Tiêu đề Straight Line Varsity Practice Sheet Solution.pdf ✅

mij‡iLv  Varsity Practice Sheet 1 03 mij‡iLv Straight Line 1. (1, 150) we›`yi Kv‡Z©mxq ̄’vbvsK wb‡Pi †KvbwU?    3  2  1 2     – 3 2  1 2    3  2  – 1 2     – 3 2  – 1 2 DËi:     – 3 2  1 2 e ̈vL ̈v: x = rcos = 1  cos150 = cos (2  90 – 30) = – cos30 = – 3 2 Ges y = rsin = 1.sin150 = sin(2  90 – 30) = sin30 = 1 2  Kv‡Z©mxq ̄’vbv1⁄4 (x, y)      – 3 2  1 2 2.    – 3 2  2  3 2 2 we›`ywUi †cvjvi ̄’vbvsK KZ?     3  7 4     3  5 4     3  3 4     3   4 DËi:     3  3 4 e ̈vL ̈v: r =    – 3 2 2 2 +    3 2 2 2 = 18 4 + 18 4 = 36 4 = 3  =  – tan–1     y x =  – tan–1       3 2 2 –3 2 2 =  – tan–1 1 =  –  4 = 3 4  †cvjvi ̄’vbvsK     3  3 4 3. (– 2  – 2) we›`yi †cvjvi ̄’vbvsK †KvbwU?     2  – 3 2     2  3 2     2  – 5 4     2  5 4 DËi:     2  5 4 e ̈vL ̈v: r = x 2 + y2 = ( – 2) 2 + (– 2) 2 = 4 = 2  =  + tan–1     y x =  + tan–1      – 2 – 2 =  + tan–1 1 =  +  4 = 5 4 4. (1, – 1) we›`ywUi †cvjvi ̄’vbvsK †KvbwU? ( 2, 45) ( 2, 135) ( 2, 225) ( 2, 315) DËi: ( 2, 315) e ̈vL ̈v: r = (1) 2 + (–1) 2 = 2  = 2 – tan–1     y x = 2 – tan–1     –1 1 = 2 – tan–1 1 = 2 –  4 = 7 4 = 315  (r, ) = ( 2, 315) 5. ( 3, 1) we›`ywUi †cvjvi ̄’vbvsK †KvbwU? (2, 30) (2, 45) (2, 60) (2, 90) DËi: (2, 30) e ̈vL ̈v: r = ( 3) 2 + 12 = 4 = 2  = tan–1     y x = tan–1     1 3 = 30  (r, ) = (2, 30) 6. y 2 = 1 + 2x mgxKiYwUi †cvjvi mgxKiY †KvbwU? 2rcos2  = 1 2rsin2  = 1 2rcos2 2 = 1 2rsin2 2 = 1 DËi: 2rsin2 2 = 1 e ̈vL ̈v: y 2 = 1 + 2x  x 2 + y2 = x 2 + 1 + 2x  x 2 + 2x + 1 = x2 + y2  (x + 1)2 = x 2 + y2
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)