07 00 MT A.. Attempt ANY FIVE of the following : (i) Slope of the line (m) 4 y intercept of the line (c) 0 By slope intercept form, The equation of the line is y m + c y (4) + (0) y 4 MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model Answer Paper Ma. Marks : 40 The equation of the given line is y 4 (ii) If the terminal arm lies on the positive Y-ais, then angle made is 90º or 70º. (iii) Equation of a line parallel to Y-ais and passing through point (, 0) is. (iv) tan 5 + tan sec + 5 sec + 5 sec sec 6 sec 4 [Taking square roots] (v) y 5 ( + ) Comparing with the equation of a line in slope point form, y y m ( ) m 5 Slope of the line y 5 ( + ) is 5 (vi) + 90º [Given] tan 4 [Given]
/ MT cot tan [ cot tan (90 )] cot 4 A.. Solve ANY FOUR of the following : (i) (Analytical figure) P.4 cm R P.4 cm R mark for circle mark for tangent (ii) The terminal arm passes through P (5, ) 5 and y r y (5)( ) 5 44 69 r units
/ MT Let the angle be sin y r cos r 5 tan y 5 cosec sec cot r y r y 5 5 (iii) P (, 4), Q (, 6), R (8, ), S (0, k) Line PQ is parallel to line RS Slope of line PQ Slope of line RS 6 4 k 0 8 k 4 k k 4 + k 5 (iv) Value of k is 5. L (Analytical figure) L.6 cm M O N.6 cm A N M A mark for drawing circle mark for drawing tangent
4 / MT (v) sin + sin² [Given] sin sin² sin cos sin + cos sin cos sin cos 4 [Squaring both sides] cos cos 4 cos² + cos 4 sin + cos cos sin (vi) Let,A (0, 5) (, y ) B (5, 6) (, y ) The line passes through points A and B The equation of the line by two point form is y y y y 0 0 5 y 5 5 6 5 y 5 5 (y 5) 5y 5 5y + 5 y + 5 5 [Dividing throughout by 5] y + 5 is the equation of the line passing through (0, 5) 5 and (5, 6) A.. Solve ANY THREE of the following : (i) (Analytical figure) L K 60º 55º 7 cm M
5 / MT L O K 60º 55º 7 cm M mark for triangle mark for perpendicular bisectors mark for circumcircle (ii) sec cos sec cos sin + cos sin + sin + 4 sin 4
6 / MT sin 4 4 sin 4 sin cosec sin cosec is in IV quadrant cosec cos ec ( ) cos ec ( ) cos ec cos ec cos ec cos ec [Taking square roots] (iii) Let, A, 5 (, y ) B, k (, y ) C 4, 0 5 (, y ) Points A, B and C are collinear Slope of line AB Slope of line BC y y y y k 0 k 5 k 0 4 5 k 0
7 / MT k 0 0 k k k k + k 4k k 4 The value of k is 4 (iv) a cos + b sin m (a cos + b sin ) m...(i) [Squaring both sides] (a sin b cos ) n (a sin b cos ) n...(ii) [Squaring both sides] Adding (i) and (ii), (a cos + b sin ) + (a sin b cos ) m + n a cos + ab cos. sin + b sin + a sin ab sin. cos + b cos m + n a cos + b cos + a sin + b sin m + n cos (a + b ) + sin (a + b ) m + n (a + b ) (cos + sin ) m + n a + b m + n [ sin + cos ] (v) Let, A (, ), B ( 9, 6), C (, 4), D (6, 9) Slope of a line Slope of side AB Slope of line AB Slope of line CD Slope of line CD y y 6 9 ( ) 5 9 5 8 5 8 9 4 6 ( ) 5 6 5 8
8 / MT Slope of line AB and slope of line CD are equal. line AB line CD The line joining (, ) and ( 9, 6) is parallel to the line joining (, 4) and (6, 9). A.4. Solve ANY TWO of the following : (i) L.H.S. cos ec cos ec cos ec cos ec cos ec cos ec cos ec cos ec cos ec cot cos ec cot cos ec cot cot cot cos ec cot cos ec sin cos sin tan tan cos sec tan (sec tan )(sec tan ) (sec tan ) sec tan sec tan tan sec sec tan sec tan R.H.S. cosec cosec + sec + tan
9 / MT (ii) (Analytical figure) S S 4.9 cm 95º R 5.9 cm N 4.9 cm I 95º R 5.9 cm N mark for triangle mark for angle bisectors mark for perpendicular mark for incircle (iii) L.H.S. tan cot cot tan sin cos cos sin cos sin sin cos sin cos cos sin sin cos cos sin sin cos sin cos cos(sin cos) sin(cos sin) sin cos cos(sin cos) sin(sin cos) sin cos sin cos cos sin
0 / MT sin cos sin cos cos sin sin cos sin sin cos cos sin cos cos sin sin sin. cos cos cos. sin sin. cos cos. sin [ sin + cos ] sin. cos cos. sin cos. sin cos sin sec. cosec + R.H.S. tan cot + cot tan sec, cos ec cos sin + sec. cosec. A.5. Solve ANY TWO of the following : (i) (Analytical figure) U R 5.8 cm 5. cm S 4.5 cm H V S S S S4 S 5
/ MT U R 5.8 cm 5. cm S 4.5 cm H V S S S S 4 mark for drawing Analytical figure mark for SHR mark for constructing 5 congruent parts mark for constructing VS 5 S HS S mark for constructing UVS RHS S 5 (ii) A 0º 60º E Seg AB represents a tower. 60º 0º B C D Let A be a position of an observer. D and C are the initial and final position of the car. EAD and EAC are angles of depression
/ MT m EAD m ADB 0º [Converse of alternate angles m EAC m ACB 60º test] The car took 6 sec. to travel from D to C Let the speed of car be units/seconds Distance speed time CD 6 CD 6 units...(i) In right angled ABC tan 60 AB BC [By definition] AB BC AB BC...(ii) In right angled ABD, tan 0 AB BD AB BD AB BD...(iii) BC BD [From (ii) and (iii)] BC BD BC BC + CD [ B - C - D] BC BC CD BC CD BC 6 [From (i)] BC 6 BC Time Dis tance Speed BC seconds The time taken by the car to reach the foot of the tower is seconds.
/ MT (iii) seg AD is the median of seg BC D is midpoint of seg BC D + y + y, + +( 8), 8, 0, (, 5) By two point form, The equation of median AD y y y y 5 5 ( ) y 4 4 ( 5) 5 5 + y 4 4 + 5 5 y 4 6 9 9 ( 5) 6 (y 4) 9 45 6y 4 9 6y 45 + 4 0 9 6y 0 y 7 0 [Dividing throughout by ] The equation of median AD is y 7 0 Slope of line AC y y 8 4 5 4
4 / MT Slope of parallel lines are equal Slope of the line parallel to line AC is The line passes through B (, ) The equation of the line parallel to line AC passing through point B by the slope point form is y y m ( ) y ( ) [ ( )] y + ( + ) y + + 9 y + 9 0 y + 7 0 The equation of the line parallel to AC passing through point B is y + 7 0