Tiêu đề Straight line Engineering Question Bank Solution (HSC 26).pdf ✅

03 mij‡iLv Straight Line WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. GKwU mij‡iLvi x I y Aÿ Øviv LwÐZ AskØq abvZ¥K Ges g~jwe›`y †_‡K Aw1⁄4Z j¤^, mij‡iLv‡K mgwØLwÐZ K‡i| j‡¤^i •`N ̈© 7 GKK n‡j, mij‡iLvi mgxKiY wbY©q Ki| [BUET 23-24] mgvavb: awi, mij‡iLvi mgxKiY: xcos + ysin = p GLv‡b, p = 7 †h‡nZz, g~jwe›`y †_‡K Aw1⁄4Z j¤^, mij‡iLv‡K mgwØLwÐZ K‡i|   = 45 Y X 7 O  xcos45 + ysin45 = 7  x 2 + y 2 = 7  x + y – 7 2 = 0 (Ans.) 2. (k, – 3k), (5, k) Ges (– k, 2) we›`yÎq Øviv Drcbœ wÎfz‡Ri †ÿÎdj 28 eM©GKK, †hLv‡b k GKwU c~Y© msL ̈v| wÎfz‡Ri j¤^we›`yi ̄’vbv1⁄4 wbY©q Ki| [BUET 22-23] mgvavb: (k, – 3k), (5, k) Ges (– k, 2) cÖkœg‡Z, 1 2     k – 3k 5 k – k 2 k – 3k = 28  k 2 + 10 + 3k2 + 15k + k2 – 2k = 56  5k2 + 13k – 46 = 0  k = 2, – 23 5 wKš‘ (k  Z)  k = 2  ̄’vbv1⁄4 A(2, – 6); B(5, 2); C(– 2, 2) B C A O(x, y) awi, j¤^we›`y O(x, y)| AB I OC ci ̄úi j¤^ nIqvq, mAB  mOC = – 1  2 + 6 5 – 2  y – 2 x + 2 = – 1  3x + 8y = 10 .....(i) Avevi, AC I OB ci ̄úi j¤^ nIqvq, mAC  mOB = – 1  2 + 6 – 2 – 2  y – 2 x – 5 = – 1  x – 2y = 1 ....(ii) (i) I (ii) mgvavb K‡i, wb‡Y©q j¤^we›`y(x, y)      2  1 2 (Ans.) 3. (a – 1)x + y = a ........ (i) (2, 2)  (b, 0) ........... (ii) †iLvØq (K) j¤^ n‡j a I b Gi m¤úK© Kx? (L) mgvšÍivj n‡j a I b Gi m¤úK© Kx? [BUET 21-22] mgvavb: †`Iqv Av‡Q, (a – 1)x + y = a  y = – (a – 1)x + a GB †iLvi Xvj, m1 = – (a – 1) Ges (2, 2) Ges (b, 0) we›`y؇qi ms‡hvMKvix †iLvi Xvj m2 = 2 – 0 2 – b = 2 2 – b (K) †iLvØq j¤^ n‡j, m1m2 = – 1  – (a – 1) × 2 2 – b = – 1  2a – 2 = 2 – b  2a + b = 4 (Ans.) (L) †iLvØq mgvšÍivj n‡j, m1 = m2  – (a – 1) = 2 2 – b  (a – 1)(2 – b) = – 2  2a – ab – 2 + b = – 2  2a + b – ab = 0 (Ans.) 4. g~j we›`y n‡Z xsec – ycosec = k Ges xcos – ysin = kcos2 †iLv؇qi j¤^ `~iZ¡ h_vμ‡g 2 cm Ges 3 cm| k Gi gvb wbY©q Ki| [BUET 19-20] mgvavb: g~jwe›`y (0, 0) †_‡K xsec – ycosec – k = 0 †iLvi Dci j¤^`~iZ¡,       0 × sec – 0 × cosec – k sec2  + cosec2  = 2  k 2 1 cos2  + 1 sin2  = 4 [eM© K‡i]

mij‡iLv  Engineering Question Bank Solution (HSC 26) 3 Avevi, 2x – 3y + 4 = 3x – 2y + 1 [– wPý wb‡q]  x + y = 3  x 3 + y 3 = 1............. (ii)  R(3, 0), S(0, 3) (i) I (ii) Gi †Q`we›`y  (1, 2) PS Gi Xvj = 3 – 0 0 + 1 = 3  PS Gi mgvšÍivj I (1, 2) we›`yMvgx †iLv, y – 2 = 3(x – 1)  3x – y = 1 (Ans.) 9. 3x + 4y = 11 Ges 12x – 5y = 2 †iLv؇qi AšÍfz©3 m~2‡Kv‡Yi mgwØLЇKi mgxKiY wbY©q Ki| [BUET 06-07; KUET 13-14, 06-07; CUET 13-14, 07-08] mgvavb: wb‡Y©q mgwØLÐK, 3x + 4y – 11 3 2 + 42 =  12x – 5y – 2 122 + 52 †h‡nZz, a1a2 + b1b2 > 0  (– ve) wb‡q m~2‡Kv‡Yi mgwØLÐK cvIqv hv‡e|  3x + 4y – 11 5 = – 12x – 5y – 2 13  39x + 52y – 143 = – 60x + 25y + 10  99x + 27y – 153 = 0  11x + 3y – 17 = 0 (Ans.) 10. y = x mij‡iLv wfwËK P(5, 6) we›`yi cÖwZwe‡¤^i ̄’vbv1⁄4 wbY©q Ki| [BUET 01-02] mgvavb: x – y = 0 .....(i) †iLvi Dci j¤^ P(5, 6) we›`yMvgx †iLvi mgxKiY, x + y =5 + 6  x + y – 11 = 0 ...... (ii) (i) I (ii) †iLvi †Q`we›`y    11 2  11 2 awi, cÖwZwe¤^ (, )       + 5 2 = 11 2   + 6 2 = 11 2  (, )  (6,5) (Ans.) 11. (1, 2), (4, 4) Ges (2, 8) h_vμ‡g wÎfzR ABC Gi evû·qi ga ̈we›`y| ABC wÎfzRwUi †ÿÎdj wbY©q Ki| [BUET 01-02] mgvavb: awi, D  (1, 2); E  (4, 4); F  (2, 8)  DEF = 1 2       1 4 2 2 4 8 1 1 1 = 1 2 × 16 = 8 eM© GKK †h‡nZz D, E, F nj ga ̈we›`y †m‡nZz, ABC = 4 × DEF = 4 × 8 eM© GKK = 32 eM© GKK (Ans.) 12. hw` 2x + by + 4 = 0, 4x – y – 2b = 0 Ges 3x + y – 1 = 0 †iLvÎq mgwe›`y nq, Z‡e b Gi gvb ̧‡jv wbY©q Ki| [BUET 01-02; KUET 09-10] mgvavb: mgwe›`y n‡j,       2 4 3 b – 1 1 4 – 2b – 1 = 0 2(1 + 2b) – b(– 4 + 6b) + 4(4 + 3) = 0  2 + 4b + 4b – 6b2 + 28 = 0  3b2 – 4b – 15 = 0  b = 3, – 5 3 (Ans.) 13. A(2, 1) I B(5, 2) we›`y؇qi ms‡hvRK †iLv‡K mg‡Kv‡Y mgwØLwÐZ K‡i Giƒc †iLvi mgxKiY wbY©q Ki| [BUET 96-97] mgvavb: AB Gi ga ̈we›`y     2 + 5 2  1 + 2 2      7 2  3 2 AB Gi Xvj = 2 – 1 5 – 2 = 1 3  j¤^ †iLvi Xvj = – 3  wb‡Y©q †iLvi mgxKiY, y – 3 2 = – 3     x – 7 2  3x + y – 12 = 0 (Ans.) 14. ABC wÎfz‡Ri kxl©we›`yÎq A(6, 2), B(– 3, 8) Ges C(– 5, – 3) n‡j, A we›`y w`‡q AwZμgKvix D”PZv wb‡`©kK †iLvi mgxKiY wbY©q Ki| [BUET 96-97; KUET 04-05] mgvavb: B(– 3, 8) D C(– 5, – 3) A(6, 2) BC †iLvi mgxKiY, y – 8 – 3 – 8 = x + 3 – 5 + 3  11x – 2y + 49 = 0 ........... (i) A(6, 2) we›`yMvgx I (i) bs †iLvi Dci j¤^ †iLvi mgxKiY, 2x + 11y = (2  6) + (11  2)  2x + 11y = 34 (Ans.) 15. ABC wÎfz‡Ri fi‡K›`a G     0 1 3 Ges Bnvi `yBwU kxl©we›`y A(– 1, 0) Ges B(1, 0)| †`LvI †h, AD ga ̈gv BC evûi Dci j¤^| [BUET 95-96] mgvavb: B(1, 0) D C(x, y) G( ) 0 1 3 A(– 1, 0) E(0, 0) 2 1