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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)

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