MT - GEOMETRY - SEMI PRELIM - I : PAPER - 5

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Cameron Atkins
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1 07 00 MT MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) Slope of the line (m) 0 y intercept of the line (c) By slope intercept form, The equation of the line is y mx + c y (0)x + y The equation of the given line is y (ii) sec [Given] But, sec 60 sec sec 60 60º (iii) Slope of the line (m) Its y-intercepts (c) 5 Equation of the line by slope-intercept form, y mx + c y x + 5 (iv) sin 4 cos 0 sin 4 cos sin cos 4 tan 4 (v) (x + ) y y (x + ) Comparing with the equation of a line in slope point form, y y m (x x ) m Slope of the line (x + ) y is.

2 / MT (vi) 0º [Given] sin sin ( 0) sin 0 sin ( 0).. Solve NY FOUR of the following : (i) (nalytical figure) O.9 cm M O.9 cm M (ii) The terminal arm passes through P (, ) x and y r x y () +( ) mark for circle mark for tangent r units

3 Let the angle be sin y r cos x r tan y x / MT cosec r y sec r x cot x y (iii) Let (, 4) (x, y ) and m 5 The equation of the line passing through and having slope 5 by slope point form is, y y m (x x ) y 4 5 (x ) y 4 5x 5 5x y x y 0 The equation of the line passing through the points (, 4) and having slope 5 is 5x y 0. (iv) L (nalytical figure) L.6 cm M.6 cm O N N M mark for drawing circle mark for drawing tangent

4 4 / MT (v) tan + tan [Given] tan tan 4 [Squaring both sides] tan + tan. tan tan 4 tan + + tan + tan 4 tan 4 tan + tan (vi) Let, (, 5) (x, y ) B (4, 7) (x, y ) The line passes through points and B The equation of the line by two point form is x x x x y y y y x ( ) 4 y 5 5 ( 7) x 7 y 5 (x + ) 7 (y 5) x + 6 7y + 5 7y x y x y x 7 7 [Dividing throughout by 7] y x is the equation of the line passing through 7 7 (, 5) and (4, 7)

5 5 / MT.. Solve NY THREE of the following : (i) (nalytical figure).9 cm P.9 cm 8.8 cm Q B.9 cm P.9 cm M 8.8 cm Q B mark for circle mark for perpendicular bisector mark for tangents (ii) tan sin sin cos cos cos sin cos 9 cos...(i)

6 6 / MT sin + cos sin cos [From (i)] sin sin cos sin cos...(ii) [From (i) and (ii)] (iii) Let, (k, ), B (, 4), C ( k +, ) Points, B and C are collinear Slope of line B Slope of line BC 4 k ( 4) ( k ) 7 k 4 k 7 k k 7 ( k ) ( k) 7k k 7k + k 4 7 9k k k 9 The value of k is.

7 7 / MT (iv) L.H.S. tan sec sec tan tan(sec + ) (sec ) tan tan sec sec (sec ) tan sec sec sec (sec ) tan [ + tan sec ] sec sec (sec ) tan sec(sec + ) (sec ) tan sec tan sec tan sin cos cos cos sin tan, sec cos sin cos cos sin cosec R.H.S. tan sec + + sec + tan cosec (v) Let, (x, ) (x, y ) B (8, ) (x, y ) Slope of line B [Given] 4 y y Slope of line B x x ( ) 8 x 8 x 9 8 x

8 8 / MT (8 x) x 6 x 4 6 x x x 4 The value of x is Solve NY TWO of the following : (i) sec + tan p sin cos cos p sin cos p ( sin) cos p sin sin cos p sin cos sin sin p sin sin sin sin p sin + sin p sin + sin p [By Componendo-Dividendo] p sin p p sin p p p + sin [By Invertendo] (ii) (nalytical figure) U 5 cm 0º T M 7 cm S

9 9 / MT U 5 cm I 0º T 7 cm S mark for triangle mark for angle bisectors mark for perpendicular mark for incircle (iii) sin + cos cos sin cos. cos ( sin ) ( + sin ) cos sin sin cos By theorem on equal ratios, + sin cos cos ( sin) cos sin sin cos sin cos cos ( sin) cos sin Dividing the numerator and denominator of R.H.S. by cos cos + sin cos cos cos + sin ( sin) cos + sin cos cos + sin sin cos cos sin cos + sin + cos sec tan

10 0 / MT.5. Solve NY TWO of the following : (i) (nalytical figure) T T E E 4.9 cm 4.9 cm 0º H 6. cm M 0º 6. cm H 4 M (ii) mark for drawing analytical figure mark for MT mark for constructing 7 congruent parts mark for constructing H 5 M 7 mark for constructing EH TM 7 E D x 60º 0º 60 m 60 m Let E be the position of the cloud and let BC represent the surface B C of the lake. Let be the point of observer and let F be the reflection of the cloud x EC CF Let EC CF x m BCD is a rectangle [By definition] F B CD 60 m [Opposite sides of rectangle]

11 / MT EC ED + DC [E - D - C] x ED + 60 ED (x 60)m lso, DF DC + CF [D - C - F] DF (60 + x) DF (x + 60) m In right angled DE, tan 0º ED [By definition] D x 60 D D x 60 m In right angled DF, DF tan 60º [By definition] D x + 60 (x 60) (x 60) x + 60 x 80 x + 60 x x x 40 x 0 The height of the cloud above the lake is 0 m. (iii) Let point P be the point of intersection of lines 4x + y + 0 and 6x + 5y x + y + 0 4x + y...(i) Multiplying throughout by we get, x + 9y 6...(ii) 6x + 5y x + 5y 6...(iii) Multiplying throughout by we get, x 0y...(iv) dding (ii) and (iv), x + 9y 6 x 0y y 6 y 6 Substituting y 6 in equation (i), 4x + ( 6) 4x 8

12 / MT 4x + 8 4x 6 x 6 4 x 4 P (4, 6) Let Q be the point of intersection of lines 4x y 7 0 and x + y x y 7 0 4x y 7...(v) x + y x + y 5...(vi) dding equation (v) and (vi), 4x y 7 x + y 5 6x x x in equation (v), 4 y 7 y 7 8 y 9 y Q (, ) The equation of line PQ by two point from, y y x x y y x x y ( 6) 6 ( ) x 4 4 y 6 6 x 4 y 6 x 4 (y + 6) (x 4) y + x + x + y + 0 x + y 0 The required equation of line is x + y 0.

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 3

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 3 07 00 MT.. ttempt NY FIVE of the following : (i) Slope of the line (m) intercept of the line (c) 3 B slope intercept form, The equation of the line is m + c ( ) + 3 + 3 The equation of the given line is