Tiêu đề XI - maths - chapter 10 - sol the straight line-i.pdf ✅

Nishith Multimedia India (Pvt.) Ltd. 68 IX-Mathematics e-Techno Text Book 1. Key: 1 2. Key: 1 Hint: AB + BC = AC (or) ABC 3. Key: 1 4. Key: 1 5. Key: 2 6. Key: 1, 3 7. Key: 1 8. Key: 1 9. Key: 3 10. Key: 4 11. Key: 4,4,2,(1,3) 12. Key: 3 13. Key: 2 A point P divides the straight line joining two points A x ,y  1 1  and B x ,y  2 2  internally in the ratio m : n is mx nx my ny 2 1 2 1 , m n m n           Since P divides the line segment joining points A (0,0) and B(9,12) internally in 1 : 2 ratio 9 1 0 2 9 x 3 3 3        , 12 1 2 0 12 y 4 3 3       The coordinates of point P are (3, 4). 14. Key: 1,4 1) Is correct since distance of (a,b) from B and C are equal 2) Is false since 2 2 AB AC  3) Is incorrect since 2 2 AB AC  4) Is correct since   2 2 2 b AB 1 a b b a a             and   2 2 2 b AC 1 a b a b a             Note that 2 2 AB AC  . 15. Required point = b a b a a b b a b a a b         , a b a b                   2 2 2 2 a b a 2ab b , a b a b             16. Key: 1 Let P is the point on x - axis which divides the line segment joining points in the ratio  :1, then, 5 3 6 4 x ,y 1 1            But for any point on x axis, its y- coordinate = 0 6 4 2 0 1 3          (OR) The x - axis divides the line segment joining the points (x1, y1) and (x2, y2) in the ratio –y1 : y2 = –(–4) : 6 = 4 :6 = 2 : 3
Nishith Multimedia India (Pvt.) Ltd. 69 IX-Mathematics e-Techno Text Book 17. Key: 1 18. Key: 1 Given AP = PQ = QB .........(1) Let R be the midpoint of PQ such that PR = RQ .........(2) Now AR = AP + PR AR = QB + QR = BR [ from (1) and (2)]  R is the midpoint of AB R = 2 3 5 1 , 2 2          1 ,3 2       . 19. Key: 4, 3, 1, 2 Hint: area of 1 1 p 1 2p 1 2p 3 p 1 2 ABC 2p 3p 2 2 2 1 3 2p 1          20. Key: 3 Let the ratio be m : n. P (x, y) = (x1 + t(x2 – x1), y1 + t(y2 – y1)) and A (x1, y1), B(x2, y2) Now, P(x1 + t(x2 – x1), y1 + t(y2 – y1)) =           m n my ny , m n mx2 nx1 2 1 m n mx nx x t(x x ) 2 1 1 2 1        x1 (m + n) + t (x2 – x1) (m + n) = mx2 + nx1  mx1 + nx1 + t (x2 – x1) (m + n) = mx2 + nx1      t m n x x mx mx   2 1 2 1       t m n x x m x x   2 1 2 1     m = mt + nt  nt = m(1 – t) 1 t t n m    21. Key: 3 The coordinates of the point of division are              1 2 1 2 1 2 1 1 2 1 1 t x tx 1 t y ty x t x x , y t y y , 1 t t 1 t t                    Clearly, these are the coordinates of the point dividing the join of x ,y 1 1  and x ,y 2 2  in the ratio t : (1 – t) For the internal division, we must have t > 0 and 1– t > 0    0 t 1. 1. The point to which the origin has to be shifted to eliminate x and y terms in the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is 2 2 hf bg gh af , ab h ab h           . 1 2( 2) (3/2) ( 7/2)(3/2) 2(1/2) ab h gh af 1; 2( 2) (3/2) (3/2)(1/2) 2( 7/2) ab h hf bg 2 2 2 2                 2. Key: 1 f g , ( 1, 1) h h            3. The point to which the origin has to be shifted to eliminate x and y terms in the equation ax2 + by2 + 2gx + 2fy + c = 0 is g f , a b        .
Nishith Multimedia India (Pvt.) Ltd. 70 IX-Mathematics e-Techno Text Book ( 1, 2) 1 2 , 1 1 b f , a g                     4. Key : 2,4 Sol: Here a = 9, b = 3, h =  3 angle of rotation 1 2h 1 2 3 1 1 tan tan 2 a b 2 9 3                     = 1 1 1 tan 2 12 3           n 5 when 1 2 12 12          5. Key - 4 Here a = 2, h = 3 2 , b = 3   1 1 1 3 1 2h 1 1 2 2 tan tan tan 3 2 a b 2 2 3 2                            1 2 2 6 6               6. Key: 4 Comparing the given equation with 2 2 ax 2hxy by 2gx 2fy c 0       we get a = 3, h = - 2, b = - 2, g = -3/2, f = - 1 required point is 2 2 hf bg gh af 2 3 3 3 1 3 , , , ab h ab h 6 4 6 4 10 5                                7. Here a = 1, b = -1, h = 0, g = 1, f = 2 Required point is   2 2 hf bg gh af 0 1 0 2 , , 1,2 ab h ab h 1 0 1 0                          8. Key: 4 Here a = 1, b = 1, h = 0, g = -a, f = -2a reqiored point is 2 2 hf bg gh af , ab h ab h           =   a 0 2a , a,2a 1 1         . 9. Key: 3,1,4,2 Sol: Let (x,y) be the new coordinates of (x,y) a) (x,y) = (x- h, y - k) = (7-1, 5-2) = (6,3) b) (x,y) = (-3,-1+1-2) = (-4,-1) c) (x,y) = (0-1,5-2) = (-1,3) d) (x,y) = (-1,-1,-2-2) = (-2,-4) 10. Key: 1 11. The first degree terms are missing. Hence, the new origin is ( 2,1) 1/2 1/2 , 1/2 1 h g , h f                    
Nishith Multimedia India (Pvt.) Ltd. 71 IX-Mathematics e-Techno Text Book The new equation is 1 xy ( 2) 1 6 0 xy 4 2         c = 4 12. New coordinates are (x –2, y + 1) 2(x – 2)2 + 7(y + 1)2 + 8(x – 2) – 14 (y + 1) + 15 = 0  2x2 + 7y2 = 0 13. Key: 1 Sol: (x,y) = x cos y sin ,xsin y cos         = 3 1 1 3 h. 2 3. ,4. 2 3. 2 2 2 2         =  3,5 14.  The transformed equation is 2(x – 2)2 + 4 (x – 2)(y + 3) + 5 (y + 3)2 – 4(x – 2) – 22 (y + 3) + 7 = 0  2(x2 – 4x + 4) + 4 (xy + 3x – 2y – 6) + 5 (y2 + 6y + 9) – 4x + 8 – 22y – 66 + 7 = 0  2x2 + 5y2 – 8x + 8 + 4xy +12x – 8y – 24 + 30y + 45 – 4x+ 8 – 22y – 59 =0  2x2 + 5y2 + 4xy – 22 = 0 15. Key: 3 16. Key: 2 17. New coordinates are (x + 2, y –2). The transformed equation is 2(x + 2)2 + (y – 2)2 – 8(x + 2) + 4(y – 2) + 1 = 0 2x2 + 8x +8 + y2 – 4y + 4 – 8x – 16 +4y – 8 +1 =0  2x2 + y2 – 11 = 0 18. Key: 3,2,4,1 19. Key: 2 2 2 y 4y 8x 12 y 4y 4 8x 8             2     y 2 8 x 1 If the origin is shifted to (1,2), then the equation becomes y2 = 8x But the given equation is y2 = 4ax  a = 2 20. Let (x, y) be the original coordinates of P. x = cos  + 1 and y = cos  + 1  (x, y) =         2 , 2cos 2 2cos2 2   2 1 cos2 2cos     21. Key: 2,4,1,3 Cumulative Assignment Date: 18-08-2010 1. Key: 1 The point to which the origin has to be shifted to eliminate x and y terms in the equation ax2 + by2 + 2gx + 2fy + c = 0 is g f , a b        . ( 2,1) 7 7 , 2 4 b f , a g                   2. Key: 2 The point to which the origin has to be shifted to eliminate x and y terms in the equation ax2 + by2 + 2gx + 2fy + c = 0 is g f , a b        .