Nội dung text Straight Line CQ & MCQ Practice Sheet Solution (HSC 26).pdf
mij‡iLv CQ & MCQ Practice Sheet Solution (HSC 26) 1 03 mij‡iLv Straight Line Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 1 1 1 1 1 1 1 2 2022 1 1 1 1 1 1 1 1 2 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 5 4 4 5 4 4 4 8 4 2022 5 4 6 5 5 5 5 4 4 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| Y X O B(– 1, 5) P(1, – 1) A(3, – 1) [XvKv †evW©- Õ23] (K) DÏxc‡Ki AB mij‡iLvwU Y Aÿ‡K †h we›`y‡Z †Q` K‡i Zvi wbY©q Ki| (L) P we›`yMvgx Ges AB mij‡iLvi mv‡_ 45 †KvY Drcbœ K‡i Giƒc mij‡iLv؇qi mgxKiY wbY©q Ki| (M) AB Gi Dci j¤^‡iLvi mgxKiY wbY©q Ki hv P we›`y †_‡K 2 GKK `~‡i Aew ̄’Z| mgvavb: (K) GLv‡b, A(3, – 1) Ges B(– 1, 5) we›`yMvgx AB mij‡iLvi mgxKiY, x – 3 3 – (– 1) = y – (– 1) – 1 – 5 x – 3 4 = y + 1 – 6 – 6x + 18 = 4y + 4 – 6x + 18 – 4y – 4 = 0 – 6x – 4y + 14 = 0 3x + 2y – 7 = 0 3x + 2y = 7 x 7 3 + y 7 2 = 1 Y A‡ÿi †Q`we›`y 0 7 2 (Ans.) (L) ÔKÕ n‡Z cÖvß, AB mij‡iLvi mgxKiY, 3x + 2y – 7 = 0 ......(i) 2y = – 3x + 7 y = – 3 2 x + 7 2 AB †iLvi Xvj = – 3 2 P(1, – 1) we›`yMvgx †h‡Kv‡bv †iLvi mgxKiY, y – (– 1) = m(x – 1) y + 1 = m(x – 1).....(ii) †h‡nZz (i) bs †iLvi mv‡_ (ii) bs †iLv 45 †KvY Drcbœ K‡i †m‡nZz, tan45 = m + 3 2 1 – 3 2 m 1 = 2m + 3 2 2 – 3m 2 1 = 2m + 3 2 – 3m 2 – 3m = (2m + 3) (+) wPý wb‡q, 2 – 3m = 2m + 3 m = – 1 5 Ges (–) wPý wb‡q, 2 – 3m = – 2m – 3 m = 5 (ii) bs mgxKi‡Y m Gi gvb ewm‡q cvB, m = – 1 5 n‡j,
2 Higher Math 1st Paper Chapter-3 y + 1 = – 1 5 (x – 1) 5y + 5 = – (x – 1) x + 5y + 4 = 0 (Ans.) m = 5 n‡j, y + 1 = 5(x – 1) y + 1 = 5x – 5 5x – y – 6 = 0 (Ans.) (M) ÔKÕ n‡Z cÖvß, AB †iLvi mgxKiY, 3x + 2y – 7 = 0 AB †iLvi j¤^ †iLvi mgxKiY, 2x – 3y + k = 0 ...(i) P(1, – 1) we›`y n‡Z (i) bs †iLvi j¤^`~iZ¡, 2 1 – 3(– 1) + k 2 2 + (– 3) 2 = 2 2 + 3 + k 4 + 9 = 2 5 + k 13 = 2 (5 + k) 2 13 = 4 (5 + k2 ) = 4 13 5 + k = 2 13 k = 2 13 – 5 k Gi gvb (i) bs mgxKi‡Y ewm‡q cvB, 2x – 3y 2 13 – 5 = 0 wb‡Y©q j¤^ †iLvi mgxKiY, 2x – 3y 2 13 – 5 = 0 (Ans.) 2| `„k ̈Kí-1: B(1, – 1) C(6, 2) A(– 2, 4) Y D Y X X O AD || BC, ACD = 90 [ivRkvnx †evW©- Õ23] `„k ̈Kí-2: A(8, 3) Ges B = (p, q), AB Gi j¤^ mgwØLÛ‡Ki mgxKiY y = – 2x + 4. (K) x – 2y + 1 = 0 Ges 3x – y + 5 = 0 mij‡iLv؇qi AšÍfz©3 m~2‡KvY wbY©q Ki| (L) `„k ̈Kí-1 n‡Z D we›`yi ̄’vbvsK wbY©q Ki| (M) `„k ̈Kí-2 e ̈envi K‡i p Ges q Gi gvb wbY©q Ki| mgvavb: (K) cÖ`Ë mgxKiYØq, x – 2y + 1 = 0 .....(i) 2y = x + 1 y = 1 2 x + 1 2 †iLvwUi Xvj, m1 = 1 2 Ges 3x – y + 5 = 0 .....(ii) y = 3x + 5 †iLvwUi Xvj, m2 = 3 †iLv؇qi ga ̈eZ©x m~2‡KvY n‡j, tan = m1 – m2 1 + m1m2 tan = 1 2 – 3 1 + 1 2 3 = 1 – 6 2 2 + 3 2 = – 5 5 = |– 1| = 1 tan = 1 = tan–1 (1) = 45(Ans.) (L) GLv‡b, B(1, – 1) Ges C(6, 2) BC †iLvi mgxKiY, x – 1 1 – 6 = y – (– 1) – 1 – 2 x – 1 – 5 = y + 1 – 3 3x – 3 = 5y + 5 3x – 5y – 8 = 0 †h‡nZzAD || BC A(– 2, 4) we›`yMvgx 3x – 5y – 8 = 0 †iLvi mgvšÍivj AD †iLvi mgxKiY, 3x – 5y = 3 (– 2) – 5 4 3x – 5y + 26 = 0....(i) Avevi, A(– 2, 4) Ges C(6, 2) we›`yMvgx AC †iLvi mgxKiY, x – (– 2) – 2 – 6 = y – 4 4 – 2 x + 2 – 8 = y – 4 2 2x + 4 = – 8y + 32 2x + 8y – 28 = 0 x + 4y – 14 = 0 Avevi, AC CD x + 4y – 14 = 0 †iLvi Dci j¤^ C(6, 2) we›`yMvgx CD †iLvi mgxKiY, 4x – y = 4 6 – 1 2 4x – y – 22 = 0 y = 4x – 22 ...(ii) D we›`ywU n‡e AD I CD †iLvi †Q`we›`y| y Gi gvb (i) bs mgxKi‡Y ewm‡q, 3x – 20x + 110 + 26 = 0 x = 8 y = 4 8 – 22 [(ii) bs n‡Z] = 10 wb‡Y©q we›`y D(8, 10) (Ans.)
mij‡iLv CQ & MCQ Practice Sheet Solution (HSC 26) 3 (M) GLv‡b, AB Gi j¤^ wØLÛ‡Ki mgxKiY, y = – 2x + 4 .....(i) Xvj = – 2 AB †iLvi Xvj = 1 2 †`Iqv Av‡Q, A(8, 3) Ges B(p, q) Xvj, q – 3 p – 8 = 1 2 2q – 6 = p – 8 p – 2q – 2 = 0 .....(ii) (i) bs n‡Z 2x + y – 4 = 0 †iLvi j¤^ A(8, 3) we›`yMvgx AB †iLvi mgxKiY, x – 2y = 8 – 6 x – 2y – 2 = 0 x = 2y + 2.....(iii) AB †iLv I j¤^ mgwØLÛ‡Ki †Q`we›`y AB †iLvi ga ̈we›`y| x Gi gvb (i) bs ewm‡q, y = – 2 (2y + 2) + 4 4y + y + 4 – 4 = 0 y = 0 x = 2 0 + 2 = 2 [(iii) bs n‡Z] AB †iLvi ga ̈we›`yi ̄’vbvsK 8 + p 2 3 + q 2 8 + p 2 = 2 Ges 3 + q 2 = 0 8 + p = 4; 3 + q = 0 p = – 4 Ges q = – 3 (Ans.) 3| B(0, 7) C(– 4, 5) A(5, 0) Y D Y X X [h‡kvi †evW©- Õ23] (K) AB mij‡iLvi Aÿ؇qi ga ̈eZ©x LwÛZvs‡ki wÎLÛb we›`y wbY©q Ki| (L) (7, 9) we›`yMvgx Ges AB †iLvi mv‡_ 45 †KvY Drcbœ K‡i Giƒc mij‡iLvi mgxKiY wbY©q Ki| (M) D we›`yi ̄’vbvsK wbY©q Ki| mgvavb: (K) GLv‡b, AB mij‡iLvwU x Aÿ‡K A(5, 0) we›`y‡Z Ges y Aÿ‡K B(0, 7) we›`y‡Z †Q` K‡i‡Q| B(0, 7) A(5, 0) Y Y X X O Q P awi, P I Q we›`yØq AB †iLvi mgwÎLÛb we›`y| Zvn‡j, P we›`y AB †iLv‡K 1 : 2 Abycv‡Z AšÍwe©f3 K‡i| P we›`yi ̄’vbvsK = 1 0 + 2 5 1 + 2 1 7 + 2 0 1 + 2 = 10 3 7 3 Avevi, Q we›`yAB †iLv‡Z 2 : 1 Abycv‡Z AšÍwef©3 K‡i| Q we›`yi ̄’vbvsK = 2 0 + 1 5 2 + 1 2 7 + 1 0 2 + 1 = 5 3 14 3 wb‡Y©q wÎLÛb we›`yØq 10 3 7 3 , 5 3 14 3 (Ans.) (L) GLv‡b, A(5, 0) Ges B(0, 7) AB †iLvi mgxKiY, x – 5 5 – 0 = y – 0 0 – 7 x – 5 5 = y – 7 – 7x + 35 = 5y 7x + 5y – 35 = 0 .....(i) 5y = – 7x + 35 y = – 7 5 + 7 AB †iLvi Xvj = – 7 5 (7, 9) we›`yMvgx m Xvj wewkó †h‡Kv‡bv †iLvi mgxKiY, y – 9 = m(x – 7) .....(ii) (ii) bs †iLvwU (i) bs †iLvi mv‡_ 45 †KvY Drcbœ K‡i, tan45 = m + 7 5 1 – m . 7 5 1 = 5m + 7 5 5 – 7m 5 1 = 5m + 7 5 – 7m 5 – 7m = (5m + 7) (+) wPý wb‡q cvB, 5 – 7m = 5m + 7 12m = – 2 m = – 1 6 (–) wPý wb‡q cvB, 5 – 7m = – 5m – 7 7m – 5m = 5 + 7 2m = 12 m = 6 m Gi gvb (ii) bs mgxKi‡Y ewm‡q cvB, m = – 1 6 n‡j, y – 9 = – 1 6 (x – 7)
4 Higher Math 1st Paper Chapter-3 x + 6y – 54 – 7 = 0 x + 6y – 61 = 0 (Ans.) m = 6 n‡j, y – 9 = 6(x – 7) y – 9 = 6x – 42 6x – y – 42 + 9 = 0 6x – y – 33 = 0 (Ans.) (M) ÔLÕ n‡Z cÖvß, AB †iLvi mgxKiY, 7x + 5y – 35 = 0 .....(i) Ges AB †iLvi Xvj = – 7 5 Avevi, AD CD CD †iLvi Xvj = 5 7 (– 4, 5) we›`yMvgx Ges 5 7 Xvj wewkó †iLvi mgxKiY, y – 5 = 5 7 (x + 4) 7y – 35 = 5x + 20 5x – 7y + 55 = 0 ......(ii) GLb, (i) I (ii) bs †iLvi †Q`we›`yB n‡e D we›`yi ̄’vbvsK| (i) 7 + (ii) 5 K‡i cvB, 49x + 35y – 245 = 0 25x – 35y + 275 = 0 74x + 30 = 0 x = – 15 37 x Gi gvb (i) bs mgxKi‡Y ewm‡q cvB, 7 – 15 37 + 5y – 35 = 0 – 105 37 + 5y – 35 = 0 5y = 35 + 105 37 = 1400 37 y = 1400 37 5 y = 280 37 †Q`we›`yi ̄’vbvsK D – 15 37 280 37 (Ans.) 4| Y Y X X C B O A 4x – 3y + 12 = 0 3x – 4y = 8 [Kzwgjøv †evW©- Õ23] (K) OAB Gi †ÿÎdj wbY©q Ki| (L) Giƒc GKwU †iLvi mgxKiY wbY©q Ki hv C we›`yMvgx Ges x – y + 2 = 0 †iLvi mgvšÍivj| (M) †`LvI †h, DÏxc‡Ki †iLv؇qi AšÍf©y3 †Kv‡Yi mgwØLÛK؇qi ci ̄úi j¤^| mgvavb: (K) GLv‡b, AB †iLvi mgxKiY, 3x – 4y = 8 x 8 3 + y 8 – 4 = 1 x 8 3 + y – 2 = 1 AB †iLvwU x Aÿ‡K A 8 3 0 Ges y Aÿ‡K B(0, – 2) we›`y‡Z †Q` K‡i| OA = 8 3 GKK Ges OB = |– 2| GKK = 2 GKK OAB Gi †ÿÎdj = 1 2 OA OB = 1 2 8 3 2 eM© GKK = 8 3 eM© GKK (Ans.) (L) wPÎ Abymv‡i C we›`ywU, 4x – 3y + 12 = 0 ....(i) Ges 3x – 4y – 8 = 0 ....(ii) †iLv؇qi †Q`we›`y| (i) 4 – (ii) 3 K‡i cvB, 16x – 12y + 48 = 0 9x – 12y – 24 = 0 (–) (+) (+) we‡qvM K‡i 7x + 72 = 0 x = – 72 7 x Gi gvb (i) bs mgxKi‡Y ewm‡q cvB, 3 – 72 7 – 4y – 8 = 0 – 216 7 – 4y – 8 = 0 4y = – 216 7 – 8 = – 272 7 y = – 272 7 4 y = – 68 7 †Q`we›`yC – 72 7 – 68 7 x – y + 2 = 0 †iLvi mgvšÍivj C – 72 7 – 68 7 we›`yMvgx †iLvi mgxKiY, x – y = – 72 7 + 68 7 x – y = – 4 7 7x – 7y + 4 = 0 (Ans.) (M) cÖ`Ë †iLvØq, 3x – 4y = 8 3x – 4y – 8 = 0.....(i) Ges 4x – 3y + 12 = 0 ....(ii) (i) bs I (ii) bs †iLv؇qi AšÍfz©3 †Kv‡Yi mgwØLÛK †iLv؇qi mgxKiY, 3x – 4y – 8 3 2 + (– 4) 2 = 4x – 3y + 12 4 2 + (– 3) 2 3x – 4y – 8 = (4x – 3y + 12)