07 00 MT A.. Attempt ANY 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 (0x + ( y 0 y The equation of the given line is y (ii ( + cot ( + cos ( cos cosec ( cos [ + cot cosec ] cosec sin [sin + cos, sin cos ] sin sin (iii A (, 5 and B (4, Co-ordinates of the midpoint of line segment AB by midpoint formula is, 4 5, 6 6, (, MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model Answer Paper Max. Marks : 40 (iv + 90º [Given] (90 cosec [Given] sec cosec (90 [ sec cosec (90 ] sec a cosec sec
/ MT (v y 5 (x 7 Comparing with the equation of a line in slope point form, y y m (x x m Slope of the line y 5 (x 7 is (vi + 90º [Given] tan 4 cot tan cot 4 [Given] [ cot tan (90 ] A.. Solve ANY FOUR of the following : (i (Analytical figure O. cm R O. cm R mark for circle mark for tangent
/ MT (ii The terminal arm passes through P (, x and y r x y ( ( 4 9 r units Let the angle be sin y r cosec r y cos x r sec r x tan y x cot x y (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 Value of k is 5. (iv (Analytical figure L N.6 cm M A
4 / MT L.6 cm O N M A mark for drawing circle mark for drawing tangent (v L.H.S. sec + cosec sec, cos ec cos sin cos sin sin + cos cos. sin cos. sin [ sin + cos ] sec. cosec R.H.S. sec + cosec sec. cosec (vi Let,A (, 5 (x, y B ( 4, (x, y The line passes through points A and B The equation of the line by two point form is x x x x x ( ( 4 y y y y y ( 5 5 (
5 / MT x y 5 4 5 x y 5 (x + (y + 5 x 4 y + 0 y x 4 0 y x 4 y x 7 [Dividing throughout by ] y x 7 is the equation of the line passing through (, 5 and ( 4,. A.. Solve ANY THREE of the following : (i Analysis : In PQR, P + Q + R 80 40 + Q + 90 80 Q 80 0 Q 50º R P (Analytical figure R 40º 50º Q 6 cm P 40º 6 cm 50º Q mark for drawing PQR mark for drawing perpendicular bisectors mark for drawing circumcircle
(ii sin 4 cos 0 sin 4 cos sin 4 cos 4 tan + tan sec + 4 + 6 9 9 6 9 5 9 sec cot sec sec sec sec cot 4 + cot cosec + 4 + 9 6 6 9 6 5 6 cosec 6 / MT 5 [Taking square roots] tan 4 cosec cosec cosec cosec 5 4 [Taking square roots] (iii Let, A, 5 (x, y B, k (x, y C 4, 0 5 (x, y
7 / MT Points A, B and C are collinear Slope of line AB Slope of line BC y y y y x x x x k 0 k 4 5 5 k 0 k 0 k k k + k 4k k The value of k is 4 k 0 0 k 4 (iv L.H.S. cos A + cos A cos A cos A cos A cos A cos A cos A cos A cos A cos A sin A cos A sin A sin A cos A cos A sin A cos A sin A sin A cos cosec A cot A cosec, cot sin sin R.H.S. cosec A cot A
(v Let A (k, (x, y B (4, 5 (x, y 8 / MT Slope of line AB [Given] Slope of line AB y y x x 5 ( 4 k 5 4 k 8 4 k 4 k 6 k 6 4 k k The value of k is. A.4. Solve ANY TWO of the following : (i tan + sin m tan sin n m n (tan + sin (tan sin tan + tan.sin + sin [tan tan sin + sin ] tan + tan.sin + sin tan + tan. sin sin 4 tan.sin...(i 4 mn 4 tan sin tan sin 4 tan sin sin 4 sin cos 4 sin cos
9 / MT 4 sin sec tan sec 4 ta n. sin tan sec 4 sin tan...(ii From (i and (ii, m n 4 mn (ii (Analytical figure T T O D 0º M 5º 6.4 cm A 0º D 6.4 cm 5º A M mark for triangle mark for angle bisectors mark for perpendicular mark for incircle (iii L.H.S. cos sin tan sin cos cos sin sin sin cos cos cos sin cos sin sin cos cos cos sin + cos sin sin cos cos sin cos sin cos sin cos sin cos sin
0 / MT (cos sin(cos + cos. sin + sin (cos sin cos + sin + sin. cos + sin. cos [ sin + cos ] R.H.S. cos sin + tan sin cos + sin. cos A.5. Solve ANY TWO of the following : (i U (Analytical figure U R R 5.8 cm 5. cm 5.8 cm 5. cm S 4.5 cm H V S S S S 4.5 cm H V S4 S 5 S S S S 4 S 5 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
(ii / MT seg AB represents the tree A AB m The tree breaks at point D seg AD is the broken part of tree which then takes the position of DC AD DC D m m DCB 60º Let DB x m AD + DB AB [ A - D - B] AD + x B 60º C AD ( x m DC ( x m In right angled DBC, sin 60º DB DC x x x x [By definition] x x x x x x DB DB m 4 ( ( DB DB 4 6 4 DB 4(.7 6 DB 4.5 6 DB 5.5 m The height at which the tree is broken from the bottom by the wind is 5.5 m.
/ MT (iii seg AD is the median of seg BC D is midpoint of seg BC D x + x y + y, + +( 8, 8, 0, (, 5 By two point form, The equation of median AD x x x x y y y y x 5 5 ( y 4 4 ( 5 x 5 5 + y 4 4 + 5 x 5 y 4 6 9 9 (x 5 6 (y 4 9x 45 6y 4 9x 6y 45 + 4 0 9x 6y 0 x y 7 0 [Dividing throughout by ] The equation of median AD is x y 7 0 Slope of line AC y x y x 8 4 5 4
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 (x x y ( [x ( ] y + (x + y + x + 9 x y + 9 0 x y + 7 0 / MT The equation of the line parallel to AC passing through point B is x y + 7 0