Tiêu đề Straight Line Practice Sheet Solution (HSC FRB 25).pdf ✅

mij‡iLv  Final Revision Batch '25 1 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

mij‡iLv  Final Revision Batch '25 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)