Tiêu đề Straight-Line Varsity Practice Sheet Solution.pdf ✅

mij‡iLv  Varsity Practice Sheet Solution 1 03 mij‡iLv Straight Line weMZ mv‡j DU-G Avmv cÖkœvejx 1. (1, 1) we›`yMvgx I 2x – 3y – 5 = 0 †iLvi Dci j¤^‡iLvi mgxKiY †KvbwU? [DU 23-24] 3x + 2y – 5 = 0 3x + 2y + 5 = 0 3x – 2y – 1 = 0 – 2x + 3y + 1 = 0 DËi: 3x + 2y – 5 = 0 e ̈vL ̈v: 3x + 2y = 3  1 + 2  1  3x + 2y – 5 = 0 2. (0, 2) Ges (– 2, 0) we›`yMvgx mij‡iLv x A‡ÿi abvZ¥K w`‡Ki mv‡_ Kx †KvY Drcbœ K‡i? [DU 22-23] 30 45 60 120 DËi: 45 e ̈vL ̈v: (0, 2) (– 2, 0) X Y X Y tan = 2 – 0 0 – (– 2) = 1   = 45 3. y A‡ÿi mgvšÍivj Ges 2x – 7y + 11 = 0 I x + 3y = 8 †iLv؇qi †Q`we›`y w`‡q AwZμgKvix mij‡iLvi mgxKiY †KvbwU? [DU 22-23, 17-18; JUST 19-20] 3x – 7 = 0 13x – 23 = 0 7x – 3 = 0 7x – 23 = 0 DËi: 13x – 23 = 0 e ̈vL ̈v: awi, wb‡Y©q mgxKiY: x = a 2x – 7y + 11 = 0  y = 2x + 11 7 ....... (i)  x + 3y = 8  x + 3  2x + 11 7 = 8 [(i) n‡Z]  7x + 6x + 33 = 56  13x – 23 = 0 4. y = 1 + 1 2 + x eμ‡iLv x Aÿ‡K A we›`y‡Z Ges y Aÿ‡K B we›`y‡Z †Q` Ki‡j AB mij‡iLvi mgxKiY wb‡Pi †KvbwU? [DU 22-23] x + 2y + 3 = 0 x + 2y – 3 = 0 x – 2y + 3 = 0 x – 2y – 3 = 0 DËi: x – 2y + 3 = 0 e ̈vL ̈v: x Aÿ ev y = 0 n‡j,  0 = 1 + 1 2 + x  2 + x = – 1  x = – 3  A(– 3, 0) y Aÿ ev x = 0 n‡j,  y = 1 + 1 2 + 0 = 1 + 1 2 = 3 2  B     0 3 2 AB †iLvi mgxKiY: x – 3 + y 3 2 = 1  – x + 2y = 3  x – 2y + 3 = 0 5. †Kv‡bv we›`yi †cvjvi ̄’vbv1⁄4 (c, ) n‡j, we›`ywUi Kv‡Z©mxq ̄’vbv1⁄4 KZ? [DU 21-22] (– 1, 0) (– c, 0) (c, – c) (– c, c) DËi: (– c, 0) e ̈vL ̈v: x = c.cos = – c y = c.sin = 0 6. hw` P we›`ywU x – 3y = 2 mij‡iLvi Dci Aew ̄’Z nq Ges (2, 3) I (6, – 5) we›`yØq †_‡K mg`~ieZ©x nq, Z‡e P Gi ̄’vbv1⁄4 KZ? [DU 21-22; MBSTU 19-20] (12, 4) (14, 4) (16, 4) (18, 4) DËi: (14, 4) e ̈vL ̈v: awi, P we›`ywU (x, y) x – 3y – 2 = 0 ..... (i)  (x – 2) 2 + (y – 3) 2 = (x – 6) 2 + (y + 5) 2  – 4x + 4 – 6y + 9 = – 12x + 36 + 10y + 25  8x – 16y – 48 = 0  x – 2y – 6 = 0 ....... (ii)
2  Higher Math 1st Paper Chapter-3 (i) I (ii) bs mgvavb K‡i,  x = 14, y = 4  (x,y) = (14, 4) A_ev, x – 3y = 2 mij‡iLvwU ïaygvÎ Ackb (14, 4) Øviv wm× nq| AZGe, GwUB mwVK DËi n‡e| 7. x 2 + y2 – by = 0 e„Ë Gi mgxKiY †cvjvi ̄’vbv1⁄4 Gi gva ̈‡g cÖKvk Ki‡j mgxKiYwU n‡eÑ [DU-7Clg 19-20; JnU 15-16] x = ysin r = bcos r = bsin r = b DËi: r = bsin e ̈vL ̈v: x 2 + y2 – by = 0  r 2 – brsin = 0  r = bsin 8. 2rsin2      2 = 1 Gi Kv‡Z©mxq mgxKiYÑ [DU 18-19] y 2 = 1 + 2x y 2 = 4(1 – x) y 2 = 4(1 + x) x 2 = 4(1 + y) DËi: y 2 = 1 + 2x e ̈vL ̈v: 2rsin2      2 = 1  r(1 – cos) = 1  r – rcos = 1  r = 1 + x  r 2 = (1 + x)2  x 2 + y2 = 1 + 2x + x2  y 2 = 1 + 2x 9. y = b Ges 3x – y + 1 = 0 †iLv؇qi AšÍf©y3 m~2‡Kv‡Yi gvbÑ [DU 18-19; JUST 19-20] 30 45 60 90 DËi: 60 e ̈vL ̈v: m1 = 0 = tan1  1 = 0 m2 = 3 = tan2  2 = 60  ga ̈eZ©x †KvY = 60 – 0 = 60 10. y = x + 4 n‡Z y = x †iLvi j¤^ `~iZ¡ KZ? [DU 18-19; Agri. Guccho 20-21; BSMRSTU 18-19] 2 2 2 3 2 4 2 DËi: 2 2 e ̈vL ̈v: x – y + 4 = 0 ...... (i) x – y = 0 ............ (ii)  `~iZ¡ =     4 – 0 1 2 + (– 1) 2 = 4 2 = 2 2 GKK 11. y = 2 Ges y = |x| †iLv ̧‡jv Øviv Ave× †ÿ‡Îi †ÿÎdjÑ [DU 18-19] 2 eM© GKK 4 eM© GKK 6 eM© GKK 8 eM© GKK DËi: 4 eM© GKK e ̈vL ̈v: (– 2 , 2) (2 , 2) y = – x C D B y = x O X Y Y X y = 2 OBC = 2 OBD = 2      1 2  2  2 = 4 eM© GKK 12. A(2, 5), B(5, 9) Ges D(6, 8) we›`yÎq ABCD i¤^‡mi wZbwU kxl©we›`y n‡j, PZz_© we›`yC Gi ̄’vbv1⁄4 †KvbwU? [DU 17-18; RU 22-23; KU 14-15] (4, 3) (9, 12) (4, 7) (7, 9) DËi: (9, 12) e ̈vL ̈v: A(2, 5) B(5, 9) D(6, 8) C(x, y)  C  (5 + 6 – 2, 9 + 8 – 5)  (9, 12) Shortcut: D(x3 , y3 ) C(x, y) B(x2 , y2 B(x ) 1 , y1 ) x = x2 + x3 – x1; y = y2 + y3 – y1 A_©vr, †Kv‡bv i¤^m ev mvgvšÍwi‡Ki †h kx‡l©i ̄’vbv1⁄4 †ei Ki‡Z n‡e, Zvi †_‡K Kv‡Q Aew ̄’Z kx‡l©i fzR/†KvwU †hvM K‡i wecixZ kx‡l©i fzR/†KvwU we‡qvM Ki‡Z n‡e| 13. 2x + 3y – 4 = 0 Ges xcos + ysin = p GKB mij‡iLv wb‡`©k Ki‡j p Gi gvbÑ [DU 16-17] 1 13 2 13 3 13 4 13 DËi: 4 13