Tiêu đề Straight Line Engineering Practice Sheet Solution.pdf ✅

mij‡iLv  Engineering Practice Sheet Solution 1 03 mij‡iLv Straight Line WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1| GKwU mij‡iLv x I y A‡ÿi †hvM‡evaK w`‡Ki mv‡_ mgvb Ask †Q` K‡i| g~j we›`y †_‡K H mij‡iLvi j¤^ `~iZ¡ 7 GKK n‡j, mij‡iLvi mgxKiY wbY©q Ki| [BUET 23-24] mgvavb: GLv‡b, OA = OB   = 45  mij‡iLvi j¤^ AvK...wZi mgxKiY: xcos45 + ysin45 = 7  x 2 + y 2 = 7  x + y = 7 2 O  A● ●P ●B 2| (k, – 3k), (5, k) Ges (– k, 2) we›`yÎq Øviv Drcbœ wÎfz‡Ri †ÿÎdj 28 (twenty eight) 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¤^ e‡j, 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 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 x sec – y cosec = k Ges x cos – y sin = k cos2 †iLv؇qi j¤^ `~iZ¡ h_vμ‡g 2 cm Ges 3 cm| k Gi gvb wbY©q Ki| [BUET 19-20; MIST 19-20] mgvavb: g~jwe›`y (0, 0) †_‡K x sec – y cosec – k = 0 †iLvi Dci j¤^`~iZ¡,      0 × sec – 0 × cosec – k sec2  + cosec2  = 2  k 2 1 cos2  + 1 sin2  = 4 [eM© K‡i]  k 2 sin2  + cos2  cos2  × sin2  = 4
2  Higher Math 1st Paper Chapter-3  k 2 cos2  . sin2  = 4  4k2 sin2  cos2  = 16  (k sin 2) 2 = 42 ............ (i) g~j we›`y (0, 0) †_‡K x cos – y sin – k cos2 = 0 †iLvi Dci j¤^ `~iZ¡,       0 × cos – 0 × sin – k cos 2 cos2  + sin2  = 3      – k cos2 1 = 3  (k cos2) 2 = 32 ........... (ii) (i) + (ii)  k 2 (sin2 2 + cos2 2) = 25  k =  5 (Ans.) 5| A (1, 2) kxl© wewkó e‡M©i GKwU KY© 3x – 4y – 6 = 0 n‡j, A we›`yMvgx evû؇qi mgxKiY wbY©q Ki| [BUET 18-19; MIST 18-19] mgvavb: A (1, 2) we›`y Øviv 3x – 4y – 6 = 0 †iLv wm× bq| †iLvi Xvj, m1 = 3 4 , tan 45 =  m1 – m2 1 + m1m2  1 =  3 4 – m2 1 + 3m2 4  1 =  3 – 4m2 4 4 + 3m2 4 A (1, 2) 45  1 =  3 – 4m2 4 + 3m2 (+) wb‡q, m2 = – 1 7 (–) wb‡q, m2 = 7 myZivs, wb‡Y©q †iLv, (y – 2) = 7(x – 1) Ges (y – 2) = – 1 7 (x – 1) (Ans.) 6| `ywU mij‡iLv (–1, 2) we›`y w`‡q hvq Ges Zviv 3x – y + 7 = 0 †iLvi mv‡_ 45 †KvY Drcbœ K‡i| †iLv `ywUi mgxKiY wbY©q Ki Ges Zv‡`i mgxKiY n‡Z †`LvI †h, Zviv ci ̄úi j¤^fv‡e Ae ̄’vb K‡i| [BUET 16-17] mgvavb: awi, mij‡iLvØq, y – 2 = m(x + 1); 3x – y + 7 = 0 Gi Xvj = 3 tan45 =  m – 3 1 + 3m  1 =  m – 3 1 + 3m  1 + 3m =  (m – 3) (+)  m – 3 = 1 + 3m  m = –2 (–) – m + 3 = 1 + 3m  m = 1 2 m = – 2 n‡j, †iLvwU y – 2 = – 2 (x + 1)  2x + y = 0 m = 1 2 n‡j, †iLvwU y – 2 = 1 2 (x + 1)  x – 2y + 5 = 0 †h‡nZz, †iLv؇qi Xvj –2 I 1 2 Ges m1.m2 = – 2. 1 2 = –1 myZivs, †iLvØq ci ̄úi j¤^| (Showed) 7| GKwU mij‡iLv (1, 4) we›`y w`‡q hvq Ges Aÿ؇qi mv‡_ cÖ_g PZzf©v‡M 8 eM© GKK †ÿÎdj wewkó GKwU wÎfzR MVb K‡i Zvi mgxKiY wbY©q Ki| [BUET 15-16] mgvavb: GLv‡b, 1 2 ab = 8  ab = 16 .......... (i) Avevi, x a + y b = 1  bx + ay = 16  b + 4a = 16 [†h‡nZz †iLvwU (1, 4) we›`yMvgx]  b = 4 (4 – a) ............ (ii) (i) I (ii) bs n‡Z, 4a (4 – a) = 16  4a – a 2 = 4  a 2 – 4a + 4 = 0  a = 2  b = 4 (4 – 2) = 8  wb‡Y©q mij‡iLvi mgxKiY: x 2 + y 8 = 1 (Ans.) 8| 2x – 3y + 4 = 0 Ges 2y – 3x – 1 = 0 mij‡iLv؇qi ga ̈eZ©x †KvY ̧‡jvi mgwØLÐKmg~n h_vμ‡g x Aÿ‡K P, R Ges y Aÿ‡K Q, S we›`y‡Z †Q` K‡i| GKwU mij‡iLvi mgxKiY wbY©q Ki hv mgwØLÐKmg~‡ni †Q`we›`y w`‡q hvq Ges PS †iLvi mgvšÍivj| [BUET 14-15] mgvavb: 2x – 3y + 4 2 2 + (–3) 2 =  – 3x + 2y – 1 (–3) 2 + 22  2x – 3y + 4 = – 3x + 2y – 1 [+ wPý wb‡q]  5x + 5 = 5y  x – 1 + y 1 = 1 ...... (i)  P (– 1,0), Q(0,1) 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.)
mij‡iLv  Engineering Practice Sheet Solution 3 9| 3x + 4y = 11 Ges 12x – 5y = 2 †iLv؇qi AšÍfz©3 m~2‡Kv‡Yi mgwØLЇKi mgxKiY wbY©q Ki| [BUET 06-07] 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| 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.) 11| 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.) 12| (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.) 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 awi, C (x, y) fi‡K‡›`ai m~Î n‡Z, x + 1 – 1 3 = 0  x = 0 y + 0 + 0 3 = 1 3  y = 3  AD †iLvi Xvj = 1 3 – 0 0 + 1 = 1 3  BC †iLvi Xvj = 0 – 3 0 + 1 = – 3  AD †iLvi Xvj × BC †iLvi Xvj = 1 3 × (– 3 ) = –1  AD  BC (Showed)
4  Higher Math 1st Paper Chapter-3 weMZ mv‡j KUET-G Avmv cÖkœvejx 16| LM mij‡iLvwU g~jwe›`y n‡Z 5 GKK `~ieZ©x Ges x I y Aÿ‡K h_vμ‡g A I B we›`y‡Z †Q` K‡i| g~jwe›`y n‡Z LM Gi Dci Aw1⁄4Z j¤^ y A‡ÿi †hvM‡evaK w`‡Ki mv‡_  3 †KvY Drcbœ K‡i| C we›`yi ̄’vbv1⁄4 (– 1, – 2) n‡j ABC wÎfz‡Ri †ÿÎdj wbY©q Ki| [KUET 19-20] mgvavb: y y x x LM LM B 5 5  3  3 C(–1,–2) A A   LM 1g PZzf©v‡M Dfq Aÿ‡K †Q` Ki‡j,  = 90 – 60 = 30 j‡¤^i •`N© ̈ p = 5 GKK  LM = x cos + y sin = p  x cos30 + y sin30 = 5  x 10 3 + y 10 = 1  A     10 3  0 , B(0, 10), C(–1, – 2)  ABC = 1 2       10 3 0 1 0 10 1 –1 –2 1 = 5 + 20 3 eM© GKK (Ans.) LM 2q PZzf©v‡M Dfq Aÿ‡K †Q` Ki‡j,  = 90 + 60 = 150  LM = x cos150 + y sin150 = p  x – 10 3 + y 10 = 1  A     – 10 3  0 , B(0, 10), C(–1, –2)  ABC = 1 2      –  10 3 0 1 0 10 1 –1 –2 1 = 20 3 – 5 eM© GKK (Ans.) 17| hw` 3x + by + 1 = 0 Ges ax + 6y + 1 = 0 mij †iLvØq (5, 4) we›`y‡Z †Q` K‡i, Z‡e a Ges b Gi gvb wbY©q Ki| hw` cÖ_g †iLvwU x Aÿ‡K A we›`y‡Z Ges wØZxq †iLvwU y Aÿ‡K B we›`y‡Z †Q` K‡i, Z‡e AB mij‡iLvi mgxKiY wbY©q Ki| [KUET 19-20] mgvavb: †iLvØq (5, 4) we›`yMvgx e‡j, 3.5 + b.4 + 1 = 0  b = – 4 a.5 + 6.4 + 1 = 0  a = – 5 cÖ_g †iLv: 3x – 4y + 1 = 0  x –1 3 + y 1 4 = 1  A     – 1 3  0 wØZxq †iLv: – 5x + 6y + 1 = 0  x 1 5 + y – 1 6 = 1  B     0  – 1 6 AB Gi mgxKiY, y – 0 = – 1 6 – 0 0 + 1 3     x + 1 3  – 6y = 3x + 1  3x + 6y + 1 = 0 (Ans.) 18| 3x + 4y = 11 Ges 12x – 5y = 2 †iLv؇qi AšÍe©Z©x m~2‡Kv‡Yi mgwØLЇKi mgxKiY wbY©q Ki| [KUET 13-14, 06-07,; BUET 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.) 19| 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| [KUET 09-10; BUET 01-02] 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.) 20| 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| [KUET 04-05; BUET 96-97] mgvavb: B(–3, 8) D C(–5, –3) A(6, 2)