry =

–257

cos =xr

=–2425

sec =rx

=–2524

1

tan =yx

=–7–24 =

724

cot =xy =

–24–7 =

247

(iii) Let A (2, 3) (x1, y1), B (4, 5) (x2, y2) The equation of line AB by two point form is,

x – xx – x

1

1 2=

y – yy – y

1

1 2½

x – 22 – 4 =

y – 33 – 5 ½

x – 2–2 =

y – 3–2 ½

x – 2 = y – 3 x – y – 2 + 3 = 0 x – y + 1 = 0 ½ The equation of the line passing through the points (2, 3) and

(4, 5) is x – y + 1 = 0.

(iv)

1 mark for circle1 mark for tangent

N M•

•L

A

O

3.6 cm N

M•

•L

(Analytical figure)

•A

O

3.6 c

m

PAPER - 24 / MT

(v) tan A +1

tan A = 2 ½

1tan A

tan A

2

= 4 [Squaring both sides]

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

tan2 A + 21

tan A = 2 ½

(vi) Let A (3, 4) (x1, y1) and m = 5The equation of the line passing through A and having slope5 by 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 1

The equation of the line passing through the points (3, 4) andhaving slope 5 is 5x – y – 11 = 0.

A.3. Solve ANY THREE of the following :(i)

E 6 cm

7.5 cm

L

M

(Analytical figure)

m

l

PAPER - 25 / MT

E 6 cm

7.5 cm

L

M

O

(ii) 3 sin – 4 cos = 0 3 sin = 4 cos

sincos

=

43

½

tan =43

1 + tan2 = sec2

1 +43

2

= sec2 ½

1 +169

= sec2

9 16

9

= sec2

259

= sec2

sec =53 [Taking square roots] ½

1 mark for triangle1 mark for perpendicular bisectors1 mark for circumcircle

PAPER - 26 / MT

cot =1

tan

=1

43

cot =34 ½

1 + cot2 = cosec2 ½

1 +34

2

= cosec2

1 +9

16= cosec2

16 9

16

= cosec2

2516

= cosec2

cosec =54 [Taking square roots] ½

(iii) 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 line AB =6 – 1

–9 – (–1) ½

=5

– 9 1

=5– 8

Slope of line AB =–58

½

Slope of line CD =9 – 146 – (–2)

=–5

6 2

Slope of line CD =–58

½

PAPER - 27 / MT

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 1

joining (– 2, 14) and (6, 9).

(iv) sec + tan = p

1 sin

cos cos

= p ½

1 sin

cos

= p

(1 sin )

cos

2

2 = p2 ½

1 sin1 – sin

2

2 = p2sin cos 1

cos 1 – sin

2 2

2 2 ½

1 sin

1 sin 1 – sin

2

= p2

1 sin1 – sin

= p2 ½

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

2

2p – 1p + 1 = sin [By Invertendo] ½

(v) Let, A (1, 2) (x1, y1)

B 1 , 32 (x2, y2)

C (0, k) (x3, y3) Points A, B and C are collinear

PAPER - 28 / MT

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 ½ k = 1 + 3 k = 4 ½ The value of k is 4

A.4. Solve ANY TWO of the following :

(i) 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

½

PAPER - 29 / MT

=2

sin ½

= 2 cosec = R.H.S. ½

tan sec + 1+

sec + 1 tan

= 2 cosec

(ii)

(iii) sin2 + cos2 = 1 cos2 = 1 – sin2 ½ cos . cos = (1 – sin ) (1 + sin )

cos

1 – sin

=1 sin

cos

½By theorem on equal ratios,

1 + sin – coscos – (1 – sin )

=

cos1 – sin

=

1 sincos

½

1 sin – coscos – (1 – sin )

=cos

1 – sin

Dividing the numerator and denominator of R.H.S. by cos ½

D E5.8 cm

5.8 cm

F

×× ••65º

I

D E5.8 cm

5.8 cm

F

65º

(Analytical figure)

••

××

I

1 mark for triangle1 mark for angle bisectors1 mark for perpendicular1 mark for incircle

PAPER - 210 / MT

1 + sin – coscos – 1 + sin

=

coscos

(1 – sin )cos

1 + sin – coscos – 1 + sin

=

11 sin–

cos cos

1

sin – cos + 1sin + cos – 1

=

1sec – tan 1

A.5. Solve ANY TWO of the following :(i)

U

R

S H V

5.2 cm

4.5 cm

S1

S2

S3

S4

S5

××

5.8

cm

•

•

½ mark for drawingAnalytical figure

1 mark for SHR½ mark for constructing

5 congruent parts1½ mark for constructingVS5S HS3S1½ mark for constructing

UVS RHS

(Analytical figure)U

R

S H V

5.2 cm

4.5 cmS1

S2

S3

S4

S5

5.8

cm

PAPER - 211 / MT(ii) seg AB represents the tree

AB = 12 mThe tree breaks at point Dseg AD is the broken part of treewhich then takes the position of DC ½

AD = DCm DCB = 60ºLet DB = x m

AD + DB = AB [ A - D - B] ½ AD + x = 12 AD = (12 – x) m DC = (12 – x) m

In right angled DBC,

sin 60º =DBDC [By definition] ½

32

=x

12 – x 3 12 – x = 2x

12 3 – 3 x = 2x

12 3 = 2x 3 x

x 2 3 = 12 3 ½

x =12 32 3

DB =12 32 3

m

DB =

12 3 2 – 3

2 3 2 – 3 1

DB = 24 3 – 12 (3)

(2) – 322

DB =24 3 – 36

4 – 3 1

DB =24 (1.73) – 36

1 DB = 41.52 – 36 DB = 5.52 m 1 The height at which the tree is broken from the bottom by the

wind is 5.52 m.

A

B C60º

D12 m

PAPER - 212 / MT

(iii) P (– 2, 4), Q (4, 8), R (10, 5), S (4, 1)

Slope of a line =y – yx – x

2 1

2 1½

Slope of line PQ =8 – 4

4 – (–2)

= 4

4 2

=46 ½

Slope of line PQ =23

Slope of line RS =1 – 54 – 10 ½

=– 4– 6

Slope of line RS =23 ½

Slope of line PQ = Slope of line RS line PQ || line RS ........(i)

Slope of line QR =5 – 8

10 – 4

=– 36 ½

Slope of line QR =–12

Slope of line PS =1 – 4

4 – (–2) ½

= –3

4 2

=– 36 ½

Slope of line PS =–12

Slope of line QR = Slope of line PS line QR || line PS ........(ii) ½

In PQRS,side PQ || side RS [From (i)]side QR || side PS [From (ii)]

PQRS is a parallelogram [By definition] 1

P (– 2, 4) S (4, 1)

R (10, 5)Q (4, 8)