Title 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
mij‡iLv  Varsity Practice Sheet Solution 3 e ̈vL ̈v: 2x + 3y = 4 .................. (i) xcos + ysin = p ....... (ii) mgxKiYØq GKB mij‡iLv wb‡`©k Ki‡j,  2 cos = 3 sin = 4 p  cos = p 2 ; sin = 3p 4  sin2 + cos2 = 1  9p2 16 + p 2 4 = 1  p = 4 13 14. 5x – 7y – 15 = 0 mij‡iLvi Dci j¤^ Ges (2, – 3) we›`yMvgx mij‡iLvi mgxKiY n‡eÑ [DU 16-17, 10-11; JnU 16-17; CU 15-16] 7x – 5y – 29 = 0 5x – 7y – 31 = 0 5x + 7y + 11 = 0 7x + 5y + 1 = 0 DËi: 7x + 5y + 1 = 0 e ̈vL ̈v: 7x + 5y = 7  2 + 5  (– 3)  7x + 5y + 1 = 0 15. x = a Ges 3x – y + 1 = 0 †iLv؇qi ga ̈eZ©x m~2‡Kv‡Yi gvbÑ [DU 16-17] 30 45 60 75 DËi: 30 e ̈vL ̈v: m1 =  = tan1  1 = 90 m2 = 3 = tan2  2 = 60  ga ̈eZ©x m~2‡KvY = 90 – 60 = 30 16. †Kv‡bv we›`yi †cvjvi ̄’vbv1⁄4 (3, 150) n‡j, H we›`yi Kv‡Z©mxq ̄’vbv1⁄4Ñ [DU 15-16; JU 14-15]    3 3  2  3 2    3 3  2  – 3 2     – 3 3 2  3 2     – 3 3 2  – 3 2 DËi:     – 3 3 2  3 2 e ̈vL ̈v: x = 3cos(150) = – 3 3 2 y = 3sin(150) = 3 2  Kv‡Z©mxq ̄’vbv1⁄4     – 3 3 2  3 2 17. y = – 5x + 9 †iLvi mv‡_ j¤^ †iLvi bwZÑ [DU 14-15] 5 – 5 1 5 – 1 5 DËi: 1 5 e ̈vL ̈v: m1m2 = – 1  (– 5)m2 = – 1  m2 = 1 5 18. hw` (– 5, 1), (4, 5) Ges (7, – 4) GKwU wÎfz‡Ri kxl©we›`y nq, Zvn‡j wÎfz‡Ri †ÿÎdj KZ? [DU 14-15, 03-04, 01-02; JU 17-18; CU 12-13; RU 06-07] 48 1 2 46 1 2 50 71 1 2 DËi: 46 1 2 e ̈vL ̈v:  = 1 2       – 5 4 7 1 5 – 4 1 1 1 = 1 2       – 9 – 3 7 – 4 9 – 4 0 0 1     r1 = r1 – r2 r2 = r2 – r3 = 1 2 |(– 93)| eM© GKK = 93 2 eM© GKK = 46 1 2 eM© GKK 19. 3x + 5y = 2, 2x + 3y = 0, ax + by + 1 = 0 mgwe›`yMvgx n‡j, a Ges b Gi m¤úK©Ñ [DU 14-15] 4a – 6b = 1 4a – 6b = 2 6a – 4b = 1 6a – 4b = 2 DËi: 6a – 4b = 1 e ̈vL ̈v:      3 2 a 5 3 b – 2 0 1 = 0  – 2(2b – 3a) + (9 – 10) = 0 [3 bs Kjvg eivei we ̄Ívi K‡i]  – 4b + 6a = 1  6a – 4b = 1 20. (1, – 1) Ges (8, 6) we›`y؇qi ms‡hvM †iLvsk‡K †h we›`ywU 3 : 4 Abycv‡Z AšÍwe©f3 K‡i Gi ̄’vbv1⁄4 KZ? [DU 14-15, 10-11; JU 15-16, 14-15] (4, – 2) (– 4, – 2) (– 4, 2) (4, 2) DËi: (4, 2)
4  Higher Math 1st Paper Chapter-3 e ̈vL ̈v: x = 3  8 + 4  1 3 + 4 = 4 y = 3  6 + 4  (– 1) 3 + 4 = 2  wb‡Y©q ̄’vbv1⁄4 (4, 2) 4 (1, – 1) (8, 6) 3 (x, y) Note: (x1, y1) I (x2, y2) we›`y؇qi ms‡hvRK †iLvsk‡K (x, y) we›`ywU m1 : m2 Abycv‡Z, AšÍwe©f3 Ki‡j, x = m1x2 + m2x1 m1 + m2 ; y = m1y2 + m2y1 m1 + m2 ewnwe©f3 Ki‡j, x = m1x2 – m2x1 m1 – m2 ; y = m1y2 – m2y1 m1 – m2 21. (3, – 1) Ges (5, 2) we›`y؇qi ms‡hvMKvix mij‡iLv‡K 3 : 4 Abycv‡Z ewnt ̄’fv‡e wef3Kvix we›`yi ̄’vbv1⁄4Ñ [DU 13-14; JU 18-19, 09-10]     17 3  3     27 7  2 7     27 4  4 3 None DËi: None e ̈vL ̈v: x = 3  5 – 4  3 3 – 4 = – 3 y = 3  2 – 4  (– 1) 3 – 4 = – 10  (x, y)  (– 3, – 10) 4 (x, y) 3 (3, – 1) (5, 2) 22. x + y = 3 Ges y – x = 1 mij‡iLv؇qi †Q`we›`yMvgx x A‡ÿi mgvšÍivj mij‡iLvi mgxKiYÑ [DU 13-14] y = 2 2y = 3 x = 1 x + 3 = 0 DËi: y = 2 e ̈vL ̈v: x + y = 3 (+) y – x = 1 2y = 4  y = 2 23. (2, – 1), (a + 1, a – 3) I (a + 2, a) we›`y wZbwU mg‡iL n‡j, a Gi gvbÑ [DU 11-12; JU 14-15] 4 2 1 4 1 2 DËi: 1 2 e ̈vL ̈v: m1 = m2 a – 3 + 1 a + 1 – 2 = a – a + 3 a + 2 – a – 1  a – 2 a – 1 = 3 1  a – 2 = 3a – 3  2a = 1  a = 1 2 24. (1, – 1) we›`yMvgx Ges 2x – 3y +1 = 0 †iLvi Dci j¤^ †iLvi mgxKiY †KvbwU? [DU 11-12, 09-10, 07-08, 03-04, 02-03; RU 23-24, 08-09] 3x + 2y – 1 = 0 3x – 2y – 1 = 0 – 3x + 2y + 1 = 0 2x + 3y + 1 = 0 DËi: 3x + 2y – 1 = 0 e ̈vL ̈v: 3x + 2y = 3  1 + 2  (– 1)  3x + 2y – 1 = 0 25. y = 3x + 7 Ges 3y – x – 8 = 0 †iLv؇qi AšÍf©y3 m~2‡KvY- [DU 08-09; RU 23-24; JnU 15-16] tan–1 1 tan–11 2 tan–14 3 tan–12 3 DËi: tan–14 3 e ̈vL ̈v: m1 = 3; m2 = 1 3  tan =  3 – 1 3 1 + 3  1 3  tan=  8 3 2   = tan–14 3 [⸪ m~2‡KvY] weMZ mv‡j GST-G Avmv cÖkœvejx 1. 3x + 4y = 12 †iLvwU x I y Aÿ‡K h_vμ‡g A I B we›`y‡Z †Q` Ki‡j, g~jwe›`y n‡Z AB †iLvi Dci j¤^ `~iZ¡ KZ? [GST 23-24] 12 5 5 12 12 25 7 5 DËi: 12 5 e ̈vL ̈v: (0, 0) n‡Z 3x + 4y – 12 = 0 Gi j¤^ `~iZ¡ = |0 + 0 – 12| 3 2 + 42 = 12 5 GKK 2. `yBwU mij‡iLvi ga ̈eZ©x m~2‡KvY 45 Ges GKwU mij‡iLvi Xvj 1 2 n‡j, Aci mij‡iLvi bwZ KZ? [GST 23-24] 1 3 – 1 3 1 2 – 1 2 DËi: – 1 3