v Ë_«bŠu «roi² *«j)«

Size: px
Start display at page:

Download "v Ë_«bŠu «roi² *«j)«"

Transcription

1 v Ë_«bŠu «roi² *«j)«- 1 -

2 - 2 -

3 bšu «U¹u² W HB «Ÿu{u*« W bi*«.1... bon bšu ««b¼ bšu «ÂU bžu *«««di « bšu «WÝ«œ w to ÃU²% U wð UJ¹b «Èu² *«.2... wð UJ¹b «Èu² *«w UOŁ«bŠù«ÂUE½ Èu² *«w 5²DI½ 5Ð bf³ « WMOF W³ MÐ WLOI² WFD ro Ið roi² *«j)«qo.3... y Ë x w v Ë_«Wł b «s W UF «W œuf*« roi² *«j)«qo U¼b UFðË ULOI² *«Í «uð roi² *«j)«w œuf* WHK²<«uB « UOŁ«bŠù«Í u bš Í «u¹ Íc «roi² *«W œuf UNÐ d1 w² «jim «ÈbŠ Ë tko ÂuKF*«rOI² *«W œuf y u s tfdi Ë tko ÂuKF*«rOI² *«W œuf ² ukf 5²DIMÐ d1 Íc «roi² *«W œuf s¹ u;«s ÁUFDI rkž «roi² *«W œuf

4 43... roi² *«j)«w œuf* W¹œuLF «ub « W¹œuLF «ub «œu ¹ W¹œuLF «ub «v «roi² *«W œuf* W UF «ub «q¹u% roi² jš sž WDI½ bfð.5... ax + by + c «bi*«uý« ÂuKF roi² vkž WDI½ s UM «œulf «uþ FÞUI² 5LOI² 5Ð 5²¹Ë«e «whbm U² œuf W ö)«.6... WO½U «WOÝ«b «bšu «sž WI³ W;.7... wð«c «.ui² «WK¾Ý UÐUł Ë U³¹ b² «UÐUł.8... U KDB*«œd.9... lł«d*«

5 W bi*«.1 bon d?i? s ro?i²? *«j)«v? Ë_«bŠu «v «b? «Íe¹eŽ p?ð U³?Šd? Èu?²?*«w roi?²? *«j)«hzu?bš i?fð UN?O? UF? d? c?² MÝ w?² «Ë WOŁ«b?Š W?ÝbM¼ ifð W?Ý«œ w? ol?f?²mý U?L? W?¹u½U? «WKŠd*«w U??N?²?Ý«œ p o³?ý w²? «Ë wð UJ¹b «W?O½U? ½ô«Ë W?O?FO?³D «Âu?KF «w U?IO?³D²? «s do?? U?N w² «Ë t Èd?š_«hzU?B?)«ÆU¼dOžË WOD)«W d³ «Ë UM¹U³²*«Ë WOD)«U öfk w½uo³ «qo L² U bšu ««b¼ 2.1 vkž ΫœU ÊuJð Ê bšu «Ác¼ WÝ«œ s UN²½ô«bFÐ «b «Íe¹eŽ pm l u²¹ Ê ÆtŽUÐ œb%ë y Ë x t¹ u 0 wð UJ¹b «Èu² *«dfð -1 ÆÈu² *«w jim «5Fð -2 ÆtOKŽ UIO³Dð Íd&Ë Èu² *«w 5²DI½ 5Ð W U *«Êu½U o²að -3 ÆWMOF bžuið œb;«jim «s WŽuL rýdð -4 ÆtOKŽ ÊU²F «Ë ÊU²DI½ XLKŽ «Áœb%Ë roi² jš qo dfð -5 Ë dš U?LOI?² Í «u¹ ro?i²? qo b&ë U?L¼b U?FðË 5LOI?² Í «uð Ëd?ý œb% -6 ÆÁb UF¹ ÆWHK² ubð roi² *«j)«ôœuf o²að -7 ÆÂuKF roi² sž U WDI½ bfð Êu½U o²að -8 Æ5FÞUI² 5LOI² 5Ð 5²¹Ë«e «whbm w² œuf o²að -9 Æo³Ý U vkž WOIO³Dð qzu q%

6 bšu «ÂU 3.1 j)«qo? Ë wð UJ¹b «Èu²? *«w¼ W?FÐ ÂU? v b?šu «Ác¼ Èu²? Ÿ u?²¹ j³ðd¹ë ÆrO?I?²? jš s?ž WDI½ b?fðë roi?²? *«j)«w œu?f* W?HK²?<«uB? «Ë roi?²? *«qo? w½u «r?i «U v Ë_«W?FÐ _««b¼_uð wð UJ¹b «Èu²? *«UNM Ë_«r?I «u?b «Y U «r?i «oi? ¹ 5Š w œu? «Ë f U)«5 b?n «oi? O? ro?i²? *«j)«bf?ð lð«d «r I «wðq¹ Ϋd?Oš Ë s U? «Ë lðu «5 b?n «roi?² *«j)«w œu?f* WHK?²<«ÆdýUF «Ë lýu² «Ë s U «WŁö ««b¼_«oi O roi² jš sž WDI½ bžu *«««di «4.1 hzu?b?š u?š W?O?LKF? «pðu? ukf? «dłùë b?šu «w W?O?ÝU?Ý_«rO?¼U?H*«XO?³??² 5O U² «5Fłd*UÐ W½UF²Ýô«pMJ1 tðuio³dðë roi² *«1- Kindle, J.H., Theory and Problems of Plane and Solid Analytic Geometry, Schaum P.Bo., New York, Loney, S.L., The Elements of Coordinate Geometry, Macmillan and Co., London, bšu «WÝ«œ w to ÃU²% U 5.1 UM½S? «c?n Ë d?i*««c¼ s U?N W?O U?² ««b?šuk ÎU? U¼ ÎU?ÝU?Ý b?šu «Ác¼ d?³?²?fð UL? ÆUN?O œ «u «wð«c «.u?i² «WK¾?Ý Ë U³¹ b?² «q ÐË WO? «Ë WÝ«œ U?N²Ý«bÐ p?bm½ Ë p²³?²j w d «u?²ð b w² «Èd?š_«lł«d*«s WI?O³Dð qzu? Ë U³¹ bð q Ð p? BM½ u'«w¾onð pm VKD²¹ «c¼ë W U)«Ë W U?F «U³²J*«s U¼dOž Ë wý«b «e d*«w³²j ÆUN pln olf¹ë œu*ufo²ý«vkž bžu ¹ Íc «`¹d*«wÝ«b «- 6 -

7 wð UJ¹b «Èu² *«.2 Èu?² *«Âu?NH? X d?fðë WOŁ«b?Šù«WÝb?MN «œu³? ÂUF «ro?kf² «qš«d? w XÝ œ UOM M*«uD?) w½uo³ «qo? L² «w p X b? ²Ý«Ë t?o jim «5OFð W?OHO? Ë wł«bšù«wd³ðd*«w?o?ýu?ý_«ro¼u?h*«w?fł«d?0 bm³ ««c¼ w Âu?IMÝË ÆW?O½ü«ôœU?F*«iFÐ qš w Ë Æw½UO³ «Èu² *UÐ wð UJ¹b «Èu² *«w UOŁ«bŠù«ÂUE½ 1.2 UN¹bFÐ b?$ UM½S ÎU Uð ÎUMOO?Fð Èu² *«w p 1 x 1, y 1 qo WDI½ 5O?F² t½ rkfð W DIM «w? 5 FÞU?I²? Ë yy, xx q? Èu?² *«w? 5²ÐUŁË s¹b? UF?²? 5L?O?I²?? sž Æq _«WDI½ vl ð w² «(0,0) 0 Æ(1) qja «de½«vl? ¹Ë y (1) qja «u yy, wý d «j)«wl ¹ UL x u xx wi _«j)«vl ¹ Æ UOŁ«bŠù«Í u ÎUF Ê«u;«- 7 -

8 U?L? WDIMK x 1 wł«bšù«y u sž p q Èu?² *«w WDI½ Í b?fð vl ¹Ë b??fð Ê (1) q JA? «w k Šô WD?IMK y wł«b??šù«x u?? s?ž W DIM? «b?f?ð vl?? ¹ u?¼ x u????????? s?ž U¼b?????FÐ Ê Ë PA = OB = x 1 u¼ y u????????? sž p W?DI?M «wł«bšù«wðq¹ YO? Ð 5Ýu 5Ð ULN?F{uÐ ÎUF 5OŁ«b?Šù«sŽ d³f½ë ÆPB = OA = y 1 «U?³?łu? x wł«b?šù«d³²?f¹ë x 1, y 1 U?LNMOÐ WK U? l{ë l y wł«bšù«rł ôë x U? Æ U??O «ÁU?&«w fo? «ÎU?³ U?ÝË 5L?O «ÁU?&«w Ë q _«WDI½ s «b?²ð«fo? fo «ÎU³ UÝË x u u Ë q _«WDI½ s Î «b²ð«fo «ÎU³?łu d³²fo y wł«bšù«æx u X% ÁU&«w wð Uý Ê Ë ŸUÐ W?FÐ v «Èu²? *«ÊUL?I¹ s¹ u;«ê qja «s kšöð U?L w U² «(1) Ëb'«w UL ÊuJð Èu² *«w WDI½ Í wł«bš (1) Ëb'«lЫd «lðd «Y U «lðd «w½u «lðd «Ë_«lÐd «UOŁ«bŠù« x y (+,-) (-,-) (-,+) (+,+) (x, y) - 8 -

9 w½uo³ «Èu² *«w WO U² «jim «5Ž D(5,-2), C(-4,-1), B(-1,3), A(2,5), F(8,0), E(0,6) (1) U (2) qja «ÎU¹œu?LŽ d?²½ rł x u vk?ž di¹ë UN x = 2 wł«bšù«5f½ A jim «5OF² (2,5) WDIM «5FM tokž 5 Z¹ b² «qb½ v²š y u; Vłu*«ÁU&ô«w vkž v 2 q(«æwi¹dd «fhmð Èdš_«jIM «5F½ rł Æ(2) qja «w UL F(8,0) WDIM «ÊS?? p c? Ë ø«u* Æy u?? vkž lið E(0,6) WDIM «Ê kšô Æ ø«u* x u vkž lið (x,0) w¼ x u?? vk?ž lið w² «jimk W?? U?F «u?b «Ê u?i «s?j1 o³?ý U2 Æ(0,y) w¼ y u vkž lið w² «jimk W UF «ub «Ë w{u¹d «r U?F «t?f?{«ë v W?³? ½ W¹e?Oð UJ «U?OŁ«b?Šù«ÂUE?½ ÂUEM ««c¼ vl? ¹ Æ( ) UJ¹œ w ½dH «- 9 - s

10 w½uo³ «Èu² *«w WO U² «jim «5Ž (-3,-5) (4,0), (5,3), (0,-4), (3,-2) (1) V¹ bð Èu² *«w 5²DI½ 5Ð bf³ «2.2 ÆW¹e?Oð UJ «UOŁ«b?Šù«ÂUE½ «b? ²?ÝUÐ Èu²? *«w 5²DI½ 5Ð b?f³ «œu? ¹ sj1 qja «w U?L Èu?²? *«w UL¼UMO?ŽË p 2 x 2,y 2, p 1 x 1,y 1 5²DIM «U½c?š «S W öf «s ULNMOÐ d = P 1 P 2 bf³ U Š sj1 t½s (3) d = x 2 -x y 2 -y (1) WO U² «WK¾Ý_«sŽ WÐUłùUÐ (1) W öf «W U³Ł pmj1 ø P 1 P 2 R YK *«Ÿu½ U (1) ø x 1, x 2 W ôbð P 1 R uþ U (2) ø y 1, y 2 W ôbð P 2 R uþ U (3) sj1 u?žu? O? W¹dE½ «b??²?ÝUÐË WŁö? «WK¾?Ý_«sŽ pðuðu?ł «b?²?ÝUÐ Êô«Ë Æ(1) bžui «UI²ý«(3) qja «- 10 -

11 (2) U Æ(7,3), (4,-1) 5²DIM «5Ð W U *«bł q(«d = (-1) 2 = = 5 Æ bš«ë W UI²Ý«vKŽ lið C(1,5), B(-1,1), A(-2,-1) (3) U jim «Ê X³Ł«q(«AB = -2 -(-1) = = 5 BC = = = 2 5 AC = = = 3 5 AC = AB + BC Ê Í Æ bš«ë W UI²Ý«vKŽ lið Àö «jim «Ê wmf¹ «c¼ë ÊuJ¹ U bmž A, B, C jim «s Z²M¹ Íc «qja «U ø A C < A B + B C (4) U Æ5 U «ÍËU ² YK * ˃ C(-8,-2) Ë B(-11,3) Ë A(3,8) jim «Ê 5Ð

12 q(«ab = = 221 BC = = 34 AC = = 221 Æ5 U «ÍËU ² ABC YK *U Ê AB = AC Ê Í (2) V¹ bð ÆqOD² ˃ w¼ D(4,2), C(-2,2), B(-2,-2), A(4,-2) jim «Ê s¼dð -1 ˃ w??¼ C (2a + 3 a, 5a), B(2a, 6a), A(2a, 4a) j??i??m?? «Ê X??³??Ł -2 ÆŸö{_«ÍËU ² YK WMOF W³ MÐ WLOI² WFD ro Ið 3.2 W?LO?I²? *«WFD?I «r Ið w² «WDI?M «woł«bš œu? ¹ô bžu? bm³ ««c¼ w o²?amý Æ UDF W³ MÐ 5² ukf 5²DI½ 5Ð WK «u «5²D?IM «5Ð WK? «u «W??L??O??I??²?? *«W???FDI «vk?ž lið P(x,y) WDI?M «Ê d?? P WDIM «U?OŁ«bŠ U?L¼ UL? r W?³ M UÐ W?FDI «r Ið U?N½ Ë P 2 x 2,y 2 Ë P 1 x 1,y 1 ø r W³ M «Ë P 1 Ë P 2 5²DIM «UOŁ«bŠ W ôbð Ë P 1 PQ 5 K *«rýd½ë Èu?²? *«w Àö «jim «5F?½ «R ««c¼ sž WÐU?łö Ê s b QðË qja «v de½«(4) qja «w `{u u¼ UL PP 2 R

13 P 1 Q = x - x 1 QP = y - y 1 PR = x 2 - x P 2 R = y 2 - y (4) qja «Ê kšöð Í «u??² «Â«b???²??ÝUÐ ø PP 2 R, P 1 PQ 5 K?*«U¹«Ë 5Ð W? ö??f «U? Ê p s Z²M¹Ë ÊUNÐUA² 5 K *«ÊS w U² UÐË W¹ËU ² 5 K *«w dþum²*«u¹«ëe «P 1 P = PQ PP 2 P 2 R = QP 1 RP = r vkž qb ½ W³ M «Ë UOŁ«bŠùUÐ «uþ_«ro sž i¹uf² UÐË r = y - y 1 y 2 - y = x - x 1 x 2 - x... (1) s vkž qb ½ (1)

14 r x 2 - x = x - x 1 rx 2 - rx = x - x 1 x(1+r) = x 1 + rx 2 x = x 1 + rx r y = y 1 + ry r vkž qb ½ q *UÐË x 1 + rx r, y 1 + ry r w¼ WFDI «ro Ið WDI½ Ê Í W U??(«Ác¼ w Ë P 2 Ë P 1 5²DIM «5Ð lið P WDIM «Ê (4) qja «s kšöð Í s P 1 P 2 WFDI «œ«b?² «vkž P XF Ë «U? qš«b «s ÎULO?Ið ro I?² «wl ½ dog² W³ UÝ W³ M «d³²f½ W U(«Ác¼ w Ë Ã U)«s ÎU?LO Ið vl ¹ ro I² «ÊS UNO²Nł s Æ(5) qja «w `{u u¼ UL WLOI² *«WFDI «ULNO r IMð s¹ck «s¹ e'«uo ÁU&«(5) qja «- 14 -

15 v P s œuf½ rł P 2 v P 1 s do? ½ UM½ à U)«s ro I?² «W UŠ w kšô W?³? M «ÊuJð w U?² UÐË d?o? «ÁU?&«d?O?G½ UM½ Í Ã U?)«s W?FDI «Í e?ł b¹b??² P 2 w¼ W U(«Ác¼ w ro I² «WDI½ ÊuJðË ÆW³ UÝ x 1 - rx r, y 1 - ry r Ë P 1 (2,3) (5) U 5²?DIM? «5Ð WK? «u «W???F?DI «r????i?ð w² «W?DIM? «woł«b????š b???ł Æ 3 W³ MÐ Ã U)«s Ë qš«b «s P 2 (-4,-2) 5 q(«ul¼ qš«b «s ro I² «WDI½ UOŁ«bŠ y = (-2) = 9 8, x = (-4) = Æ Ë 9 8 w¼ qš«b «s ro I² «WDI½ UL¼ à U)«s ro I² «WDI½ UOŁ«bŠ»

16 x = (-4) = 11 y = (-2) = 10.5 Æ 11 Ë 10.5 w¼ Ã U)«s ro I² «WDI½ Ë P 1 x 1,y 1 5²DIM «5Ð WK «u «W?LOI²? *«WFDI «nobmð WDI½ woł«b?š bł (6) U Ær=1 Æ P 2 x 2,y 2 q(«w¼ WLOI² *«WFDI «UNÐ r IMð w² «W³ M «Ê b$ ro I² «WDIM W UF «ub «w i¹uf² UÐ x = x 1 + x 2 2, y = y 1 + y 2 2 Æ(-3,2) Ë (5,4) (7) U 5²DIM «5Ð WK «u «WFDI «nb²m woł«bš bł q(«- 16 -

17 x = x 1 + x 2 2 y = y 1 + y 2 2 = = = 1 = 3 W??FD?I «ÊU??L????Ið 5?²K «P 2 x 2,y 2 P 1 x 1,y 1 (8) U 5²D?IM «U??OŁ«b??Š b??ł ÆW¹ËU ² ÂU WŁöŁ v «B(9,7) Ë A(3,-1) 5²DIM «5Ð WK «u «WLOI² *«r 1 = AP 1 P 1 B = 1 2 P 1 WDIM «œu ¹ô x 1 = (9) = 5, y 1 = (7) = x 2 = 3 + 2(9) r 2 = AP 2 P 2 B = 2 1 = 7 Ë y 1 = (7) P 2 WDIM «œu ¹ù = 13 3 (3) V¹ bð dd «UOŁ«b?Š UL UNO? dd w dþ bš«(2, -1) WDIM «Ë dz«œ e d (3,5) X½U «-1 ødšü«r??imð P 2 (9,-7) Ë P 1 (-1,13) 5²DIM? «5Ð WK «u «W?L?O??I?²? *«W?F?DI «Ê s¼dð -2 ÆW³ M «fhmð à U)«s Ë qš«b «s P 4 (29,-47) Ë P 3 (5,1) 5²DIM UÐ

18 P 2 (-1,-3) P 1 (4,-2) (1) wð«c «.ui² «WK¾Ý wð UJ¹b «Èu² *«w WO U² «jim «5Ž -1 (8,-2), (-4,0), (0,7), (-3,-6), (3.5,6) WO U² «jim «Ã«Ë 5Ð W U *«bł -2» P 2 (9,5) P 2 (-c,-d) P 1 (5,2) P 1 (a,b) ÆŸö{ Í «u² ˃ w¼ D(1,2), C(4,3), B(1,0), A(-2,-1) jim «Ê X³Ł -3 à w¼ týëƒ UOŁ«bŠ Íc «YK *«Ê s¼dð -4 ÆŸö{_«ÍËU ² ÊuJ¹ C 7, 1 + 3, B 3, 1 + 3, A 5, 1-3 5²D?IM «5Ð WK «u? «W?L??O??I??²?? *«W??FDI «ÊU??L???I?ð 5²K «5²?DIM «woł«b??š b??ł -5 Æ 5 W³ MÐ Ã U)«s Ë qš«b «s P 2 (2,-13), P 1 (-7, -4) 4 jim «týëƒ Íc «YK *«w WDÝu²*«ULOI² *«w öð WDI½ woł«bš bł -6 P 1 (6,8) ÆC(2,5), B(9,-3), A(4,7) w¼ tžö{ UHB²M UOŁ«bŠ Íc «YK *«Ëƒ UOŁ«bŠ bł -7 5²DI?M «5Ð WK «u «W?L??O?I?²?? *«W?FDI «r??ið ÆP 2 WDIM «woł«bš b qš«b «s Æ(2,-3), (5,2), (-2,1) (2,9) 3 7 WDIM «X½U? «-8 W³ MÐ P 2 x 2,y

19 roi² *«j)«qo.3 5³½Ë y Ë x w v Ë_«W?ł b «s W œu?f*«hzub?š ifð bm³ ««c¼ w? df?²mý w j)««c¼ ÁU?&«b¹b?% W?O?H?O 5³?MÝ UL? U?Lz«œ ro?i?²?? jš u¼ w½u?o?³ «U¼UM M Ê Æwð UJ¹b «Èu² *«y Ë x w v Ë_«Wł b «s W UF «W œuf*«1.3 W UF «UN² œuf Ê y Ë x w v Ë_«Wł b «s WO½ü«ôœUFLK p²ý«œ s rkfð w¼ ax + by + c = 0... (1) W²ÐUŁ UOL a, b, c YOŠ ÆÎULz«œ roi² jš u¼ W œuf*«ácn w½uo³ «j)«ê wk¹ ULO 5³MÝË P 1 x 1,y 1 Àö «jim «Ê dh½ ÎUL?OI² ÎUDš q 9 (1) W œuf*«ê U³Łù Æ(6) qja «w UL W œuf*«vm M vkž lið P 3 x 3,y 3 P 2 x 2,y

20 (6) qja «Í «u¹ ÎULO?I² P 1 s rýd½ë x u vkž P 3 C, P 2 B, P 1 A blž_«em½ ÆVOðd² «vkž C B w P 3 C, P 2 B s¹œulf «ÎUFÞU Ë x W œu?f*«oi?% U?NðU?OŁ«b?ŠS? Ê (1) W œu?f*«vm M vkž lið Àö? «jim «Ê U0 u Í ax 1 + by 1 + c = 0... (2) ax 2 + by 2 + c = 0... (3) ax 3 + by 3 + c = 0... (4) Ê Z²M¹ (3) w dþ s (2) w dþ ÕdDÐË a x 2 - x 1 + b y 2 - y 1 = 0... (5) Ê Z²M¹ (4) w dþ s (2) w dþ ÕdDÐË a x 3 - x 1 + b y 3 - y 1 = 0... (6) Ê b& (6) Ë (5) s Ë x 2 - x 1 x 3 - x 1 = y 2 - y 1 y 3 - y 1... (7) Æ[ (6) Ë (5) s x 3 - x 1 x 2 - x 1 s q WLO œu ¹SÐ p X³Ł ]

21 Ê b& (6) qja «w WMO³*«WLOI² *«ldi ««uþ sž (7) w i¹uf² UÐË P 1 B P 1 C = P 2 B P 3 C ÆC > UO Ω B > UO Ê U0Ë ÊUNÐUA² P 1 C P 3, P 1 B P 2 ÊU K *U Ê ÆqLF UÐ ULNCFÐ vkž ÊUI³DM P 1 C, P 1 B 5FKC «Ê U0Ë ÆÎUC¹ ULNCFÐ vkž 5I³DM P 1 P 3, P 1 P 2 ÊUFKC «ÊuJ¹ Ê V ¹ Ê ÆP 1 P 2 WFDI «W UI²Ý«vKŽ lið P 3 WDIM «Ê«Í vkž ÊuJð (1) W œu?f*«v?m M vkž W?F?? «Ë Èd?š WDI½ Í Ê U?³Ł s?j1 q *UÐË ÆP 1 P 2 W UI²Ý«ÆÎULz«œ roi² jš u¼ (1) W œuf*«vm M Ê Í roi² *«j)«qo 2.3 w y Ë x s¹d?o?g?²*«w v Ë_«W?ł b «ôœu?f? s? œb?ž U?OM M XL?Ý «w UMOF? U¼U&«c ²¹ U?NM ö Ê Ë roi²? jš UNM ö Ê kšö½ wð UJ¹b? «Èu² *«U?N?FMB¹ w² «W?³?łu*«W¹Ë«e UÐ ro?i?²? *«ÁU?&«b¹b?% vkž U?O?{U¹ `KD «b? Ë ÆÈu?²? *«Æx u l roi² *«(1) n¹dfð x u; Vłu*«ÁU?&ô«l roi² *«UN?FMB¹ W³łu W?¹Ë«dG w¼ roi²? *«qo W¹Ë«-1 ÆWŽU UIŽ b{ ÁU&«w WÝUI Æ(Tangent) tko W¹Ë«qþ u¼ roi² *«j)«qo -2 wk¹ ULŽ Vł Ë (7) qja «v «de½«øtko u¼ U Ë øab roi² *«qo W¹Ë«U øtko u¼ U Ë øcd roi² *«qo W¹Ë«U w U?² UÐË θ 1 UN?ÝUO? w² «Ë < x BA w¼ AB qo W?¹Ë«Ê błë p½ bð ô u¼ AB roi² *«qo m ÊS

22 m 1 = tan θ 1 qo? m 2 ÊS w U² UÐË θ 2 UNÝU?O w² «Ë < x CD w¼ CD roi² *«W¹Ë«Ê Ë u¼ CD roi² *«m 2 = tan θ 2 (7) qja «P 2 x 2,y 2 Ë P 1 x 1,y 1 q?? 5²D?IMÐ U*«rO??I??²?? *«q?o?? Ê U?³?Ł sj1ë ÍËU ¹ m = tan θ = y 2 -y 1 x 2 -x 1... (1) w U??L? A w x u?? w ö?o P 2 Ë P 1 5²DIM U?Ð U*«rO?I?²? *«UML?Ý «S? θ sj m = tan θ ÊuJ¹ tko ÊS? w U² UÐË θ ÊuJ¹ tko? W¹Ë«UO? ÊS (8) qja «Æ ø«u* < P 2 P 1 D Ê b& P 1 D P 2 UO ÍËU ð YK *«s Ë m = tan θ = y 2 - y 1 x 2 - x 1 p qkž

23 (8) qja «Æ(3,5) Ë (1,4) 5²DIM UÐ U*«rOI² *«qo bł u¼ roi² *«qo Ê b$ (1) m = = 1 2 Êu½UI «Â«b ²ÝUÐ (9) U q(«æ(5,9) (-8,-4) 5²DIM UÐ d1 t½ XLKŽ «tko W¹Ë«Ë roi² *«qo s ö bł m = tan θ = = 1 θ = tan -1 1 = 45 Ê Æ"1 ÍËU ¹ (tan.) UNKþ w² «W¹Ë«e «θ" wmfð θ = tan -1 1 U³F «Ê d cð Æ45 w¼ 1 ÍËU ¹ UNKþ W³łu W¹Ë«dG Ê U K *U Š s rkfðë (10) U q(«æθ UNÝUOIÐ qo*«w¹ë«sž di*««c¼ w d³fmý WEŠö

24 WO U² «jim «Ã«Ë QÐ d9 w² «ULOI² *«s qj t²¹ë«ë qo*«bł (4) V¹ bð Æ(-11, 10) Ë (-11, 4)» Æ(10, -3) Ë (14, -7) Æ(14, 6) Ë (8, 6) à t?²? œu?f? X?ODŽ««rO?I??²? *«j)«q?o? b?& n?o? Êü«ÕdD¹ Íc ««R?? «Ë ax + by + c = 0... t² œuf Íc «roi² *«j)«qo b& no Í øw UF «UNð ubð ø(1) q????? 5?²D?I½ Í t?????o?kž c?????šq½ë Èu?????²??????*«w r?o????i?????²??????*«j)«r?ýd½ ÆP 2 x 2,y 2 Ë P 1 x 1,y 1 u¼ tko ÊS roi² *«j)«vkž ÊU²F «Ë P 2 Ë P 1 5²DIM «Ê U0Ë m = y 2 - y 1 x 2 - x 1... (2) Æ(9) qja «de½«(9) qja «Í (1) roi² *«j)«w œuf ÊUII% P 2 Ë P 1 5²DIM «ÊS p c Ë

25 ax 1 + by 1 + c = 0... (3) ax 2 + by 2 + c = 0... (4) vkž qb ½ (4) W œuf*«w dþ s (3) W œuf*«w dþ ÕdDÐË a x 2 - x 1 + b y 2 - y 1 = 0 Z²M¹ UNM Ë p s b Qð y 2 - y 1 x 2 - x 1 = - a b... (5) Ê b& (5) Ë (2) 5Ð Ê U m = - a b... (6) u¼ ax + by + c = 0... (1) dd «fh½ w U½uJ¹ Ê V ¹ y - a b = - x q UF y q UF Ë x roi² *«qo ÊS «cnðë s¹dog²*«êu¹u ¹ s¹ck «s¹b(«ê kšô Æ(6) W œuf*«â«b ²ÝUÐ roi² *«j)«qo œu ¹ q³ (1) W œuf*«s (11) U ø 2y + 4 = -5x (2) 2x - 3y + 6 = 0 (1) 5LOI² *«s q qo U q(«u¼ j)«qol Ê ªW œuf*«s bš«ë dþ w y Ë x Ê«dOG²*«(1) m = = 2 3 W œuf*«s bš«ë dþ w y 2y + 5x + 4 = 0 m = Ë x s¹dog²*«lc½ (2) qo*«êujo

26 (5) V¹ bð WO U² «ULOI² *«s q qo bł y - x - 4 = 0-3 7x = 3y x = 9-4 y + 5 = 0-3 U¼b UFðË ULOI² *«Í «uð 3.3 5O U² «5 «R «sž Vł Ë (10) qja «w sf9 ø L 2 Ë L 1 5LOI² *«s q qo W¹Ë«U øα, θ 2, θ 1 U¹«Ëe «5Ð W öf «U wk¹ U 5³ð 5 «R «sž WÐUłù«Ê Æθ 1 Æθ 2 w¼ L 1 w¼ L 2 qo W¹Ë«qO W¹Ë«(10) qja «WD?I½ C, x u? l L 1 lþuið WDI½ B, L 2 Ë L 1 lþuið WDI½ A X½U ««Ë ABC YK??LK W?³?? M UÐ W?ł U??š W¹Ë«d?³??²?Fð W¹Ë«e «ÊS? x u?? l lþu?ið θ 2 L 2 ÊuJ¹ w U² UÐË θ 2 = θ 1 + α α = θ Ë 2 θ

27 tan α = tan θ 2 θ 1 Ê 5²¹Ë«5Ð ËdH «qþ d cð = tan θ 2 tan θ tan θ 2 tan θ 1 5?ÐË L 2 Ë L 1 tan α = 5LOI² *«5Ð α m 2 - m m 1 m 2... (1) œu(«w¹ë«e «qþ 5Ð W öf «œb% ø 5¹ «u² 5LOI² 5Ð ub;«α (1) W œuf*«w¹ë«e ««bi U Êü«Ë ÊS w U² UÐË ÆdHB «ÍËU ð ULNMOÐ ub;«α W¹Ë«e «ÊS ÊULOI² *«È «uð «ÆULNOKO tan α = tan0 = m 2 - m m 2 m 1 = 0 Æm 2 - m 1 = 0 tm Ë tan0 = 0 Ê_ Æm 2 w½u?? «qo? ÍËU?? ¹ qja «w 5³? u¼ UL? θ 2 m 1 m 1 = m 2 Ë_«qO? ÊU?? «ÊU¹ «u?²¹ 5?L?O?I??²? *«Ê Í w½u «qo? W¹Ë«ÍËU ð θ 1 Ë_«qO W¹Ë«ÊS? w U² UÐË Æ(11) (11) qja «Õd?ý«Í «u?² «Ë U¹Ë«e «dþu?mð vkž œu?l?²?žôuð W??O?²M «fh½ v? «u? u «pmj1ë Æ p

28 90 ÊuJð U?L?NM?OÐ u?b?;«w¹ë«e «ÊS? L 2 Ë L 1 ÊU?L?O?I?²? *«b? U?Fð «U? ÊS w U² UÐË tan α = tan 90 = m 2 - m 1 Ω WMOF dož WOL 1 + m 2 m 1 Æ tan 90 d cð 1 + m 2 m 1 = 0 Ê_ Æ-1 ÍËU ¹ ULNOKO»d{ q UŠ ÊU «Ê«b UF²¹ L 2 Ë L 1 m 1 m 2 = -1 UNM Ë 5LOI² *«Ê Í 5²DIM UÐ U*«rO?I²? *«Í «u¹ B(1,4) Ë A(5,6) 5²DIM UÐ U*«rO?I²? *«Ê s¼dð (12) U ÆD(4,2) Ë C(10,5) = 1 2 Ë_«rOI² *«qo = 1 2 w½u «roi² *«qo ÆÊU¹ «u² ÊULOI² *U Ê ÊU¹ËU ² 5KO*«Ê U0Ë q(«(13) U UL¼U² œuf s¹ck «5LOI² *«Ê XLKŽ «a WLO bł 3x - 2y + 7 = 0 ax + 5y + 4 = 0 Ê«b UF² (2) ÊU¹ «u² (1) m 1 = -3-2 = Ë_«rOI² *«qo q(«

29 m 2 = -a 5 w½u «roi² *«qo 3 m 1 = m = -a 5 Æm 1 m 2 = -1 ÊuJ¹ Í «u² «W UŠ w (1) Í a = -7.5 UNM Ë ÊuJ¹ b UF² «W UŠ w (2) Í 2 x -a 5 = -1 UNM Ë a = Æ bš«ë W UI²Ý«vKŽ lið C(6,7), B(3,3), A(0,-1) jim «Ê X³Ł -1-3 u¼ B Ë A w U*«rOI² *«qo 0-3 = u¼ C B w U*«rOI² *«qo 6-3 = 4 3 ÆÊU¹ «u² BC AB 5LOI² *«Ê Í Æ bš«ë W UI²Ý«vKŽ ULN Ê B Æ bš«ë W UI²Ý«vKŽ C WDIM «w ÊU d²a ULN½ U0Ë Ë B Ë A Àö «jim «Ê Í (14) U q(«(15) 5O U² «5LOI² *«5Ð œu(«w¹ë«e «bł 2x - 3y + 4 = 0, x + 5y + 7 = 0 U

30 Æ p s b Qð 1 5 tan α = 2 w½u «roi² *«qo Ë 3 m 2 - m m 1 m 2 = Ë_«rOI² *«qo = 1 q(«æ tan 45 WLO d cð α = 45 Ê ÊU½uJ¹ (6) a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0 5LOI² *«Ê 5Ð V¹ bð a 1 = b 1 ÊU «5¹ «u² -1 a 2 b 2 a 1 a 2 + b 1 b 2 = 0 ÊU «s¹b UF² -2 U*«rO???I??²?? *«vk?ž Íœu??L??Ž B(-1,2) Ë A(2,4) ˃ w¼ D(5,6), C(10,1), B(3,0), A(-2,5) (1) wð«c «.ui² «WK¾Ý 5²D?IM UÐ U*«r?O??I??²?? *«Ê 5Ð ÆD(1,2) Ë C(-1,5) 5²DIM UÐ ji?m «Ê qo*«â«b??????²??ýuð X?³Ł Y U?? «lkc? «Í «u¹ YK?? w 5FK?{ Í wh?b??²m? 5Ð q «u «ro??i??²??*«ê X³Ł rzu??? YK??? ˃ w¼ L 1 qo? ÊU? Ë 60 C(2,4), B(4,8), A(8,6) Æ5F ÆtHB½ ÍËU ¹Ë jim? «Ê qo*«â«b?????²???ýuð 5Ð ÆW¹Ë«e «ÍËU ð L 2 Ë L 1 5L?OI?² *«5Ð u?b;«w¹ë«e «X½U? «-5 Æ ÍËU ¹ L 2 qo Ê s¼d³ 1 ÍËU ¹

31 roi² *«j)«w œuf* WHK²<«uB «.4 w v Ë_«W??ł b «s W?? U??F «W œu??f*«ê (1.3) bm³? «w d? c?ð U?L?? UM²??³Ł ub «vkž w² «Ë x,y s¹dog²*«æw²ðuł UOL a, b, c YOŠ ÎULOI² ÎUDš q 9 ax + by + c = 0...(1) ub «vkž (1) W œuf*«wðu² sj1 (a 0) x + b a y + c a = 0... (2) w²?l?o? XL?KŽ «ro?i?²? *«j)«w œu?f? W?ÐU?²? sj1 t½ (2) W œu?f*«s kšô c j)«w œuf? WÐU² sj 5²ÐU? «WLO? b¹b ² 5Þd?ý UMODŽ «S 5²ÐU «a, b a wk¹ U?LO? o²amýë r?oi?²? jš Í W œuf? 5O?F² ÊUÞd?ý UM ek¹ t½s? tokžë ÆrO?I²? *«ÆrOI² *«j)«w œuf* WHK²<«uB «UOŁ«bŠù«Í u bš Í «u¹ Íc «roi² *«W œuf 1.4 ÆL 2 Ë L 1 5LOI² *«kšôë (12) qja «v «de½ (12) qja «- 31 -

32 b??& noj «b??šu «s b «b?i0 tm?ž b?f?³¹ë x u?? Í «u¹ L 1 Ê XLKŽ «øt² œuf øb& «UL UNM qj y wł«bšù«5žë L 1 vkž WF «u «jim «s ΫœbŽ cš lið WDI½ q v?kž o³dm¹ «c¼ë b u¼ jim «Ác?N y wł«bšù«ê«b?łë p½ bð ô ÆÈdš_«jIM «s œbž Í cšqð p s b Qð ÆL 1 vkž Æ5ÞdAK W O²½ Uł L 1 vkž WF «u «jim «lol' y = b Êu Ê Æx u Í «u¹ L 1-1 Æx u sž b «bi0 bf³¹ L 1-2 b b?fð vkžë x u? Í «u¹ Íc «roi² *«W œu?f w¼ y = b... (1) W œuf*«ë ÆtM «bšë «U?L tm UÐd?²I? Ë t U¹ «u? vi³¹ YO? Ð x u? ÁU&UÐ L 1 ro?i²?*«d% «øb = 0 U bmž Àb ¹ «U øb vkž kšöð w¼ x u W œuf Ê v «c¼ s XK uð q¼ øy = 0... (2) Íc «L 2 roi² *«W œuf WÐU² Êü«pMJ1 q *UÐË Æy u Í «u¹ -1 Æy u sž a «bi0 bf³¹ -2 x = a... (3) ub UÐ Æx = 0 w¼ y u W œuf ÊS w U² UÐË (16) U W UŠ q w roi² *«W œuf bł Æ «bšë 5 «bi0 tmž bf³¹ë x u Í «u¹ ÊU «-1 Æ «bšë (-3) «bi0 tmž bf³¹ë y u Í «u¹ ÊU «-2 q(«- 32 -

33 Æy = 5 w¼ roi² *«W œuf -1 Æx = -3 w¼ roi² *«W œuf -2 w² «W? öf «b?$ë UNÐ d1 w² «jim «ÈbŠ Ë tko ÂuKF*«rOI² *«W œuf 2.4 Ê rkž «L 1 roi² *«j)«w œuf œu ¹»uKD*«Æm ÍËU ¹ L 1 roi² *«qo -1 ÆP 1 x 1, x 2 WDIM UÐ d1 L 1 roi² *«-2 P(x, y) q? tok?ž WF? «Ë WDI½ Í c?šq½ j)«w œu?f œu? ¹ù u¼ L 1 Æ(2) Ë (1) roi² *«qo Ê b$ (13) m = y - y 1 x - x 1 y - y 1 = m x - x 1... (1) 5ÞdA «Â«b ²ÝUÐ UNOOŁ«bŠ 5Ð jðdð qja «Â«b ²ÝUÐ vkž qb ½ UNM Ë ÆWÐuKD*«rOI² *«j)«w œuf w¼ Ác¼Ë (13) qja «- 33 -

34 (17) U Æm = 3 tko Ë A(2,-1) WDIM UÐ d1 Íc «roi² *«W œuf bł q(«w¼ roi² *«j)«w œuf Ê b$ (1) W œuf*«w i¹uf² UÐ y - (-1) = 3 (x - 2) y + 1 = 3 (x - 2) y - 3 x = 0 5²DIM UÐ U*«W?FDI «nb?²m0 d1 Íc «Ë m = -1 x 1 = y 1 = = 2 = 1 tko? Íc «roi?² *«W œu?f b?ł WFDI «nb²m* x 1 WFDI «nb²m* y 1 Æ(2, 1) (18) U Æ(3, 0), (2, 1) wł«bšù«wł«bšù«q(«w¼ nb²m*«wdi½ Í ÊuJð roi² *«j)«w œuf Ê y - 1 = -1 (x - 2) y - 1 = -x + 2 y + x - 3 = 0 y u s tfdi Ë tko ÂuKF*«rOI² *«W œuf 3.4 Ê rkž «L 1 roi² *«j)«w œuf œu ¹uKD*«- 34 -

35 WD?I½ sž y u l A tfþuið WDI½ bf³ð œb ¹ y u s L 1 b?$ë P(x,y) q? tokž W?F «Ë WDI½ Í c?šq½ Æ(2) (1) ø(14) ÆOA = b u¼ (13) L 1 Æm ÍËU ¹ roi² *«qo (1) Æb ÍËU ¹ y u s tfdi (2) roi² *«ldi qja «w ldi*«ë ro?i² *«j?)«w œuf œu? ¹ù Æ O q _«5ÞdA «Â«b ²ÝUÐ UNOOŁ«bŠ 5Ð jðdð w² «W öf «qja «w A WDIM «woł«bš 5OFð pmj1 q¼ x = 0 u¼ A WDIMK x y = b u¼ A WDIMK y wł«bšù«wł«bšù«a(0,b) Ê A(0,b) WDIM UÐ d1 L 1 da «v u% (2) da «ÊS w U² UÐË w¼ roi² *«j)«w œufl Ê y - b = m (x - 0) y = mx + b... (1) (14) qja «- 35 -

36 Æ4 ÍËU ¹ y u s tfdi Ë 2 3 ÍËU ¹ tko Íc «roi² *«W œuf bł w¼ roi² *«W œuf Ê b$ (1) y = 2 3 x + 4 W œuf*«â«b ²ÝUÐ (19) U q(«íëu ¹ y u s t?fdi Ë 120 ÍËU?ð tko W¹Ë«Íc «ro?i²? *«W œuf? bł ÍËU ¹ roi² *«qo m = tan 120 = -tan 60 = - 3 w¼ roi² *«W œuf Ê y = - 3 x - 6 (20) U Æ -6 q(«- 36 -

37 (21) U p s b??łë y = mx + b u?b «w 2y - 3x - 4 = 0 ro?i?²? *«W œu?f? l{ Æy u s tfdi Ë roi² *«qo q(«ê Ë (+1) tk U?F? Ë U?NM dþ w li¹ ji? y d?og?²*«ê b?& (1) W œu?f*«kšô Ædšü«dD «w ÊUFI¹ b XÐU «b(«ë x dog²*«ub «w UNF{Ë sj1 UDF*«W œuf*u Ê y x - 2 = 0 y = 3 2 x + 2 Ë Æ 3 2 u¼ tko Ê Ë 2 u¼ y u s roi² *«ldi Ê b$ UNM Ë (7) V¹ bð Æ2x - 3y + 6 =0 roi² *«vkž ÍœuLŽË (-2, 3) WDIM UÐ U*«rOI² *«W œuf bł -1 (- 5 ²DIM UÐ U *«r O I² *«Í «u ¹ Íc «Ë (2, -3) W D IM UÐ U *«r OI² *«W œu F b ł -2 Æ2, 2), (4, 1) 5² ukf 5²DIMÐ d1 Íc «roi² *«W œuf 4.4»uKD*«Ê rkž «L 1 roi² *«j)«w œuf œu ¹ ÆP 1 x 1, y 1 WDIM UÐ d1 L 1 roi² *«-1 ÆP 2 x 2, y 2 WDIM UÐ d1 L 1 roi² *«

38 WDI½ Í c??šq½ P 1 Ë P 2 5²DIM UÐ d1 Íc «L 1 ro?i?²? *«j?)«w œu?f? œu? ¹ù «b??²?ÝUÐ x,y 5Ð W? öf «b?$ë (15) qja «w U?L P(x, y) q? t?okž WF? «Ë W? UŁ øp 1 Ë P 2 m = y 2 - y 1 x 2 - x 1 UOŁ«bŠ W ôbð L 1 Æ(2) Ë (1) roi² *«qo U ÍËU ¹ m qo*«5þda «(15) qja «sj1 t½s? P 1 P 2 5²? ukf? 5²DIMÐ d1 r?oi?²? *«Ë ÎU? Ëd?F? `³? qo*«ê U0 ÊuJ² 5²DIM «ÈbŠ «b ²ÝUÐ W œuf*«wðu² y - y 1 = y 2 - y 1 x 2 - x 1 x - x 1... (1) y - y 2 = y 2 - y 1 x 2 - x 1 x - x 2... (2) Ë

39 Æ(2, 5) Ë (1, 3) m = 5-3 5²DIM UÐ U*«rOI² *«W œuf bł u¼ roi² *«qo 2-1 = 2 w¼ roi² *«W œuf Ê (22) U q(«y - 2x - 1 = 0 y - 3 = 2 (x - 1) Ë A(1,2) j IM? «t Ý˃ Íc «Y K? *«w AD j Ýu??²*«r O??I?²??*«W œu?f?? b?ł ÆBC lkc «nb²m w¼ D Ê XLKŽ « Æ(16) qja «w UL ABC, (23) U C(-3, -5) B(2, 3) YK *«rýd½ w¼ D WDIM «q(«d - 1 2, -1 Í u¼ AD jýu²*«roi² *«qo m = 2 - (-1) = 3 = w¼ AD jýu²*«roi² *«W œuf

40 y - 2 = 1 2 (x -1) 2y - 4 = x -1 2y - x - 3 = 0 (16) qja «(8) V¹ bð roi?² *«W œu?f b? AC lkc «nb²m w¼ E WDIM «X½U «(23) U *«w ÆBE jýu²*«s¹ u;«s ÁUFDI rkž «roi² *«W œuf 5.4 Ê rkž «L 1 roi² *«W œuf œu ¹»uKD*«Æa ÍËU ¹ x u s tfdi

41 Æ(17) Æb ÍËU ¹ y u s tfdi -2 qja «w UL VOðd² «vkž B, A w s¹ u;«ldi¹ L 1 roi² *«(17) qja «øa,b y - 0 = b - 0 øb, A s q UOŁ«bŠ U A(a, 0), B(0, b) 5²DIM UÐ d1 Íc «roi² *«W œuf U W œuf*«x a + y b 0 - a (x - a) y(-a) = bx - ab UNM Ë bx + ay = ab Ê Í vkž qb ½ ab = 1... (1) ÆWÐuKD*«L 1 vkž WL I UÐ roi² *«W œuf w¼ë

42 Æ-7 ÍËU ¹ y u s Ë 5 ÍËU ¹ x w¼ roi² *«W œuf Ê b$ u s tfdi Íc «roi² *«W œuf bł x a + y b = 1 W œuf*«â«b ²ÝUÐ x 5 + y -7 = 1 7x - 5y = 35 7x - 5y - 35 = 0 (24) U q(«æs¹ u;«s 5x + 2y - 10 = 0 roi² *«ULNFDI¹ 5²K «5²FDI «uþ bł wk¹ UL x a + y b = 1 5FDI*«u v 5x + 2y - 10 = 0 x 2 + y 5 W œuf*«u ½ (25) U Ë_«q(«Ædšü«dD «w s¹dog²*«ë dþ w okd*«b(«lc½ -1 5x + 2y = 10 Æ10 vkž W œuf*«w dþ WL IÐ 1 okd*«b(«qf$ -2 = 1... (1) x y 10 = 1 UNM Ë

43 Æ5 ÍËU ¹ y u s WFDI «uþë 2 Æy = 0 WÐUłô«øx ÍËU ¹ x u l roi² *«lþuið WDIM y Ê b$ (1) W œuf*«s u s WFDI «uþ wł«bšô«u vkž qb ½ 5x + 2y - 10 = 0 roi² *«W œuf w y = 0 l{uð 5x + 2(0) - 10 = 0 Æx u s WFDI «w¼ë x = 2 UNM Ë vkž qb M roi² *«W œuf w x = 0 l{uð y u s WFDI «b$ q *UÐË 5(0) + 2y - 20 = 0 Æy u s WFDI «w¼ë y = 5 UNM Ë øs¹ u;«s 5²FDI «uþ b¹b ² qný_«u¼ 5K(«Í w½u «q(«5²¹ëu ² Ë 5²?³łu 5²FD ldi¹ë A(7,11) (9) V¹ bð WDIM UÐ d1 Íc «roi² *«W œuf bł Æs¹ u;«s roi² *«j)«w œuf* W¹œuLF «ub «6.4 WOH?O Ë roi²? *«j)«w œuf* W¹œuL?F «ub «UI²?ý«WOHO? bm³ ««c¼ w Y ³MÝ ÆW¹œuLF «ub «v roi² *«j)«w œuf* W UF «ub «q¹u% W¹œuLF «ub «œu ¹ Ê rkž «L 1 roi² *«j)«w œuf œu ¹»uKD*«ÆL 1 vkž q _«WDI½ s UM «œulf «uþ P -1 (18) qja «w U?L? b,a 5²?FDI «s¹ u?;«s ldi¹ L ÆP qo W¹Ë«w¼ θ -2 ro?i²? *«Ê d?h½

44 ø θ t² œuf ÊuJ² x a + y = 1... (1) b tko W¹Ë«Ë P œulf «uþë b,a 5²FDI «5Ð W öž œu ¹«pMJ1 q¼ Ê (18) qja «s kšô P YK *«w a = cos θ : OFE ø«u* P = sin θ : OFD YK *«w b (18) qja «vkž qb ½ (1) a = cosθ P 1 W œuf*«w b, 1 a vkž qb% 5² öf «s Ë Ë 1 b = sinθ P sž i¹uf² UÐË

45 x cosθ + y sinθ = 1 p p ÊuJ¹ P w W œuf*«w dþ»dcð x cosθ + y sinθ = p x cosθ + y sinθ - p = 0... (2) ÆrOI² *«j K W¹œuLF «ub «w¼ (2) W œuf*«ë 6 ÍËU??¹ q _«WDI½ s t??okž UM «œu??l?f «uþ Íc «ro??i?²?? *«W œu?f?? b?ł (26) U Æ45 w¼ œulf «qo W¹Ë«Ë «bšë ÊuJ¹ (2) W œuf*«w p = 6, θ = 45 sž i¹uf² UÐ x cos 45 + y sin 45-6 = 0 x + y = 0 Í Æ sin 45 = cos 45 = 1 2 Ê d cð q(«ê XLKŽ «t² œuf V² «rł WO U² «ôu(«s q w roi² *«rý «(10) V¹ bð P = 3, θ = 4π 3 = 240 à P = 4, θ = π 6 = 30 P = 7, θ = 7π 6 = 315 œ P = 8, θ = 2π 3 = 120» W¹œuLF «ub «v «roi² *«W œuf* W UF «ub «q¹u% w¼ roi² *«j K W UF «W œuf*«ê rkfð ax + by + c = 0... (1)

46 w¼ t W¹œuLF «ub «Ê Ë x cos θ + y sin θ - p = 0... (2) w s¹œb?f «XЫuŁ ÊS ro?i² *«j)«fh?½ Êö 9 (2) Ë (1) ÊU²? œuf*«x½u? «S cosθ = ax + ± a 2 + b 2 p Ê U ] w p cosθ a = sinθ b a, sinθ = ± a 2 + b 2 ÊuJð ax + by + c = 0 Æc Uý n U ð Uý tðua?² W³ UÝ c by + ± a 2 + b 2 a 2 + b 2 = -p c Ê Í W³ÝUM² ÊuJð 5² œuf*«= k... (3) ÆW²ÐUŁ W³ ½ k YOŠ vkž qb% (3) VÝUM² «s cos θ = ak, sin θ = bk, -p = ck cosθ 2 + sinθ 2 = k 2 a 2 + b 2, -p = ø«u* 1 = k 2 a 2 + b 2 K = 1 ± a 2 + b 2 (3) w k sž i¹uf² UÐË b c ± a 2 + b 2 ± a 2 + b 2 W UF «W œuflk W¹œuLF «ub U «cðë c = 0... (4) ± a 2 + b 2 c'«uý ÊuJð Ê Ëd?{ WEŠö l wk¹ UL p Ë j Ð ubð (4) Æ[(2) W œuf*«w P Uý l W œuf*«wðu² sj1ë okd*«b(«u?ý ÊuJð YO Ð W UF «W œu?f*«œëbš Vðd½ -1 Æ a 2 + b 2 Æ(2) vkž W UF «W œuf*«œëbš r I½ -2 Æsinθ u¼ y q UF Ë cosθ u¼ x q UF ÊuJ¹ czbmžë

47 (27) U ÆW¹œuLF «ub «w 3 x + y = 6 W œuf*«v² «q(«u³ UÝ (6) okd*«b(«êuj¹ YO Ð W œuf*«vðd½ Æa = 3, b = 1, c = -6 UNO Ë 3 x + y - 6 = 0 w¼ roi² LK W¹œuLF «ub U Ê 3 x + y = x y - 3 = 0 x cos 30 + y sin 30-3 = 0 W¹Ë«Ê Ë 3 ÍËU ¹ ro?i²?*«vkž e d*«s UM «œul?f «uþ Ê kšö½ UM¼ s Æ30 ÍËU ð tko (28) U ÆW¹œuLF «ub «w x + y + 8 = 0 W œuf*«v² «q(«u³ UÝ (8) okd*«b(«êuj¹ YO Ð W œuf*«vðd½ a = -1, b = -1, c = -6 UNO Ë -x -y -8 = 0 w¼ roi² LK W¹œuLF «ub «- 47 -

48 -x - y = x y = 0 x cos y sin = 0 θ = 225 Ê Ë P = 4 2 œulf «uþ Ê kšô b?ł rł x - 3y + 10 = 0 (11) V¹ bð ro?i²? *«vkž q _«WD?I½ s UM «œul?f «uþ b?ł ÆœuLF «qo W¹Ë«Æx =4 Æax + by + c = 0 ro?i?²?*«w öð WDI½ w ÍËU???????²??? Y?K???? ˃ w¼ roi² *«Í «u¹ë P(5, -3) roi² *«Í «u¹ë (h, k) x a - y b = 1 5Ð q «u «r?o?i?²?? *«vkž Íœu??L?ŽË P(2, -3) (3) wð«c «.ui² «WK¾Ý WDIM UÐ d1 Íc «roi² *«W œuf bł -1 WDIM UÐ d1 Íc «roi² *«W œuf bł -2 ro?i²? *«b? U?F¹ Íc «roi?²? *«W œu?f b?ł Æx P 3 (-1,4), P 2 (-1,-1), P 1 (3,1) u l doš_«-3 jim? «Ê s¼dð -4 ÆtðbŽU vkž tý s UM «œulf «W œuf błë 5 U «WDIM? UÐ U*«rO?I??²? *«W œu?f?? b?ł ÆB(-4, 2) Ë A(5, 6) 5²DIM «u?b;«e?'«r Ið WDIM «Ác¼ X½U? «P(-3, 4) WDIM UÐ U*«rO?I²? *«W œuf? bł -6 Æ 2 W³ MÐ s¹ u;«5ð tm

49 5OŁ«bŠù«s¹ u?;«s ÊUFDI¹Ë P(4, 5) WDIM UÐ Ê«d1 s¹ck «5LO?I² *«w² œu?f bł -7 5²O U² «5² U(«w 5²FD ÆÊU²³łu Ë «bi*«w ÊU²¹ËU ² ÊU²FDI «Æ Uýù«w ÊU²HK² Ë «bi*«w ÊU²¹ËU ² ÊU²FDI «c¼ qo?? W¹Ë«Ë 10 U? ro?i?²??? vkž q _«WDI½ s UM? «œu?l?f «uþ ÊU? «-8 ÆrOI² LK W¹œuLF «ub «W œuf b 300 œulf «5x + 12y - 52 = 0 ro??i??²? *«v?kž q _«WDI½ s? UM «œu??l?f «u?þ Ê X³Ł -9 Æ4 ÍËU ¹ p s Z²M²?Ý«Ë W¹œuLF? «ub UÐ 16x - 12y + 75 = 0 roi?² *«W œuf? V² «-10 ÆœuLF ««c¼ qo W¹Ë«Ë q _«WDI½ s tokž UM «œulf «uþ Ë x - 2y + 1 = 0 5 LO I ² *«lþu I ð W D I M Ð d 1 Íc «r O I ² *«W œu F b ł -11 YO Ð 3x - 8y + 9 = 0 Æ- 4 3 = tko ÊuJ¹» Æq _«WDIMÐ d1 r O?I²? *«v KŽ 40 W ¹Ë«eÐ U L?N M q q O1 ÊU L O?I²? r Ý «P(3, 4) W DIM «s -12 Æx - y = 2 Æ5LOI² *«s¹c¼ w² œuf bł

50 roi² jš sž WDI½ bfð.5 t?²? ö?f ax + by + c «b?i*«u?ý Ÿu?{u bð Í œu?ð bm³ ««c¼ w Z U?FMÝ UM¹U?³?²*«rÝ w t²?o?l¼_ë ÂuKF? roi?²? v?kž WDI½ s UM «œu?lf «ud?ð dýu?³*«wh?bm w² œu?f? Ë ro?i?²?? jš vkž WDI½ s UM «œu?l?f «uþ b? MÝ U?L? W?OD)«Æ5FÞUI² 5LOI² 5Ð 5²¹Ë«e «ax + by + c «bi*«uý«1.5 Ë ax + by + c = 0 ro?i?²? *«j)«s? W?N?ł w lið (x, y ) WDIM «Ê 5³MÝ ÆU³ UÝ Ë U³łu ax + by + c «bi*«êu vkž UMÐ tm Èdš_«WN'«w dož wð UJ¹b «Èu?² *«w WDI½ P(x, y ) Ê Ë ÂuKF*«rOI² *«u¼ LM Ê d «ÆLM roi² *«vkž WF «Ë WDI?M «w LM ro?i²? *«w ö¹ë y u? Í «u¹ YO? Ð PQ ro?i²? *«rý «P s (19) qja «ÆQ(x, y ) ÊS roi² *«j)«vkž lið Q Ê U0Ë ax, by + c = 0 (19) qja «y = - ax + c... (1) b UNM Ë

51 Vłu*«Ë V U? «5¼U&ô«w y u; U?¹ «u rý PQ Ê (19) qja «s `{«Ë Èd?š_«WN?'«w lið Ë LM roi?² *«j)«s W?Nł w lið P WDIM «Êu vkž UMÐ t W?N?'«w Ë ÆU³ UÝ Ë U³łu y y Êu vkž UMÐ Í tm y > y Ë y < y «bi*«êu vkž UMÐ Í vkž qb ½ (1) W œuf*«â«b ²ÝUÐË y - y = - ax + c - y = - 1 ax + by + c b b LM ro?i?²? *«j)«s W?N?ł w lið (x, y ) WDIM «ÊS? w U?² UÐË ÆU³łu Ë U³ UÝ ax + by + c «bi*«êu vkž UMÐ tm Èdš_«LM ro?i?²??*«j)«51 vkž P(x, y ) WDIM «XF? Ë «t½ u?i «sj1 o³?ý U2 W³?łu ÊuJð ax + by + c «bi*«uý«ês? W³łu ax + by + c Uý ÊuJð W? U(«Ác¼ w Ë LM roi²? *«j)«51 vkž WF «u «P(x, y ) ro?i?²?*«j)«ul?ý vkž W?F? «u ««bi*«uý X½U Ë jim «lol' P(x, y ) jim «lo?l?' W³ U?Ý ax + by + c «bi*«ælm (29) U w¼ (20) qja «w 5³*«LM roi² *«j)«w œuf X½U «5³ 3x + 4y - 12 = 0 U ³ U?Ý t KF& U N¹«Ë U?³łu? 3x + 4y - 12 «bi*«q?f& WO U?² «jim «U?OŁ«bŠ Í Æ(-1, 7) (0, 4) (0, 0) (-3, 4) (2, 1) ÆÁ U ¹ vkž li¹ ULN¹ Ë j)«51 vkž li¹ u c*«jim «Í»

52 (20) qja «q(«b M 3x + 4y - 12 «bi*«w WDI½ q woł«bš«uf½ 3 x x 0-12 = -12 «bi*«wdfð (0, 0) WDIM «-1 2 x x 1-12 = -2 «bi*«wdfð (2, 1) WDIM «-2 3 x (-3) + 4 x 4-12 = -5 «bi*«wdfð (-3, 4) WDIM «-3 3 x x 1-12 = 7 «bi*«wdfð (5, 1) WDIM «-4 3 x x 4-12 = 4 «bi*«wdfð (0, 4) WDIM «-5 3 x (-1) + 4 x 7-12 = 13 «bi*«wdfð (-1, 7) WDIM «-6 (-1, 7), jim? «Ê Ë j)«u?? ¹ vk?ž lið (-3, 4), (2, 1), (0, 0) jim «Ê kšô» ÆtMO1 vkž lið (0, 4), (5, 1)

53 ro I?²? *«j K W ³??M UÐ q ô«w DI?½ U N?O? l Ið w² «W ID?M*«U ý«ê 5 Ð Æc (30) U XÐU «vkž bl²fð ax + by + c = 0 «bi*«w UNOOŁ«bŠ«i¹uF²Ð œb% q _«WDI½ UNO w² «WIDM*«Uý ax + by + c a x 0 + b x 0 + c = c vkž bl²fð Í j K W??³??M UÐ W?³? U? «Ë W??³?łu*«W??IDM*«w? ÊuJð q _«WDI?½ ÊS? w U??² UÐË ÆW³ UÝ Ë W³łu c q(«êu wkž UMÐ roi² *«u¼ U?L? p ÂuKF roi² vkž WDI½ s UM «œulf «uþ 2.5 w¼ W¹œuLF «ub UÐ ML ÂuKF*«rOI² *«W œuf Ê d «x cos θ + y sin θ - p = 0 u¼ ro?i?²? *«vkž Q(x, y ) WDIM «s UM «QD œu?l?f «uþ Ê Ë Æ(21) qja «w 5³

54 (21) qja «œu?lf «œ«b?²? «w ö¹ë ML roi?² *«Í «u¹ Y?O Ð RQ roi?² *«rýd½ p œu ¹ô Æ ø«u* qod² NRQD qja «ÊuJO R w q _«WDI½ s tokž UM «NR = QD = p Ê ÊuJð W¹œuLF «ub UÐ RQ roi² *«W œuf x cosθ + y sinθ - OR = 0 Æ ø«u* x cosθ + y sinθ - p - p = 0 Í t² œuf oi% UN½S roi² *«j)««c¼ vkž lið Q(x, y ) Ê U0Ë x cosθ + y sinθ - p - p = 0 p = x cosθ + y sinθ - p UNM Ë "UM ««bi*«íëu? ¹ ÂuKF roi?² vkž WM?OF WDI½ s UM «œu?lf «uþ Ê Í ÆW¹œuLF «tð ubð roi² *«W œuf w WDIM «woł«bš i¹ufð sž

55 »ukd*«p roi² *«vkž (5, 7) WDIM «s UM «œulf «uþ bł 4x + 3y - 12 = 0 W¹œuLF «ub «v roi² *«j)«w œuf u ½ 4x + 3y - 12 = 0... (1) œulf «uþ vkž qb M (1) ub «w WDIM «woł«bš«uf½ P = 4 x x = 29 5 = (31) U q(«roi² *«vkž (-2, 3) WDIM «s UM «œulf «uþ bł 8x + 15y - 34 = 0 W¹œuLF «ub «v «roi² *«W œuf u ½ 8 x + 15 y = 0 uþ vkž qb? M W¹œu?L?F «u?b UÐ ro?i²? *«W œu?f? w WDIM «woł«b?š u?f½ Í P = - 8 x (-2) + 15 x (32) U q(ukd*«p = 0 œulf «= œbfk WIKD*«WLOI «cšq½ UM½S ULz«œ Vłu œulf «uþ Ê U0Ë

56 P = = 5 17 A(1,1), B(6, -3), C(3, 1) WDIM «týëƒ Íc «YK *«WŠU bł BC = (6-3) 2 + (-3-1) 2 = 5 w¼ BC roi² *«W œuf (22) qja «s y + 3 x - 6 = Æ p s oi% -4x - 3y + 15 = 0 Í P = ÍËU ¹ BC vkž A s œulf «uþ -4 x 1-3 x = 4 2 (33) U q(«= 8 5 (22) qja «- 56 -

57 ÍËU ð WFÐd*««bŠu UÐ ABC YK *«WŠU Ê 1 2 x 5 x 8 5 = 4 (11) V¹ bð Æq _«WDI½Ë (B(-1, 1), A(2, 3) jim «týëƒ Íc «YK *«WŠU bł 5FÞUI² 5LOI² 5Ð 5²¹Ë«e «whbm U² œuf 3.5 UL¼ ULNO² œuf Ê Ë A w UFÞUIð L 2 Ë L 1 5LOI² *«Ê dh½ a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0 w `{u u¼ U?L U?LNMOÐ 5ð ub?;«5²¹ë«e «s U¹ nbm¹ PA Ê dh½ U?L Æ(23) qja «(23) qja «- 57 -

58 sž W¹Ë«e «nbm vkž W?F «Ë P q WDI½ Í ÍbFÐ Ê W¹u?² *«WÝbMN «s rkfð ÆÊU¹ËU ² ULNOFK{ PN 1 - PN 2 Ê Í Æ PAN 1, PAN 2 5 K *«oðudð «b ²ÝUÐ p s¼dð q vkž P(x, y ) WDIM «s Êô U½ UL¼Ë ÊU¹ËU?² s¹œulf «w uþ Ê U0 Êü«Ë 5L?O?I?²? *«s q W? œu?f? l{ë b?fð U?L?NM q uþ œu? ¹«sJ?1 t½s? 5L?O?I?²? *«s ÆW¹œuLF «ub UÐ W?³łu? UL?NM q w c okd*«b?(«uý«êujð Y?O Ð 5L?OI?²? *«W œuf? V²J½ ÊS t?okžë Æ5LO?I² *«s qj W?³łu*«W?N'«w lið o q _«WDI½ Ê s b? e²½ p cðë Æ(23) qja «w `{u u¼ UL W³ UÝ Ë W³łu ÊuJð WFÐ _«blž_««uþ «Uý ÊuJ¹ q _«WDI½ UNO lið w² «W¹Ë«e «U½cš «= PN 1 = + PN 2 + a 1 x + by + c 1 a b2 1 = + a 2 x + b 2 y + c 2 a b2 2 Í «b??³???²??ý«um?mjl??o??? W¹Ë««Ác¼ nb?m vkž W???F?? «Ë WD?I½ W¹ P(x, y ) Ê U0Ë ub UÐ W¹Ë«e «nbm W œuf wkž qb M (x, y ) 5OŁ«bŠôUÐ (x, y) 5OŁ«bŠô«a 1 x + by + c 1 a b2 1 = a x + b y + c a b (1) ÊuJ¹ q _«WDI½ UNO lið ô w² «W¹Ë«e «U½cš «U - PN 1 = + PN 2 - a 1 x + b 1 y + c 1 a b2 1 = + a 2 x + b 2 y + c 2 a b2 2 Í ÊuJ¹ W¹Ë«e «Ác¼ nbm W œuf ÊS w U² UÐË - a 1 x + b 1 y + c 1 a b2 1 = a 2 x + b 2 y + c 2 a b (2)

59 a 1 x + b 1 y + c 1 a b2 1 ÆL 2 Ë L 1 WO U² «W œuf*«w (2) Ë (1) 5² œuf*«wðu² sj1ë = ± a 2 x + b 2 y + c 2 a b (3) 5LOI² 5Ð W¹Ë«e «whbm W œuf w¼ Ác¼Ë 5LOI² *«5Ð 5²¹Ë«e «whbm w² œuf bł 3x - 4y + 7 = 0 Ë 12x - 5y - 8 = 0 Í U³łu ULNM q w okd*«b(«êuj¹ YO Ð 5² œuf*«v²j½ 3x - 4y + 7 = 0 Ë -12x - 5y - 8 = 0 w¼ q _«WDI½ rcð w² «W¹Ë«e «nbm W œuf Ê 3x - 4y x + 5y + 8 = Í 13 (3x - 4y + 7) = 5 (12x + 5y + 8) 99x - 77y + 51 = 0 vkž qb ½ UB²šô«bFÐË 3x - 4y = - wn dšü«nbm*«w œuf U -12x + 5y x + 27y = 0 vkž qb ½ UB²šô«bFÐË Æ5² U(«w UB²šô«WOKLŽ s b Qð (34) U q(«- 59 -

60 5LOI² *«5Ð ub;«w¹ë«e «nbm W œuf bł x + y + 2 = 0 ø«u* x + y + 2 = 0 Ë Æ(1, 3) 7x - y - 8 = 0 WDIM «rcð w² «Ë ub «w 5² œuf*«lc½ Ë -7x + y + 8 = 0 Z²MO 5² œuf*«s qj d ¹_«dD «w (1, 3) WDIM «woł«bš uf½ W?IDM w Í 5?L?O?I?²? *«s qj W?³?łu*«W?ID?M*«w lið Ë = 4 (1, 3) WDIM «Ê Í (35) U q(«æq _«WDI½ rcð w² «W¹Ë«e «w lið WDIM U Ê q _«WDI½ w¼ W¹Ë«e «Ác¼ nbm W œuf Ê x + y + 2 = -7 + y WÐuKD*«W œuf*«vkž qb ½ UB²šô«bFÐË Æp s b Qð 6x = 2y + 1 = 0 (13) V¹ bð 5LOI² *«5Ð W¹Ë«e «nbm W œuf bł 2x - y + 5 = 0 Ë 4x + 2y - 7 = 0 ÆULNOFK{ 5Ð (1, -1) WDIM «rcð w² «Ë

61 (4) wð«c «.ui² «WK¾Ý 5x + 2y = 1 roi² LK W³ M UÐ q _«WDI½ WNł w lið WO U² «jim «Í -1 ø A(3, 9), B(5, -11), C(4, 7), D(-3, 0) w¼ tžö{ ôœuf Íc «YK *«qš«œ lið q _«WDI½ Ê 5Ð -2 y - 6x = 6, 3x + 2y = 1, 5x - 3y + 11 = 0 5LOI² *«vkž 5 UM «s¹œulf «w uþ Ê s¼dð -3 24x + 7y = 70, 4x - 3y = 2 ÆA(3, 4), B(-3, 5), C(2, -7) jim «týëƒ Íc «YK *«WŠU bł -4 jim «týëƒ Íc «YK LK WKš«b «U¹«Ëe «UHBM ôœuf bł -5 A(-1, 2), B(0, 0), C(4, 4) w¼ tžö{ ôœuf Íc «YK LK WKš«b «U¹«Ëe «UHBM lþuið WDI½Ë ôœuf bł -5 4x - 3y - 65 = 0, 7x - 24y + 55 = 0, 3x + 4y - 5 = 0 qb? M U¹d³?ł UL?NO?² œuf? q ½ UM½S s?lo?i²? 5Dš lþu?ið WDI½ œu ¹ô t½ d? cð Æ lþui² «WDI½ 5OŁ«bŠ vkž 5LOI² *«lþuið WDIMÐ d1 Íc «roi² *«W œuf bł -6 2x + 5y - 9 = 0, x - 3y + 1 = 0 Æ 5 ÍËU ¹ q ô«wdi½ sž ÁbFÐ Íc «Ë jim «týëƒ Íc «ABC w BC lkc «vkž A s UM «œulf «uþ bł -7 ÆYK *«WŠU bł rł ÆA(-1, 4), B(1, -4), C(5, 4) roi² *«sž q _«WDI½ bf³ W¹œbF «WLOI «qf& w² «k Æ «bšë 3 WLO bł -8 ÍËU ð y + 5 = k (x - 3) b?ł rł 2 ÍËU? ¹ x u?? s t?fdi? Ë m = 3 4 tko? Íc «ro?i?²? *«W œu?f? b?ł -9 Æx u Ë roi² *««c¼ 5Ð ub;«œu(«w¹ë«ek nbm*«roi² *«W œuf

62 W ö)«.6 W?OÝU?Ý_«hzU?B?)«U?NKL 0 X?Dž WFÐ W?O??Oz U U?? b?šu «Ác¼ w XÝ œ WO U² «UIM «w UNBO Kð sj1 w² «Ë roi² *«j K U?OŁ«b??Šô«ÂUE½Ë wð UJ?¹b «Èu?²??*«Âu?N?H??.b?Ið - YO??Š wð UJ¹b «Èu??²? *«5O U² «5½u½UI «Ë Èu² *«w jim «5OFð WOHO Ë t¹ b UF²*«WOð UJ¹b «Èu² *«w P 2 x 2, y 2, P 1 x 1, y 1 5²DIM «5Ð d bf³ «Êu½U d = x 1 - x y 1 - y 2 2 5?²D?IM? «5?Ð WK? «u «W??????L???????O??????I??????²?????? *«W??????F?DI? «r?o????????????ið Êu?½U??????» ø UDF W³ MÐ Ã U)«s Ë qš«b «s P 2 x 2, y 2, P 1 x 1, y 1-1 x 1 + rx r, y 1 + ry r qš«b «s ro I² «WDI½ x 1 - rx r, y 1 - ry r à U)«s ro I² «WDI½ s¹do?g²*«w v Ë_«Wł b «s W œu?f*«vm M Ê U³Ł - to? Ë roi²? *«j)«qo Ê UM?OÐ m = tan θ tko Ë θ ro?i²? *«qo? W¹Ë«n¹d?Fð -Ë ro?i²?? jš u¼ x, y a Í «uð wþdý b¹b?% - UL Æ u¼ ax + by + c = 0 u¼ tð u Íc «roi?² *«qo b U??L???NMOÐ u???b??;«α W¹Ë«e «q?þ œu?? ¹«o¹d?Þ sž U??L¼b??? U??FðË 5?L??O??I??²????? m tanα = 2 - m Æ m 1 m 2 Æm 1 = m 2 ÊU «Í UL¼öO ÈËU ð «ÊULOI² *«È «u²¹ Æm 1 m 2 = -1 Í 1 ÍËU ¹ ULNOKO»d{ q UŠ ÊU «ÊULOI² *«b UF²¹» w?¼ x 1, y 1 ÆXÐUŁ a YOŠ y = a w¼ x -2 roi² *«j)«w œuf* WHK²<«uB «-3 u Í «u¹ Íc «roi² *«W œuf ÆXÐUŁ b YOŠ x = b w¼ y u Í «u¹ Íc «roi² *«W œuf» W?DI?M U?Ð d1ë m t?ko??????? Íc «r?o??????i???????²???????*«w œu???????f?????? Ã

63 y = mx + b w¼ b w¼ u¼ y y - y 1 = m x 1, y 1 u s tfdi Ë m tko Íc «roi² *«W œuf œ x 2, y 2, x 1, y 1 y - y 2 = y 2 - y 1 x 2 - x 1 x - x 2 5²DIM UÐ d1 Íc «roi² *«W œuf ¼ w?¼ a, b 5²???FD?I «y Ë x s¹ u????;«s l?di¹ Íc? «ro???i????²????*«w œu????f??? Ë Æ x a + y b = 1 p qo W¹Ë«Ë p q _«WDI½ s tokž UM «ulf «uþ ÊU «roi² *«W œuf œu?l?f «uþë ax + by + c Æxcosθ + y sinθ - p = 0 w¼ θ ÍËU ð w¼ ax + by + c = 0 W UF «roi² * W œuf* W¹œuLF «ub «Õ ax + by + c = 0 a 2 + b 2 wh?bm W œuf? UI?²ý«v?KŽ ËöŽ ÂuKF? roi?²? jš W?N??'«w Ë ax + by + c = 0 «b?i*«uý«y Ð - t?o? Ë roi?²? jš sž WDI½ b?fð ÆU³ UÝ Ë U³łu ax + by + c ro??i?²??*«j)«w?n??ł w lið vkž W ukf? WDI½ s UM «5FÞUI² 5LOI² 5Ð 5²¹Ë«e «(x, y ) «bi*«êu vkž UMÐ tm Èdš_«WDIM «u¼ ax + by + c = 0 roi² *«vkž (x, y ) WDIM «s UM «P œulf «uþ» P = ax + by + c a 2 + b 2 5FÞUI²*«5LOI² *«5Ð 5²¹Ë«e «whbm W œuf à a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0 w¼ -4 a 1 x + b 1 y + c 1 a b2 1 = ± a 2 x + b 2 y + c 2 a b

64 WO½U «WOÝ«b «bšu «sž WI³ W;.7 v Ë_«b?Šu «l q UJ²² W?OÞËd<«ŸuD?I «di*««c¼ s W?O U² «b?šu «wðqð Ác¼ w b?²ýë W?OÞËd<«ŸuDI «Ÿ«u½ Ë wþëd?<«ldi «ÂuN?H d?f²?²ý YO?Š tm ldi «T U?J*«lDI «dz«b «W?O U?²? «W?OÞËd?<«ŸuDIK W?O?ÝU??Ý_«hzU?B?)«b?Šu «w W??O½U? «W??ł b «s W??O U?² «W?? U?F «W œu??f*«b?²??ý U?L?? bz«e «ldi? «h UM «Ax 2 + Bxy + Cy 2 + DX + Ey + F = 0 W? d?f? ö?š s W œu?f*«u?nk 9 w² «W?OÞËd?<«ŸuDI «nomb?ð WO?H?O? d?f?²ðë s¹dog² ÆUNO XЫu «vkž l{uð w² «ËdA «wð«c «.ui² «WK¾Ý UÐUł Ë U³¹ b² «UÐUł.8 (1) V¹ bð (24) qja «- 64 -

65 (2) V¹ bð CD = (4 + 2) 2 + (2-2) 2 = 6 (1) AB = (4 + 2) (-2) 2 = 6 BC = -2 - (-2) = 4 AD = (4-4) 2 + (-2-2) 2 = 4 AC = (-2-4) 2 + (-2 - (2)) 2 = = 32 ÆŸö{ Í «u² qja «BC = AD Ê Ë CD = AB Ê kšô q?ja? «Ê Í < D = 90 AC 2 = CD 2 + AD 2 = 52 p? c????? ÆqOD² (25) qja «(2) AB = (2a - 2a) 2 + (6a - 4a) 2 = 2a AC = 2a + 3a - 2a 2 + (5a - 4a) 2 = 2a CB = 2a + 3a 2a 2 + (5a - 6a) 2 = 2a

66 Ê AB = AC = CB ÆŸö{_«ÍËU ² ABC YK *«Ê Í (26) qja «(3) V¹ bð ÊuJO B(x, y) UL¼ dz«b «ddi dšü«dd «woł«bš«ê dh½ -1 Æ AB nb²m 0 Ê_ 3 = x + 2 x = = y - 1 y = 11 2 B(4, 11) Ê Í ø W³ M UÐ qš«b «s P 1 P 2 WFDI «r Ið P 3 (5, 1) WDIM «Ê dh½ -2 ÊuJO

67 29 = -1-9r 1 - r r = 1.5 Æ»uKD*«u¼Ë W?O??L? Æθ = 135 m Ê_ θ = 90 qo*«w¹ë«m = qo*«w¹ë«ê m = Æm = 0 Ê_ θ = 0 qo*«w¹ë«ê -7 - (-3) = = 4 0 m = = 0 (4) V¹ bð roi² *«qo ro?i?²? *«qo?» Æ œb dož roi² *«qo à (5) V¹ bð wk¹ UL W UF «UNð u w `³B² UNM bš«ë dþ w W œuf q œëbš lc½ -x + y - 4 = 0 (2) 7x - 3y - 2 = 0 (1) 2x + 0.y - 9 = 0 (4) 0.x + y + 5 = 0 (3) m = -a u¼ roi² *«j K W UF «ub «qo b w¼ WFÐ _«ULOI² *«qo Ê -(-1) = 1-7 (2) -3 = Æ œb dož WOL = -2 0 (4) 1 = 0 (1) (3) (6) V¹ bð m 2 = -a 2 Æ b 2 m 1 = -a 1 w½u «roi² *«qo b 1 Ë_«rOI² *«qo

68 -a 1 = -a 2 -a 1 = b 1 Æ b 1 b 2 a 2 b 2 Í «u² «dý -1 b UF² «dý -2 -a 1 -a 2 b 1 b 2 = -1 a 1 a 2 = -b 1 b 2 a 1 a 2 + -b 1 b 2 = 0 (7) V¹ bð Æ- 3 u¼ tokž œulf «qo Ê m = -2 ÂuKF*«rOI² *«qo Æ - 3 w¼ WÐuKD*«rOI² *«W œuf 2 = y - 3 3x + 2y = 0 x + 2 Í 5²DIM UÐ U*«rOI² *«qo ÍËU ¹»uKD*«rOI² *«qo -2 m = = w¼»ukd*«roi² *«W œuf = y - (-3) x + 6y + 16 = 0 x - 2 (8) UL¼ U¼UOŁ«bŠS AC lkc «nb²m E Ê U0 x = = -1, y = E(=1, -1, 5) = -1.5 Ê w¼ BE jýu²*«roi² *«W œuf 3 - (-1.5) = y (-1) x x - 3y = 0 w¼ BE W œuf Ê b$ UB²šô«bFÐË V¹ bð

69 (27) qja «Æs¹ u;«s a, a 5²³łu*«5²FDI «ldi¹ L Í t² œuf oi% wn (7, 11) roi² *«Ê dh½ w¼ roi² *«W œuf Ê x a + y a = 1 WDIM UÐ d1 L roi² *«Ê U0Ë 7 a + 11 a = 1 18 a = 1 (9) V¹ bð a = 18 x + y = 18 w¼ L roi² *«W œuf Ê

70 (28) qja «1 2 (10) V¹ bð w¼ W¹œuLF «ub UÐ roi² *«W œuf x cos 30 + y sin 30-4 = 0 3 x + 1 Í 2 y - 4 = 0 3 x + y - 8 = 0 (29) qja «- 70 -

71 w¼ W¹œuLF «ub UÐ roi² *«W œuf» x cos y sin = 0-1 Í 2 x y - 8 = 0 = 2 x - 3 y + 16 = 0 (30) qja «- 1 2 x w¼ W¹œuLF «ub UÐ roi² *«W œuf à x cos y sin = 0 3 y - 3 = 0 x + 3 y + 6 = 0 Í

72 (31) qja «w¼ W¹œuLF «ub UÐ roi² *«W œuf (5) x cos y sin = 0 1 Í 2 x y - 7 = 0 x - y = 0 (32) qja «- 72 -

73 w¼ W¹œuLF «ub UÐ roi² *«j)«w œuf x - 3y = 0 x 10-3y = 0 (11) V¹ bð -x y = 10 P = 10 œulf «uþ UNM Ë m = -1-3 = 1 3 ÂuKF*«rOI² *«qo Æ-3 ÍËU¹ p œulf «qo θ = tan -1-3 w¼ θ œulf «qo W¹Ë«Æ0 s tokž UM «œulf «uþë AB (12) lkc «uþ b$ YK *«WŠU œu ¹ô V¹ bð ÍËU ¹ AB roi² *«vkž 0 AB = (2 + 1) 2 + (3-1) 2 = 13 w¼ AB W œuf y - 3 x - 2 = (-1) = 2 3 3y - 9 = 2x - 4 2x - 3y + 5 = 0 q ô«wdi½ s UM «œulf «uþ P = 2 x 0-3 x = 5 13

74 ÍËU ð WFÐd*««bŠu UÐ AOB YK *«WŠU 1 2 x 13 x 5 13 = 2.5 (33) qja «(1) wð«c «.ui² «WK¾Ý d = (a + c) 2 + (b + d) 2 à 26» 5 (2) Æ(38, -49) w¼ à U)«s ro I² «WDI½ (-2, -9) w¼ qš«b «s ro I² «WDI½ (5) Æ(5, 3) (6) Æ(1, 6), (-5, -4), (9, -2) (7) , 34 3 (8) (3) wð«c «.ui² «WK¾Ý

75 x = 5 (1) ax + by = ah + bk (2) ax + by = a 2 (3) x + 2y - 7 = 0 (4) 9x + 4y - 0 = 0 (5) 2x + 3y - 18 = 0 (6) y - x = 1» x + y = 9 (7) 1 2 x y - 10 = 0 (8) YOŠ θ œulf «qo W¹Ë«15 4 œulf «uþ x y = 0 (10) sinθ = 3 5, cosθ = x + 3y = 29» ty - 3x = 0 (11) y = 4, x = 3 (12) (4) wð«c «.ui² «WK¾Ý A, D (1) 33.5 (3) 2x+y 5 = 2x-5y+12 29, y-x 3 = 2x+y 5, y-x 3 = 2x-5y (4)

76 W? D? I? M? «7x + y - 70 = 0, 2x + 11y - 20 = 0, 9x - 13y - 90 = 0 (5) Æ(10, 0) 2x + y - 5 = 0 (6) ÆWFÐd bšë 24 WŠU *« Ω ŸUHð ô«(7) K = , (8) 3y - x + 2 = 0, 4y - 3x + 6 = 0 (9)

77 Angle Area Axes Axis Bisector Cartesian Plane Centroid Co - ordinate Geometry Cosine Direction Distance Equation Graph Horizontal line Inclined Intercept Locus Median Midpoint Negative Origin Parallel Perpendicular Point U KDB*«œd.9 W¹Ë«WŠU ËU u nbm wð UJ¹b «Èu² *«WDÝu²*«ULOI² *«vi²k WOŁ«bŠ WÝbM¼ W¹Ë«e «ÂU9 VOł ÁU&«W U W œuf w½uoð rý jd wi jš qzu ldi wýbm¼ q YK *«w jýu² roi² nb²m*«wdi½ V UÝ q ô«wdi½ Í «u ÍœuLŽ WDI½

78 Positive Proportion Quadrant Ratio Sign Sine Straight line Tangent Triangle Vertical line Vłu VÝUMð lð W³ ½ Uý W¹Ë«e «VOł roi² jš W¹Ë«e «qþ YK wý jš

79 lł«d*«.10 WOÐdF «lł«d*«djh ««œ Ë_«e'«W?OKOK ² «W?ÝbMN «Ë q UJ² «Ë q{uh?² «ÊËdš Ë u?ž ÊU½bŽ -1 Æ1990 ÊULŽ «œ w½u? «e??'«w?okok??² «W?ÝbMN «Ë q UJ?² «Ë q{u?h?² «ÊËd?š Ë u??ž ÊU½b?Ž -2 Æ1990 ÊULŽ djh ««Ë w½u? «e?'«w?okok?²? «WÝb?MN «W¹u½U? «W?ÝbMN «ÊËd?š Ë ÊU?L??Ž b?l?š«-3 Æ1971 ÊULŽ rokf² «Ë WOÐd² «WO³Mł_«lł«d* 1- Swokowski E., Calculus with Analytic Geometry, Alternate ed., Prindle Weber and Schmidt, Thomas, G.B., Finney, R.L, Calculus and Analytic Geometry, 7th ed., Addison - Wesley Publishing Co., London, Silverman, R.A., Calculus with Analytic Geometry, Prentice - Hall inc., Englewood Cliffs, New Jersey, Kindle, J.H., Theory and Problems of Plane and Solid Analytic Geometry, Schaum P.Bo., New York, Loney, S.L., The Elements of Coordinate Geometry, Macmillan and Co., London,

An Example file... log.txt

An Example file... log.txt # ' ' Start of fie & %$ " 1 - : 5? ;., B - ( * * B - ( * * F I / 0. )- +, * ( ) 8 8 7 /. 6 )- +, 5 5 3 2( 7 7 +, 6 6 9( 3 5( ) 7-0 +, => - +< ( ) )- +, 7 / +, 5 9 (. 6 )- 0 * D>. C )- +, (A :, C 0 )- +,

More information

?¼± Ø Ø ±π ±µ ÊUe«Â ± ØµØ o«u*««d²ýô«wokc. œ«bž Ê YOŠ

?¼± Ø Ø ±π ±µ ÊUe«Â ± ØµØ o«u*««d²ýô«wokc. œ«bž Ê YOŠ Ëœ Í c«ub²ô«oo³d²«ë UOÝUÝ_«Ëb«œUIF½«ÊUJ U¹dUÐ œufý pk*«wfuł UOLOJ«r?¼± Ø Ø ±π ±µ ÊUe«Â ± ØµØ o«u*««d²ýô«wokc dj³*«qo ²K œ«bž Ê YOŠ œëb 5Ð b²*« w WO³¹ bð Ëœ oo³d²«ë UOÝUÝ_«Í c«ub²ô«œufý pk*«wfuł UOLOJ«r

More information

! " # $! % & '! , ) ( + - (. ) ( ) * + / 0 1 2 3 0 / 4 5 / 6 0 ; 8 7 < = 7 > 8 7 8 9 : Œ Š ž P P h ˆ Š ˆ Œ ˆ Š ˆ Ž Ž Ý Ü Ý Ü Ý Ž Ý ê ç è ± ¹ ¼ ¹ ä ± ¹ w ç ¹ è ¼ è Œ ¹ ± ¹ è ¹ è ä ç w ¹ ã ¼ ¹ ä ¹ ¼ ¹ ±

More information

U¼ UŁ Ë UNðU¹b% W*uF «d¼uþ UOKF «UÝ«b «W³KÞ U¼«d¹ UL WOMÞu «ÕU M «WF Uł w

U¼ UŁ Ë UNðU¹b% W*uF «d¼uþ UOKF «UÝ«b «W³KÞ U¼«d¹ UL WOMÞu «ÕU M «WF Uł w U¼ UŁ Ë UNðU¹b% W*uF «d¼uþ UOKF «UÝ«b «W³KÞ U¼«d¹ UL WOMÞu «ÕU M «WF Uł w U Ž b³ž Æœ Æ5D K WOMÞu «ÕU M «WF Uł WOÐd² «WOK UA U²Ý Æ5D K WOMÞu «ÕU M «WF Uł WOÐd² «WOK UA U²Ý U Ž b³ž Æœ WOMÞu «ÕU M «WF Uł

More information

Worksheet A VECTORS 1 G H I D E F A B C

Worksheet A VECTORS 1 G H I D E F A B C Worksheet A G H I D E F A B C The diagram shows three sets of equally-spaced parallel lines. Given that AC = p that AD = q, express the following vectors in terms of p q. a CA b AG c AB d DF e HE f AF

More information

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0 Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y

More information

h : sh +i F J a n W i m +i F D eh, 1 ; 5 i A cl m i n i sh» si N «q a : 1? ek ser P t r \. e a & im a n alaa p ( M Scanned by CamScanner

h : sh +i F J a n W i m +i F D eh, 1 ; 5 i A cl m i n i sh» si N «q a : 1? ek ser P t r \. e a & im a n alaa p ( M Scanned by CamScanner m m i s t r * j i ega>x I Bi 5 n ì r s w «s m I L nk r n A F o n n l 5 o 5 i n l D eh 1 ; 5 i A cl m i n i sh» si N «q a : 1? { D v i H R o s c q \ l o o m ( t 9 8 6) im a n alaa p ( M n h k Em l A ma

More information

1. The unit vector perpendicular to both the lines. Ans:, (2)

1. The unit vector perpendicular to both the lines. Ans:, (2) 1. The unit vector perpendicular to both the lines x 1 y 2 z 1 x 2 y 2 z 3 and 3 1 2 1 2 3 i 7j 7k i 7j 5k 99 5 3 1) 2) i 7j 5k 7i 7j k 3) 4) 5 3 99 i 7j 5k Ans:, (2) 5 3 is Solution: Consider i j k a

More information

T h e C S E T I P r o j e c t

T h e C S E T I P r o j e c t T h e P r o j e c t T H E P R O J E C T T A B L E O F C O N T E N T S A r t i c l e P a g e C o m p r e h e n s i v e A s s es s m e n t o f t h e U F O / E T I P h e n o m e n o n M a y 1 9 9 1 1 E T

More information

176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s

176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s A g la di ou s F. L. 462 E l ec tr on ic D ev el op me nt A i ng er A.W.S. 371 C. A. M. A l ex an de r 236 A d mi ni st ra ti on R. H. (M rs ) A n dr ew s P. V. 326 O p ti ca l Tr an sm is si on A p ps

More information

Udaan School Of Mathematics Class X Chapter 10 Circles Maths

Udaan School Of Mathematics Class X Chapter 10 Circles Maths Exercise 10.1 1. Fill in the blanks (i) The common point of tangent and the circle is called point of contact. (ii) A circle may have two parallel tangents. (iii) A tangent to a circle intersects it in

More information

This document has been prepared by Sunder Kidambi with the blessings of

This document has been prepared by Sunder Kidambi with the blessings of Ö À Ö Ñ Ø Ò Ñ ÒØ Ñ Ý Ò Ñ À Ö Ñ Ò Ú º Ò Ì ÝÊ À Å Ú Ø Å Ê ý Ú ÒØ º ÝÊ Ú Ý Ê Ñ º Å º ² ºÅ ý ý ý ý Ö Ð º Ñ ÒÜ Æ Å Ò Ñ Ú «Ä À ý ý This document has been prepared by Sunder Kidambi with the blessings of Ö º

More information

Parts Manual. EPIC II Critical Care Bed REF 2031

Parts Manual. EPIC II Critical Care Bed REF 2031 EPIC II Critical Care Bed REF 2031 Parts Manual For parts or technical assistance call: USA: 1-800-327-0770 2013/05 B.0 2031-109-006 REV B www.stryker.com Table of Contents English Product Labels... 4

More information

WO½œ _«WF U'«W³KÞ s tmož bmž «c «bo Qð U¹u² 0 w UFH½ù«Ê«eðù«W öž

WO½œ _«WF U'«W³KÞ s tmož bmž «c «bo Qð U¹u² 0 w UFH½ù«Ê«eðù«W öž µ œbf «±π bk *«WO½U ½ù«ÂuKF «ÀU Ðú ÕU M «WF Uł WK WO½œ _«WF U'«W³KÞ s tmož bmž «c «bo Qð U¹u² 0 w UFH½ù«Ê«eðù«W öž The Relationship Between Emotional Stability and Levels of Self- Assertion Among a Sample

More information

STRAIGHT LINES EXERCISE - 3

STRAIGHT LINES EXERCISE - 3 STRAIGHT LINES EXERCISE - 3 Q. D C (3,4) E A(, ) Mid point of A, C is B 3 E, Point D rotation of point C(3, 4) by angle 90 o about E. 3 o 3 3 i4 cis90 i 5i 3 i i 5 i 5 D, point E mid point of B & D. So

More information

". :'=: "t',.4 :; :::-':7'- --,r. "c:"" --; : I :. \ 1 :;,'I ~,:-._._'.:.:1... ~~ \..,i ... ~.. ~--~ ( L ;...3L-. ' f.':... I. -.1;':'.

. :'=: t',.4 :; :::-':7'- --,r. c: --; : I :. \ 1 :;,'I ~,:-._._'.:.:1... ~~ \..,i ... ~.. ~--~ ( L ;...3L-. ' f.':... I. -.1;':'. = 47 \ \ L 3L f \ / \ L \ \ j \ \ 6! \ j \ / w j / \ \ 4 / N L5 Dm94 O6zq 9 qmn j!!! j 3DLLE N f 3LLE Of ADL!N RALROAD ORAL OR AL AOAON N 5 5 D D 9 94 4 E ROL 2LL RLLAY RL AY 3 ER OLLL 832 876 8 76 L A

More information

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o u l d a l w a y s b e t a k e n, i n c l u d f o l

More information

O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE

O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE CCE RF CCE RR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl 50 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALE 50 00 G È.G È.G È.. Æ fioê,» ^È% / HØ È 0 S. S. L. C. EXAMINATION,

More information

UNIQUE FJORDS AND THE ROYAL CAPITALS UNIQUE FJORDS & THE NORTH CAPE & UNIQUE NORTHERN CAPITALS

UNIQUE FJORDS AND THE ROYAL CAPITALS UNIQUE FJORDS & THE NORTH CAPE & UNIQUE NORTHERN CAPITALS Q J j,. Y j, q.. Q J & j,. & x x. Q x q. ø. 2019 :. q - j Q J & 11 Y j,.. j,, q j q. : 10 x. 3 x - 1..,,. 1-10 ( ). / 2-10. : 02-06.19-12.06.19 23.06.19-03.07.19 30.06.19-10.07.19 07.07.19-17.07.19 14.07.19-24.07.19

More information

0# E % D 0 D - C AB

0# E % D 0 D - C AB 5-70,- 393 %& 44 03& / / %0& / / 405 4 90//7-90/8/3 ) /7 0% 0 - @AB 5? 07 5 >0< 98 % =< < ; 98 07 &? % B % - G %0A 0@ % F0 % 08 403 08 M3 @ K0 J? F0 4< - G @ I 0 QR 4 @ 8 >5 5 % 08 OF0 80P 0O 0N 0@ 80SP

More information

vóm Ú F Ó «Ì UO{U¹ Ó u Ú Ó½ W Ò OIO³ ÚDÓð q Ó L ÓŽ Ô «Ó ËÓ Ë ÏWDA½Ó ÏWOLOK Ú FÓð Ï UÓ U Ó O Ý ÏWO d Ú F Ó Ï UÐ U ÓI Ô

vóm Ú F Ó «Ì UO{U¹ Ó u Ú Ó½ W Ò OIO³ ÚDÓð q Ó L ÓŽ Ô «Ó ËÓ Ë ÏWDA½Ó ÏWOLOK Ú FÓð Ï UÓ U Ó O Ý ÏWO d Ú F Ó Ï UÐ U ÓI Ô äés``«vénjuôdg p oá`naén nk vóm Ú F Ó «Ì UO{U¹ Ó u Ú Ó½ W Ò OIO³ ÚDÓð q Ó L ÓŽ Ô «Ó ËÓ Ë ÏWDA½Ó ÏWOLOK Ú FÓð Ï UÓ U Ó O Ý ÏWO d Ú F Ó Ï UÐ U ÓI Ô pa qz«ë dðuł U½UO ÍuÐd² «d¹ud² «Ë Y ³K ÊU ]DI «e d ÊU ]DI

More information

P a g e 5 1 of R e p o r t P B 4 / 0 9

P a g e 5 1 of R e p o r t P B 4 / 0 9 P a g e 5 1 of R e p o r t P B 4 / 0 9 J A R T a l s o c o n c l u d e d t h a t a l t h o u g h t h e i n t e n t o f N e l s o n s r e h a b i l i t a t i o n p l a n i s t o e n h a n c e c o n n e

More information

Èb œuf «o( WLŽ«b «UÝ UL*«5OÝUÝ_«lÝU² «Ë s U «5HB «co öð

Èb œuf «o( WLŽ«b «UÝ UL*«5OÝUÝ_«lÝU² «Ë s U «5HB «co öð Èb œuf «o( WLŽ«b «UÝ UL*«5OÝUÝ_«lÝU² «Ë s U «5HB «co öð»u¹ œ«už nýu¹ Æœ Æ5D K WŠu²H*«bI «WF Uł WOLOKF² «fkðu½ WIDM d¹b bžu U²Ý µµ »U¹ œ«už nýu¹ Æœ 5OÝUÝ_«lÝU² «Ë s U «5HB «co öð Èb œuf «o( WLŽ«b «UÝ UL*«h

More information

Solutionbank M1 Edexcel AS and A Level Modular Mathematics

Solutionbank M1 Edexcel AS and A Level Modular Mathematics file://c:\users\buba\kaz\ouba\m_6_a_.html Page of Exercise A, Question A bird flies 5 km due north and then 7 km due east. How far is the bird from its original position, and in what direction? d = \ 5

More information

(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2

(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2 CIRCLE [STRAIGHT OBJECTIVE TYPE] Q. The line x y + = 0 is tangent to the circle at the point (, 5) and the centre of the circles lies on x y = 4. The radius of the circle is (A) 3 5 (B) 5 3 (C) 5 (D) 5

More information

Vectors. Teaching Learning Point. Ç, where OP. l m n

Vectors. Teaching Learning Point. Ç, where OP. l m n Vectors 9 Teaching Learning Point l A quantity that has magnitude as well as direction is called is called a vector. l A directed line segment represents a vector and is denoted y AB Å or a Æ. l Position

More information

SOLVED SUBJECTIVE EXAMPLES

SOLVED SUBJECTIVE EXAMPLES Example 1 : SOLVED SUBJECTIVE EXAMPLES Find the locus of the points of intersection of the tangents to the circle x = r cos, y = r sin at points whose parametric angles differ by /3. All such points P

More information

P a g e 3 6 of R e p o r t P B 4 / 0 9

P a g e 3 6 of R e p o r t P B 4 / 0 9 P a g e 3 6 of R e p o r t P B 4 / 0 9 p r o t e c t h um a n h e a l t h a n d p r o p e r t y fr om t h e d a n g e rs i n h e r e n t i n m i n i n g o p e r a t i o n s s u c h a s a q u a r r y. J

More information

A L A BA M A L A W R E V IE W

A L A BA M A L A W R E V IE W A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N

More information

MATHEMATICS PAPER IB COORDINATE GEOMETRY(2D &3D) AND CALCULUS. Note: This question paper consists of three sections A,B and C.

MATHEMATICS PAPER IB COORDINATE GEOMETRY(2D &3D) AND CALCULUS. Note: This question paper consists of three sections A,B and C. MATHEMATICS PAPER IB COORDINATE GEOMETRY(D &3D) AND CALCULUS. TIME : 3hrs Ma. Marks.75 Note: This question paper consists of three sections A,B and C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. 0X =0.

More information

CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY CO-ORDINATE GEOMETRY 1 To change from Cartesian coordinates to polar coordinates, for X write r cos θ and for y write r sin θ. 2 To change from polar coordinates to cartesian coordinates, for r 2 write

More information

CAT. NO /irtl,417~ S- ~ I ';, A RIDER PUBLICATION BY H. A. MIDDLETON

CAT. NO /irtl,417~ S- ~ I ';, A RIDER PUBLICATION BY H. A. MIDDLETON CAT. NO. 139-3 THIRD SUPPLEMENT I /irtl,417~ S- ~ I ';,... 0 f? BY H. A. MIDDLETON.. A RIDER PUBLICATION B36 B65 B152 B309 B319 B329 B719 D63 D77 D152 DA90 DAC32 DAF96 DC70 DC80 DCC90 DD6 DD7 DF62 DF91

More information

Day 66 Bellringer. 1. Construct a perpendicular bisector to the given lines. Page 1

Day 66 Bellringer. 1. Construct a perpendicular bisector to the given lines. Page 1 Day 66 Bellringer 1. Construct a perpendicular bisector to the given lines. a) b) HighSchoolMathTeachers@2018 Page 1 Day 66 Bellringer c) d) HighSchoolMathTeachers@2018 Page 2 Day 66 Bellringer 2. Identify

More information

Executive Committee and Officers ( )

Executive Committee and Officers ( ) Gifted and Talented International V o l u m e 2 4, N u m b e r 2, D e c e m b e r, 2 0 0 9. G i f t e d a n d T a l e n t e d I n t e r n a t i o n a2 l 4 ( 2), D e c e m b e r, 2 0 0 9. 1 T h e W o r

More information

O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE

O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE CCE PF CCE PR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl 50 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 50 00 G È.G È.G È.. Æ fioê,» ^È% / HØ È 0 S. S. L. C. EXAMINATION,

More information

I118 Graphs and Automata

I118 Graphs and Automata I8 Graphs and Automata Takako Nemoto http://www.jaist.ac.jp/ t-nemoto/teaching/203--.html April 23 0. Û. Û ÒÈÓ 2. Ø ÈÌ (a) ÏÛ Í (b) Ø Ó Ë (c) ÒÑ ÈÌ (d) ÒÌ (e) É Ö ÈÌ 3. ÈÌ (a) Î ÎÖ Í (b) ÒÌ . Û Ñ ÐÒ f

More information

ADDITIONAL MATHEMATICS

ADDITIONAL MATHEMATICS 005-CE A MATH HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 005 ADDITIONAL MATHEMATICS :00 pm 5:0 pm (½ hours) This paper must be answered in English 1. Answer ALL questions in Section A and any FOUR

More information

General Neoclassical Closure Theory: Diagonalizing the Drift Kinetic Operator

General Neoclassical Closure Theory: Diagonalizing the Drift Kinetic Operator General Neoclassical Closure Theory: Diagonalizing the Drift Kinetic Operator E. D. Held eheld@cc.usu.edu Utah State University General Neoclassical Closure Theory:Diagonalizing the Drift Kinetic Operator

More information

Mathematics Class X Past Year Paper Time: 2½ hour Total Marks: 80

Mathematics Class X Past Year Paper Time: 2½ hour Total Marks: 80 Pas Year Paper Mathematics Class X Past Year Paper - 013 Time: ½ hour Total Marks: 80 Solution SECTION A (40 marks) Sol. 1 (a) A + X B + C 6 3 4 0 X 0 4 0 0 6 6 4 4 0 X 0 8 0 0 6 4 X 0 8 4 6 X 8 0 4 10

More information

CONCURRENT LINES- PROPERTIES RELATED TO A TRIANGLE THEOREM The medians of a triangle are concurrent. Proof: Let A(x 1, y 1 ), B(x, y ), C(x 3, y 3 ) be the vertices of the triangle A(x 1, y 1 ) F E B(x,

More information

Scandinavia SUMMER / GROUPS. & Beyond 2014

Scandinavia SUMMER / GROUPS. & Beyond 2014 / & 2014 25 f f Fx f 211 9 Öæ Höf æ å f 807 ø 19 øø ä 2111 1 Z F ø 1328 H f fö F H å fö ö 149 H 1 ö f Hø ø Hf 1191 2089 ä ø å F ä 1907 ä 1599 H 1796 F ø ä Jä J ( ) ø F ø 19 ö ø 15 á Å f 2286 æ f ó ä H

More information

$%! & (, -3 / 0 4, 5 6/ 6 +7, 6 8 9/ 5 :/ 5 A BDC EF G H I EJ KL N G H I. ] ^ _ ` _ ^ a b=c o e f p a q i h f i a j k e i l _ ^ m=c n ^

$%! & (, -3 / 0 4, 5 6/ 6 +7, 6 8 9/ 5 :/ 5 A BDC EF G H I EJ KL N G H I. ] ^ _ ` _ ^ a b=c o e f p a q i h f i a j k e i l _ ^ m=c n ^ ! #" $%! & ' ( ) ) (, -. / ( 0 1#2 ' ( ) ) (, -3 / 0 4, 5 6/ 6 7, 6 8 9/ 5 :/ 5 ;=? @ A BDC EF G H I EJ KL M @C N G H I OPQ ;=R F L EI E G H A S T U S V@C N G H IDW G Q G XYU Z A [ H R C \ G ] ^ _ `

More information

Trans Web Educational Services Pvt. Ltd B 147,1st Floor, Sec-6, NOIDA, UP

Trans Web Educational Services Pvt. Ltd B 147,1st Floor, Sec-6, NOIDA, UP Solved Examples Example 1: Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4, x + 2y = 5. Method 1. Consider the equation (x + y 6) (2x + y 4) + λ 1

More information

JEE-ADVANCED MATHEMATICS. Paper-1. SECTION 1: (One or More Options Correct Type)

JEE-ADVANCED MATHEMATICS. Paper-1. SECTION 1: (One or More Options Correct Type) JEE-ADVANCED MATHEMATICS Paper- SECTION : (One or More Options Correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR

More information

= 0 1 (3 4 ) 1 (4 4) + 1 (4 3) = = + 1 = 0 = 1 = ± 1 ]

= 0 1 (3 4 ) 1 (4 4) + 1 (4 3) = = + 1 = 0 = 1 = ± 1 ] STRAIGHT LINE [STRAIGHT OBJECTIVE TYPE] Q. If the lines x + y + = 0 ; x + y + = 0 and x + y + = 0, where + =, are concurrent then (A) =, = (B) =, = ± (C) =, = ± (D*) = ±, = [Sol. Lines are x + y + = 0

More information

2. A die is rolled 3 times, the probability of getting a number larger than the previous number each time is

2. A die is rolled 3 times, the probability of getting a number larger than the previous number each time is . If P(A) = x, P = 2x, P(A B) = 2, P ( A B) = 2 3, then the value of x is (A) 5 8 5 36 6 36 36 2. A die is rolled 3 times, the probability of getting a number larger than the previous number each time

More information

Planning for Reactive Behaviors in Hide and Seek

Planning for Reactive Behaviors in Hide and Seek University of Pennsylvania ScholarlyCommons Center for Human Modeling and Simulation Department of Computer & Information Science May 1995 Planning for Reactive Behaviors in Hide and Seek Michael B. Moore

More information

-Z ONGRE::IONAL ACTION ON FY 1987 SUPPLEMENTAL 1/1

-Z ONGRE::IONAL ACTION ON FY 1987 SUPPLEMENTAL 1/1 -Z-433 6 --OGRE::OA ATO O FY 987 SUPPEMETA / APPR)PRATO RfQUEST PAY AD PROGRAM(U) DE ARTMET OF DEES AS O' D 9J8,:A:SF ED DEFS! WA-H ODM U 7 / A 25 MRGOPf RESOUTO TEST HART / / AD-A 83 96 (~Go w - %A uj

More information

s «w nmf « h K*« UM «s WMOŽ vkž WOŽUL²ł«WÝ«œ ö¼ roký bl włu½ Æœ

s «w nmf « h K*« UM «s WMOŽ vkž WOŽUL²ł«WÝ«œ ö¼ roký bl włu½ Æœ UM «s WMOŽ vkž WOŽUL²ł«WÝ«œ s «w nmf «s «w nmf «UM «s WMOŽ vkž WOŽUL²ł«WÝ«œ ö¼ roký bl włu½ Æœ h K*««u?Ý s? «w nmf «WFO³Þ vkž df² «v WÝ«b «Ác¼ bnð Ë»U?³?Ý_«p? c? Ë t?þu/ Ë Á u Ë Á UA²½«Èb YOŠ s h?zu?b?šë

More information

wh UF «U c «WOLM s U «nb «v W{Ëd «s

wh UF «U c «WOLM s U «nb «v W{Ëd «s wh UF «U c «WOLM s U «nb «v W{Ëd «s oo³d²k WLLB WDA½ Ë WMJ2 UO Oð«d²Ý wðëœ s¹uł no Qð ÊUŽd vn WLłdð ÍuÐd² «d¹ud² «Ë Y ³K ÊU ]DI «e d ÊU ]DI «s ;«b³ž W ÝR 2007 5D K tk «Â« s U «nb «v W{Ëd «s whþuf «U c

More information

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola)

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola) QUESTION BANK ON CONIC SECTION (Parabola, Ellipse & Hyperbola) Question bank on Parabola, Ellipse & Hyperbola Select the correct alternative : (Only one is correct) Q. Two mutually perpendicular tangents

More information

IB ANS. -30 = -10 k k = 3

IB ANS. -30 = -10 k k = 3 IB ANS.. Find the value of k, if the straight lines 6 0y + 3 0 and k 5y + 8 0 are parallel. Sol. Given lines are 6 0y + 3 0 and k 5y + 8 0 a b lines are parallel a b -30-0 k k 3. Find the condition for

More information

Downloaded from pdmag.info at 18: on Friday March 22nd 2019

Downloaded from pdmag.info at 18: on Friday March 22nd 2019 2 (- ( ( )*+, )%.!"# $% &" :( ( 2 * -% 345 678-397.) /0 &" &2. ) ( B(

More information

2014 CANADIAN SURF LIFESAVING CHAMPIONSHIPS PROGRAM

2014 CANADIAN SURF LIFESAVING CHAMPIONSHIPS PROGRAM 2014 CANAIAN URF LIFEAVING CHAMPIONHIP PROGRAM Lv Lv vy b v, w,, w Lv y w k Lv z by I Oy C (IOC) Cw G F T IOC z I L v F (IL) w v v IL N Mb Oz v b v T Lv y v by v C I Lv W C z I L v F Cw Lv C z Ry L v y

More information

CHAPTER 10 VECTORS POINTS TO REMEMBER

CHAPTER 10 VECTORS POINTS TO REMEMBER For more important questions visit : www4onocom CHAPTER 10 VECTORS POINTS TO REMEMBER A quantity that has magnitude as well as direction is called a vector It is denoted by a directed line segment Two

More information

I N A C O M P L E X W O R L D

I N A C O M P L E X W O R L D IS L A M I C E C O N O M I C S I N A C O M P L E X W O R L D E x p l o r a t i o n s i n A g-b eanste d S i m u l a t i o n S a m i A l-s u w a i l e m 1 4 2 9 H 2 0 0 8 I s l a m i c D e v e l o p m e

More information

On the curvatures of spacelike circular surfaces

On the curvatures of spacelike circular surfaces Kuwait J. Sci. 43 (3) pp. 50-58, 2016 Rashad A. Abdel-Baky 1,2, Yasin Ünlütürk 3,* 1 Present address: Dept. of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, P.O. Box 126300, Jeddah

More information

,,,,..,,., {. (, ),, {,.,.,..,,.,.,,....... {.. : N {, Z {, Q {, Q p { p{ {. 3, R {, C {. : ord p {. 8, (k) {.42,!() { {. 24, () { {. 24, () { {. 25,., () { {. 26,. 9, () { {. 27,. 23, '() { ( ) {. 28,

More information

1. Matrices and Determinants

1. Matrices and Determinants Important Questions 1. Matrices and Determinants Ex.1.1 (2) x 3x y Find the values of x, y, z if 2x + z 3y w = 0 7 3 2a Ex 1.1 (3) 2x 3x y If 2x + z 3y w = 3 2 find x, y, z, w 4 7 Ex 1.1 (13) 3 7 3 2 Find

More information

Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths

Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is

More information

Department of Mathematics

Department of Mathematics Department of Mathematics TIME: 3 Hours Setter: DS DATE: 03 August 2015 GRADE 12 PRELIM EXAMINATION MATHEMATICS: PAPER II Total marks: 150 Moderator: AM Name of student: PLEASE READ THE FOLLOWING INSTRUCTIONS

More information

1 / 24

1 / 24 CBSE-XII-017 EXAMINATION CBSE-X-01 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instuctions : 1. All questions are compulsory.. The question paper consists of 34 questions

More information

SYSTEM OF CIRCLES OBJECTIVES (a) Touch each other internally (b) Touch each other externally

SYSTEM OF CIRCLES OBJECTIVES (a) Touch each other internally (b) Touch each other externally SYSTEM OF CIRCLES OBJECTIVES. A circle passes through (0, 0) and (, 0) and touches the circle x + y = 9, then the centre of circle is (a) (c) 3,, (b) (d) 3,, ±. The equation of the circle having its centre

More information

1.1 Exercises, Sample Solutions

1.1 Exercises, Sample Solutions DM, Chapter, Sample Solutions. Exercises, Sample Solutions 5. Equal vectors have the same magnitude and direction. a) Opposite sides of a parallelogram are parallel and equal in length. AD BC, DC AB b)

More information

QUESTION BANK ON STRAIGHT LINE AND CIRCLE

QUESTION BANK ON STRAIGHT LINE AND CIRCLE QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,

More information

THE TRANSLATION PLANES OF ORDER 49 AND THEIR AUTOMORPHISM GROUPS

THE TRANSLATION PLANES OF ORDER 49 AND THEIR AUTOMORPHISM GROUPS MATHEMATICS OF COMPUTATION Volume 67, Number 223, July 1998, Pages 1207 1224 S 0025-5718(98)00961-2 THE TRANSLATION PLANES OF ORDER 49 AND THEIR AUTOMORPHISM GROUPS C. CHARNES AND U. DEMPWOLFF Abstract.

More information

T T V e g em D e j ) a S D } a o "m ek j g ed b m "d mq m [ d, )

T T V e g em D e j ) a S D } a o m ek j g ed b m d mq m [ d, ) . ) 6 3 ; 6 ;, G E E W T S W X D ^ L J R Y [ _ ` E ) '" " " -, 7 4-4 4-4 ; ; 7 4 4 4 4 4 ;= : " B C CA BA " ) 3D H E V U T T V e g em D e j ) a S D } a o "m ek j g ed b m "d mq m [ d, ) W X 6 G.. 6 [ X

More information

" #$ P UTS W U X [ZY \ Z _ `a \ dfe ih j mlk n p q sr t u s q e ps s t x q s y i_z { U U z W } y ~ y x t i e l US T { d ƒ ƒ ƒ j s q e uˆ ps i ˆ p q y

 #$ P UTS W U X [ZY \ Z _ `a \ dfe ih j mlk n p q sr t u s q e ps s t x q s y i_z { U U z W } y ~ y x t i e l US T { d ƒ ƒ ƒ j s q e uˆ ps i ˆ p q y " #$ +. 0. + 4 6 4 : + 4 ; 6 4 < = =@ = = =@ = =@ " #$ P UTS W U X [ZY \ Z _ `a \ dfe ih j mlk n p q sr t u s q e ps s t x q s y i_z { U U z W } y ~ y x t i e l US T { d ƒ ƒ ƒ j s q e uˆ ps i ˆ p q y h

More information

The Plan for Excellence.

The Plan for Excellence. f D T P f Ex www R-Im C: T P f Ex TABLE OF CONTENTS INTRODUCTION 3-6 LETTER FROM THE SUPERINTENDENT 3 ROLE OF DISTRICT LEADERSHIP 4 DEVELOPMENT OF RE-IMAGINE CADDO 5 QUICK REFERENCE GUIDE 6 RE-IMAGINE

More information

VECTORS ADDITIONS OF VECTORS

VECTORS ADDITIONS OF VECTORS VECTORS 1. ADDITION OF VECTORS. SCALAR PRODUCT OF VECTORS 3. VECTOR PRODUCT OF VECTORS 4. SCALAR TRIPLE PRODUCT 5. VECTOR TRIPLE PRODUCT 6. PRODUCT OF FOUR VECTORS ADDITIONS OF VECTORS Definitions and

More information

and ALTO SOLO C H, Rnnciman. Woman's Mistake f*" M>rqut4te\ ol ting about wilh a crutch 00 ac- Gr,,,,d Ri ''*' d5 L o,

and ALTO SOLO C H, Rnnciman. Woman's Mistake f* M>rqut4te\ ol ting about wilh a crutch 00 ac- Gr,,,,d Ri ''*' d5 L o, BU AK A k A AD DA AB U XXX HA HUDAY U 3 92 3 > k z j - - Y Bk 73( 3 - Q H 2 H 9 k B 3 Bk 29 K 7 k 2 B k k k k Y Y D k Y A Uk x 22 B B x k - B B B 22 A B A 27 -- Ak A - - j B B D B Q A- U A-AK B k AD A

More information

( ) 2 + ( 2 x ) 12 = 0, and explain why there is only one

( ) 2 + ( 2 x ) 12 = 0, and explain why there is only one IB Math SL Practice Problems - Algebra Alei - Desert Academy 0- SL Practice Problems Algebra Name: Date: Block: Paper No Calculator. Consider the arithmetic sequence, 5, 8,,. (a) Find u0. (b) Find the

More information

a b = a a a and that has been used here. ( )

a b = a a a and that has been used here. ( ) Review Eercise ( i j+ k) ( i+ j k) i j k = = i j+ k (( ) ( ) ) (( ) ( ) ) (( ) ( ) ) = i j+ k = ( ) i ( ( )) j+ ( ) k = j k Hence ( ) ( i j+ k) ( i+ j k) = ( ) + ( ) = 8 = Formulae for finding the vector

More information

Sm1ther AGENDA 1. CALL TO ORDER 3. ADJOURNMENT. Town of o

Sm1ther AGENDA 1. CALL TO ORDER 3. ADJOURNMENT. Town of o Sm1ther Twn COMMTTEE OF THE WHOLE COUNCL CHAMBERS, 1027 ALDOUS STREET TUESDAY, MARCH 10, 2009, AT 6:15 P.M. AGENDA 1. CALL TO ORDER 2 DELE ONS Bard Ventures Ltd. Mr. Rihard Bek, Gelgist, r Bard Ventures

More information

Higher Mathematics Skills Checklist

Higher Mathematics Skills Checklist Higher Mathematics Skills Checklist 1.1 The Straight Line (APP) I know how to find the distance between 2 points using the Distance Formula or Pythagoras I know how to find gradient from 2 points, angle

More information

STANDARDISED MOUNTINGS

STANDARDISED MOUNTINGS STANDARDISED MOUNTINGS for series 449 cylinders conforming to ISO 21287 standard Series 434 MOUNTINGS CONFORMING TO ISO 21287 - ISO 15552 - AFNOR NF ISO 15552 - DIN ISO 15552 STANDARDS applications Low

More information

Trade Patterns, Production networks, and Trade and employment in the Asia-US region

Trade Patterns, Production networks, and Trade and employment in the Asia-US region Trade Patterns, Production networks, and Trade and employment in the Asia-U region atoshi Inomata Institute of Developing Economies ETRO Development of cross-national production linkages, 1985-2005 1985

More information

Mathematics Revision Guides Vectors Page 1 of 19 Author: Mark Kudlowski M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier VECTORS

Mathematics Revision Guides Vectors Page 1 of 19 Author: Mark Kudlowski M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier VECTORS Mathematics Revision Guides Vectors Page of 9 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier VECTORS Version:.4 Date: 05-0-05 Mathematics Revision Guides Vectors Page of 9 VECTORS

More information

={ V ± v I {

={ V ± v I { v n ± EÚ + M, Æ b x I j,

More information

opposite hypotenuse adjacent hypotenuse opposite adjacent adjacent opposite hypotenuse hypotenuse opposite

opposite hypotenuse adjacent hypotenuse opposite adjacent adjacent opposite hypotenuse hypotenuse opposite 5 TRtGOhiOAMTRiC WNCTIONS D O E T F F R F l I F U A R N G TO N I l O R C G T N I T Triangle ABC bas a right angle (9Oo) at C and sides of length u, b, c. The trigonometric functions of angle A are defined

More information

LA PRISE DE CALAIS. çoys, çoys, har - dis. çoys, dis. tons, mantz, tons, Gas. c est. à ce. C est à ce. coup, c est à ce

LA PRISE DE CALAIS. çoys, çoys, har - dis. çoys, dis. tons, mantz, tons, Gas. c est. à ce. C est à ce. coup, c est à ce > ƒ? @ Z [ \ _ ' µ `. l 1 2 3 z Æ Ñ 6 = Ð l sl (~131 1606) rn % & +, l r s s, r 7 nr ss r r s s s, r s, r! " # $ s s ( ) r * s, / 0 s, r 4 r r 9;: < 10 r mnz, rz, r ns, 1 s ; j;k ns, q r s { } ~ l r mnz,

More information

B.Sc. MATHEMATICS - II YEAR

B.Sc. MATHEMATICS - II YEAR MANONMANIAM SUNDARANAR UNIVERSITY DIRECTORATE OF DISTANCE & CONTINUING EDUCATION TIRUNELVELI 671, TAMIL NADU B.Sc. MATHEMATICS - II YEAR DJMA - ANALYTICAL GEOMETRY D AND VECTOR CALCULUS (From the academic

More information

Starting with the base and moving counterclockwise, the measured side lengths are 5.5 cm, 2.4 cm, 2.9 cm, 2.5 cm, 1.3 cm, and 2.7 cm.

Starting with the base and moving counterclockwise, the measured side lengths are 5.5 cm, 2.4 cm, 2.9 cm, 2.5 cm, 1.3 cm, and 2.7 cm. Chapter 6 Geometric Vectors Chapter 6 Prerequisite Skills Chapter 6 Prerequisite Skills Question 1 Page 302 Starting with the base and moving counterclockwise, the measured side lengths are 5.5 cm, 2.4

More information

Examination paper for TFY4240 Electromagnetic theory

Examination paper for TFY4240 Electromagnetic theory Department of Physics Examination paper for TFY4240 Electromagnetic theory Academic contact during examination: Associate Professor John Ove Fjærestad Phone: 97 94 00 36 Examination date: 16 December 2015

More information

! -., THIS PAGE DECLASSIFIED IAW EQ t Fr ra _ ce, _., I B T 1CC33ti3HI QI L '14 D? 0. l d! .; ' D. o.. r l y. - - PR Pi B nt 8, HZ5 0 QL

! -., THIS PAGE DECLASSIFIED IAW EQ t Fr ra _ ce, _., I B T 1CC33ti3HI QI L '14 D? 0. l d! .; ' D. o.. r l y. - - PR Pi B nt 8, HZ5 0 QL H PAGE DECAFED AW E0 2958 UAF HORCA UD & D m \ Z c PREMNAR D FGHER BOMBER ARC o v N C o m p R C DECEMBER 956 PREPARED B HE UAF HORCA DVO N HRO UGH HE COOPERAON O F HE HORCA DVON HEADQUARER UAREUR DEPARMEN

More information

(b) g(x) = 4 + 6(x 3) (x 3) 2 (= x x 2 ) M1A1 Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3).

(b) g(x) = 4 + 6(x 3) (x 3) 2 (= x x 2 ) M1A1 Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3). Paper. Answers. (a) METHOD f (x) q x f () q 6 q 6 f() p + 8 9 5 p METHOD f(x) (x ) + 5 x + 6x q 6, p (b) g(x) + 6(x ) (x ) ( + x x ) Note: Accept any alternative form that is correct. Award A for a substitution

More information

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent Mathematics. The sides AB, BC and CA of ABC have, 4 and 5 interior points respectively on them as shown in the figure. The number of triangles that can be formed using these interior points is () 80 ()

More information

( )( ) PR PQ = QR. Mathematics Class X TOPPER SAMPLE PAPER-1 SOLUTIONS. HCF x LCM = Product of the 2 numbers 126 x LCM = 252 x 378

( )( ) PR PQ = QR. Mathematics Class X TOPPER SAMPLE PAPER-1 SOLUTIONS. HCF x LCM = Product of the 2 numbers 126 x LCM = 252 x 378 Mathematics Class X TOPPER SAMPLE PAPER- SOLUTIONS Ans HCF x LCM Product of the numbers 6 x LCM 5 x 378 LCM 756 ( Mark) Ans The zeroes are, 4 p( x) x + x 4 x 3x 4 ( Mark) Ans3 For intersecting lines: a

More information

Distance. Warm Ups. Learning Objectives I can find the distance between two points. Football Problem: Bailey. Watson

Distance. Warm Ups. Learning Objectives I can find the distance between two points. Football Problem: Bailey. Watson Distance Warm Ups Learning Objectives I can find the distance between two points. Football Problem: Bailey Watson. Find the distance between the points (, ) and (4, 5). + 4 = c 9 + 6 = c 5 = c 5 = c. Using

More information

REQUIRED MATHEMATICAL SKILLS FOR ENTERING CADETS

REQUIRED MATHEMATICAL SKILLS FOR ENTERING CADETS REQUIRED MATHEMATICAL SKILLS FOR ENTERING CADETS The Department of Applied Mathematics administers a Math Placement test to assess fundamental skills in mathematics that are necessary to begin the study

More information

SOLVED PROBLEMS. 1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l 2 + m 2 + n 2 = 0 is

SOLVED PROBLEMS. 1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l 2 + m 2 + n 2 = 0 is SOLVED PROBLEMS OBJECTIVE 1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l 2 + m 2 + n 2 = 0 is (A) π/3 (B) 2π/3 (C) π/4 (D) None of these hb : Eliminating

More information

MULTIPLE PRODUCTS OBJECTIVES. If a i j,b j k,c i k, = + = + = + then a. ( b c) ) 8 ) 6 3) 4 5). If a = 3i j+ k and b 3i j k = = +, then a. ( a b) = ) 0 ) 3) 3 4) not defined { } 3. The scalar a. ( b c)

More information

CIRCLES. ii) P lies in the circle S = 0 s 11 = 0

CIRCLES. ii) P lies in the circle S = 0 s 11 = 0 CIRCLES 1 The set of points in a plane which are at a constant distance r ( 0) from a given point C is called a circle The fixed point C is called the centre and the constant distance r is called the radius

More information

Higher Mathematics Course Notes

Higher Mathematics Course Notes Higher Mathematics Course Notes Equation of a Line (i) Collinearity: (ii) Gradient: If points are collinear then they lie on the same straight line. i.e. to show that A, B and C are collinear, show that

More information

RMT 2013 Geometry Test Solutions February 2, = 51.

RMT 2013 Geometry Test Solutions February 2, = 51. RMT 0 Geometry Test Solutions February, 0. Answer: 5 Solution: Let m A = x and m B = y. Note that we have two pairs of isosceles triangles, so m A = m ACD and m B = m BCD. Since m ACD + m BCD = m ACB,

More information

Note : This document might take a little longer time to print. more exam papers at : more exam papers at : more exam papers at : more exam papers at : more exam papers at : more exam papers at : more

More information

F l a s h-b a s e d S S D s i n E n t e r p r i s e F l a s h-b a s e d S S D s ( S o-s ltiad t e D r i v e s ) a r e b e c o m i n g a n a t t r a c

F l a s h-b a s e d S S D s i n E n t e r p r i s e F l a s h-b a s e d S S D s ( S o-s ltiad t e D r i v e s ) a r e b e c o m i n g a n a t t r a c L i f e t i m e M a n a g e m e n t o f F l a-b s ah s e d S S D s U s i n g R e c o v e r-a y w a r e D y n a m i c T h r o t t l i n g S u n g j i n L e, e T a e j i n K i m, K y u n g h o, Kainmd J

More information

page 1 Total ( )

page 1 Total ( ) A B C D E F Costs budget of [Claimant / Defendant] dated [ ] Estimated page 1 Work done / to be done Pre-action Disbs ( ) Time ( ) Disbs ( ) Time ( ) Total ( ) 1 Issue /statements of case 0.00 0.00 CMC

More information

The Evolution of Outsourcing

The Evolution of Outsourcing Uvy f R I DCmm@URI S H Pj H Pm Uvy f R I 2009 T Ev f O M L. V Uvy f R I, V99@m.m Fw wk : ://mm../ P f B Cmm Rmm C V, M L., "T Ev f O" (2009). S H Pj. P 144. ://mm..//144://mm..//144 T A b y f f by H Pm

More information