Tiêu đề Straight Line Varsity Practice Sheet Solution.pdf ✅

2  Higher Math 1st Paper Chapter-3  r 2 = (1 + rcos) 2  r = 1 + rcos r – rcos = 1  r     1 – cos2 2 = 1  r 2 sin2 2 = 1  2rsin2 2 = 1 7. rcos(– ) = k mgxKiYwUi Kv‡Z©mxq mgxKiY †KvbwU? xcos + ysin = k xcos + ysin = k xcos – ysin = k xcos – ysin = k DËi: xcos + ysin = k e ̈vL ̈v: rcos( – ) = k  rcos.cos + rsin sin= k  xcos + ysin = k 8. †Kv‡bv we›`yi †cvjvi ̄’vbvsK (c, ) n‡j we›`ywUi Kv‡Z©mxq ̄’vbvsK KZ? [DU 21-22] (–1, 0) (– c, 0) (c, – c) (– c, c) DËi: (– c, 0) e ̈vL ̈v: x = rcos = c.cos = – c y = rsin = c.sin = 0  we›`ywU (– c, 0) x  =  y 9. r = a sin †cvjvi mgxKi‡Yi Kv‡Z©mxq mgxKiY KZ? ax 2 + y2 – y = 0 x 2 + y2 + ay = 0 x 2 + y2 – ay = 0 x 2 + ay2 – y = 0 DËi: x 2 + y2 – ay = 0 e ̈vL ̈v: r = asin  r 2 = arsin [x2 + y2 = r2 , y = rsin]  x 2 + y2 = ay  x 2 + y2 – ay = 0 10. x A‡ÿi Ici Aew ̄’Z GKwU we›`y p n‡Z (1, 2) I (4, 5) we›`y `yBwUi `~iZ¡ mgvb n‡j p we›`yi ̄’vbvsK KZ? (0, 6) (0, 9) (6, 0) (9, 0) DËi: (6, 0) e ̈vL ̈v: g‡b Kwi, p we›`ywU (x, 0) kZ©g‡Z, (x – 1)2 + (0 – 2)2 = (x – 4)2 + (0 – 5)2  x 2 – 2x + 1 + 4 = x2 – 8x + 16 + 25  – 2x + 5 + 8x – 41 = 0  x = 6  p we›`yi ̄’vbvsK (6, 0) 11. y Aÿ I (7, 2) †_‡K (a, 5) we›`ywUi `~iZ¡ mgvb n‡j, a Gi gvb KZ? 29 7 29 9 27 7 29 5 DËi: 29 7 e ̈vL ̈v: y Aÿ †_‡K (a, 5) we›`yi `~iZ¡ = |a| Ges (7, 2) n‡Z (a, 5) Gi `~iZ¡ = (7 – a) 2 + (2 – 5) 2 = 7 2 – 2.7a + a2 + (–3) 2 = a 2 – 14a + 58 kZ©g‡Z, a 2 – 14a + 58 = |a|  a 2 – 14a + 58 = a2  14a = 58  a = 58 14 = 29 7 12. †Kv‡bv we›`yi †KvwU 3 Ges we›`ywUi `~iZ¡ (5, 3) †_‡K 4 GKK n‡j we›`ywUi fzR KZ? 3 A_ev 9 2 A_ev 7 3 A_ev 5 9 A_ev 1 DËi: 9 A_ev 1 e ̈vL ̈v: †`Iqv Av‡Q, we›`yi †KvwU 3| awi fzR p  we›`ywUi ̄’vbvsK (p, 3)  (p, 3) n‡Z (5, 3) we›`ywUi `~iZ¡ = (p – 5) 2 + (3 – 3) 2 kZ©g‡Z, (p – 5) 2 + (3 – 3) 2 = 4  p 2 – 10p + 25 – 16 = 0  p 2 – 10p + 9 = 0  p 2 – 9p – p + 9 = 0  p(p – 9) – 1(p – 9) = 0  (p – 9) (p – 1) = 0  p = 9 or p = 1 13. `ywU we›`yi †cvjvi ̄’vbvsK (2 3 , 90) Ges (2 5, 180) n‡j we›`y `ywUi `~iZ¡ KZ? 4 3 4 2 4 5 2 3 0 DËi: 4 2 e ̈vL ̈v: Avgiv Rvwb, `ywU †cvjvi we›`yi ga ̈eZ©x `~iZ¡, = r1 2 + r2 2 – 2r1r2 cos(1 – 2)  (2 3, 90) Ges (2 5, 180) Gi ga ̈eZ©x `~iZ¡, = (2 3) 2 + (2 5) 2 – 2.2 3.2 5 cos(180– 90) = 12 + 20 – 0 = 32 = 16  2 = 4 2 14. GKwU mgevû wÎfz‡Ri `ywU kxl©we›`yi ̄’vbvsK (2, 1) Ges (2, 5) n‡j, Z...Zxq kxl©we›`yi ̄’vbvsK KZ? (2 + 2 3, 3) (2 + 3, 3) (3 + 2 3, 3) (3 + 3, 3) DËi: (2 + 2 3, 3)
mij‡iLv  Varsity Practice Sheet 3 e ̈vL ̈v: (2, 1) Ges (2, 5) we›`y؇qi ga ̈eZ©x `~iZ¡ = 4 GKgvÎ Ackb Gi (2 + 2 3, 3) we›`ywU †_‡KB (2, 1) Ges (2, 5) Dfq we›`yi `~iZ¡ 4 GKK nq| ZvB Option Test Kivi gva ̈‡g GB AskwU mn‡RB mgvavb Kiv hvq| 15. (–3, –4) I (6, 2) we›`y؇qi ms‡hvM †iLvwU‡K Y-Aÿ‡iLv †h AbycvZ AšÍwe©f3 K‡i, Zv n‡jvÑ 2 : 1 1 : 2 2 : 3 3 : 2 DËi: 1 : 2 e ̈vL ̈v: wb‡Y©q AbycvZ =     1g we›`yi fzR 2q we›`yi fzR =     –3 6 = 1 : 2 Note: X Aÿ †h Abycv‡Z wef3 K‡i =     1g we›`yi †KvwU 2q we›`yi †KvwU Y Aÿ †h Abycv‡Z wef3 K‡i =     1g we›`yi fzR 2q we›`yi fzR we.`a.: Modulus Gi g‡a ̈ (–ve) AšÍwe©f3 (+ ve)  ewnwe©f3 16. (4, 2) Ges (– 2, – 4) we›`y؇qi ms‡hvM †iLvwU‡K x Aÿ‡iLv †h Abycv‡Z wef3 K‡i, Zv n‡jvÑ 2 : 1 1 : 2 1 : 3 3 : 1 DËi: 1 : 2 e ̈vL ̈v:     y1 y2 =     2 – 4 = 1 2 17. (–3, – 4) I (6, 2) we›`y؇qi ms‡hvM †iLvwU‡K y Aÿ‡iLv †h Abycv‡Z AšÍwe©fw3 K‡i, Zv n‡jvÑ 2 : 1 1 : 2 2 : 3 3 : 2 DËi: 1 : 2 e ̈vL ̈v:     x1 x2 =     –3 6 = 1 2 18. †h we›`y(1, 4) Ges (9, – 12) we›`y؇qi ms‡hvMKvix †iLvsk‡K AšÍt ̄’fv‡e 3 : 5 Abycv‡Z AšÍ©wef3 K‡i Zvi ̄’vbvsKÑ (4, – 2) (– 2, 4) (– 4, 2) (4, 2) DËi: (4, – 2) e ̈vL ̈v: ● ● ● m2 = 5 m1 = 3 (9, – 12) (x, y) (1, 4)  x = (3  9) + (5  1) 5 + 3 ; y = 3  (– 12) + (5  4) 3 + 5 = 27 + 5 8 = – 36 + 20 8 = 32 8 = – 16 8 = 4 = – 2  wb‡Y©q we›`yi ̄’vbvsK (x, y) = (4, – 2) 19. A(3, 4) Ges B(5, 9) we›`y؇qi ms‡hvM †iLvsk‡K †h we›`ywU 2 : 3 Abycv‡Z ewnwe©f3 K‡i Zvi ̄’vbvsK KZ? (– 3, – 8) (– 5, – 9) (– 1, – 6) (– 1, – 7) DËi: (– 1, – 6) e ̈vL ̈v: Avgiv Rvwb, ewn©wefw3i †ÿ‡Î,    m1  x2 – m2x1 m1 – m2  m1y2 – m2y1 m1 – m2 =     2  5 – 3  3 2 – 3  2  9 – 3  4 2 – 3 =     10 – 9 – 1  18 – 12 – 1 = (– 1, – 6) 20. (–2, 4) Ges (8, –10) we›`y؇qi ms‡hvRK †iLv‡K 2 : 3 Abycv‡Z ewnwe©f3Kvix we›`yi ̄’vbvsK †KvbwU? (–22, 8) (22, –8) (–22, 32) (22, –32) DËi: (–22, 32) e ̈vL ̈v:    m1  x2 – m2x1 m1 – m2  m1y2 – m2y1 m1 – m2 =     2  8 – 3  (–2) –1  2  (–10) – 3  4 –1 (–2, 4)  (x1,y1) (8, – 10)  (x2,y2) m1 : m2  2 : 3 = (–22, 32) 21. A (1, – 2) I B (– 8, 1) we›`y؇qi ms‡hvRK †iLvsk BA †K 2 : 1 Abycv‡Z AšÍwe©f3Kvix we›`yi ̄’vbv1⁄4 wb‡Pi †KvbwU? (– 5, – 1) (– 2, – 1) (– 2, 0) (– 5, 0) DËi: (– 2, – 1) e ̈vL ̈v:    m1  x2 + m2x1 m1 + m2  m1y2 + m2y1 m1 + m2 =     2  1 + 1  (–8) 2 + 1  2  (– 2) + 1  1 2 + 1 (–8, 1)  (x1, y1) (1, – 2)  (x2, y2) m1 : m2  2 : 1 = (– 2, – 1) 22. (7, – 8) we›`ywU (3, – 2) Ges (– 3, 7) we›`y؇qi ms‡hvM †iLvsk‡K †h Abycv‡Z ewn©wef3 K‡iÑ 7 : 2 2 : 5 7 : 5 9 : 7 DËi: 2 : 5 e ̈vL ̈v: R(7, – 8) 1 P(3, – 2) Q(–3, 7) ● ● ● k g‡b Kwi, cÖ`Ë we›`y `ywU P(3, – 2), Q(–3, 7) Ges R (7, – 8) we›`ywU PQ †iLvsk‡K k : 1 Abycv‡Z ewnwe©f3 K‡i|  k  (– 3) – (1  3) k – 1 = 7
4  Higher Math 1st Paper Chapter-3  – 3k – 3 = 7k – 7  – 10k = – 4  k = 4 10 = 2 5  k : 1 = 2 : 5 23. (3, 1) we›`ywU (1, – 3) I (6, 7) we›`y؇qi ms‡hvRK †iLvsk‡K †h Abycv‡Z AšÍ©wef3 K‡i Zvi ̄’vbvsK KZ? 3 : 2 5 : 2 7 : 3 2 : 3 DËi: 2 : 3 e ̈vL ̈v: Avgiv Rvwb, (x, y) =    m1  x2 + m2x1 m1 + m2  m1y2 + m2y1 m1 + m2  fzR 3 = m1  6 + m2  1 m1 + m2  6m1 + m2 = 3m1 + 3m2  6m1 – 3m1 = 3m2 – m2  3m1 = 2m2  m1 m2 = 2 3  m1 : m2 = 2 : 3 24. (3, 2), (4, 6), (– 2, 4) kxl© wewkó †ÿÎwUi †ÿÎdj KZ? 10 eM© GKK 11 eM© GKK 12 eM© GKK 13 eM© GKK DËi: 11 eM© GKK e ̈vL ̈v: wZb we›`y wewkó †ÿÎwU †KejgvÎ GKwU wÎfzR n‡e weavq wÎfz‡Ri †ÿÎdj = 1 2     3 2 4 6 2 4 3 2 = 1 2 |18 + 16 – 4 – 12 + 12 – 8| = 1 2 |22| = 11 eM© GKK 25. hw` (– 5, 1), (4, 5) Ges (7, – 4) GKwU wÎfz‡Ri kxl©we›`y nq Zvn‡j, wÎfz‡Ri †ÿÎdj KZ? 48 1 2 46 1 2 50 71 1 2 DËi: 46 1 2 e ̈vL ̈v:  =     1 2     – 5 1 4 5 7 – 4 – 5 1 =     1 2 {– 25 – 16 + 7 – 20 – 35 – 4} =     1 2 {–93} = 1 2 (93) = 93 2 = 46 1 2 eM© GKK Shortcut: Step: 1. wZbwU we›`yi †h‡Kv‡bv GKwU‡K k~b ̈ K‡i †dj‡Z n‡e| 2. k~b ̈ evbv‡Z fyR I †KvwU‡Z †h cwieZ©b n‡q‡Q Aci we›`yØq‡K GKB cwieZ©b Ki‡Z n‡e| awi, A(– 5,1) = (– 5 + 5, 1– 1) = (0, 0) B (4, 5) = (4 + 5, 5 – 1) = (9, 4) C(7, – 4) = (7 + 5, – 4 –1) = (12, – 5) (–) wÎfz‡Ri †ÿÎdj =     1 2 {9  (– 5) – (4  12)} =     1 2 (– 45 – 48) = 1 2  93 = 93 2 = 46 1 2 26. x – y + 2 = 0, x + 2y – 4 = 0 Ges 2x – y – 3 = 0 GB wZbwU †iLv Øviv MwVZ mg‡KvYx wÎfz‡Ri †ÿÎdj KZ? 25 4 11 2 15 2 11 4 DËi: 15 2 e ̈vL ̈v: x – y + 2 = 0 ....(i) x + 2y – 4 = 0 ....(ii) 2x – y – 3 = 0 ....(iii) (i) I (ii) mgvavb K‡i, x = 0, y = 2  we›`ywU (0, 2) (ii) I (ii) mgvavb K‡i, x = 2, y = 1  we›`ywU (2, 1) (i) I (iii) mgvavb K‡i, x = 5, y = 7  we›`ywU (5, 7) we›`y wZbwU Øviv MwVZ wÎfz‡Ri †ÿÎdj, =     1 2     0 2 2 1 5 7 0 2 = 1 2 {(14 + 10) – (4 + 5)}  1 2 (24 – 9) = 1 2  (15) = 15 2 27. x Gi †Kvb gv‡bi Rb ̈ (1, 2), (2, 1) I (3, x) we›`yÎq Øviv MwVZ wÎfz‡Ri †ÿÎdj k~b ̈ n‡e? 1 2 3 0 DËi: 0 e ̈vL ̈v:  =     1 2     1 2 2 1 3 x 1 2 = 1 2 (1 + 2x + 6 – x – 3 – 4) = 1 2 (x) = x 2 kZ©g‡Z, x 2 = 0 x = 0