Kopertina e librit Matematika 12

Zgjidhja e ushtrimit 1

Zgjidhja e ushtrimit 1 të mësimit 6A në librin Matematika 12 nga shtëpia botuese Mediaprint me autorë Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Bledger, Harry Smith, Geoff Staley, Robert Ward-Penny dhe Dave Wilkins.

Pyetja

Gjej pikën e mesit të segmentit që ka skaje secilin çift pikash:

  1. (4,2)(4, 2),, (6,8)(6, 8)
  2. (0,6)(0,6),, (12,2)(12, 2)
  3. (2,2),(−4,6)(2, 2), (-4, 6)
  4. (−6,4),(6,−4)(-6, 4), (6, -4)
  5. (7,−4)(7, -4),, (−3,6)(-3, 6)
  6. (−5,−5),(−11,8)(-5, -5), (-11, 8)
  7. (6a,4b),(2a,−4b)(6a, 4b), (2a, -4b)
  8. (−4u,0),(3u,−2v)(-4u, 0), (3u, -2v)
  9. (a+b,2a−b),(3a−b,−b)(a+b, 2a-b), (3a-b, -b)
  10. (42,1),(22,7)(4\sqrt{2}, 1), (2\sqrt{2}, 7)
  11. (2−3,32+43),(32+3,−2+23)(\sqrt{2}-\sqrt{3}, 3\sqrt{2}+4\sqrt{3}), (3\sqrt{2}+\sqrt{3}, -\sqrt{2}+2\sqrt{3})

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Zgjidhja

  1. (x1,y1)=(4,2),(x2,y2)=(6,8)⇒(x_1, y_1)=(4, 2), (x_2, y_2)=(6, 8) \rArr  (x1+x22,y1+y22)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) =(4+62,2+82)=(102,102)=(5,5)= \Big(\dfrac{4+6}{2}, \dfrac{2+8}{2}\Big) = \Big(\dfrac{10}{2}, \dfrac{10}{2}\Big) = (5, 5)
  2. (x1,y1)=(0,6),(x2,y2)=(12,2)⇒(x_1, y_1) = (0, 6), (x_2, y_2) = (12, 2) \rArr (x1+x22,y1+y22)=(0+122,6+22)=(122,82)=(6,4)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{0+12}{2}, \dfrac{6+2}{2}\Big) = \Big(\dfrac{12}{2}, \dfrac{8}{2}\Big) = (6, 4)
  3. (x1,y1)=(2,2),(x2,y2)=(−4,6)⇒(x_1, y_1) = (2, 2), (x_2, y_2) = (-4,6) \rArr (x1+x22,y1+y22)=(2+(−4)2,2+62)=(−22,82)=(−1,4)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{2+(-4)}{2}, \dfrac{2+6}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{8}{2}\Big) = (-1, 4)
  4. (x1,y1)=(−6,4),(x2,y2)=(6,−4)(x_1, y_1) = (-6, 4), (x_2, y_2) = (6, -4) ⇒\rArr (x1+x22,y1+y22)=(−6+62,4+(−4)2)=(02,02)=(0,0)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{-6+6}{2}, \dfrac{4+(-4)}{2}\Big) = \Big(\dfrac{0}{2}, \dfrac{0}{2}\Big) = (0, 0)
  5. (x1,y1)=(7,−4),(x2,y2)=(−3,6)⇒(x_1, y_1) = (7, -4), (x_2, y_2)=(-3,6) \rArr (x1+x22,y1+y22)=(7+(−3)2,−4+62)=(42,22)=(2,1)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{7+(-3)}{2}, \dfrac{-4+6}{2}\Big) = \Big(\dfrac{4}{2}, \dfrac{2}{2}\Big) = (2, 1)
  6. (x1,y1)=(−5,−5),(x2,y2)=(−11,8)⇒(x_1, y_1) = (-5, -5), (x_2, y_2) = (-11, 8) \rArr (x1+x22,y1+y22)=(−5+(−11)2,−4+62)=(−162,32)=(−8,32)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{-5+(-11)}{2}, \dfrac{-4+6}{2}\Big) = \Big(\dfrac{-16}{2}, \dfrac{3}{2}\Big) = \Big(-8, \dfrac{3}{2}\Big)
  7. (x1,y1)=(6a,4b),(x2,y2)=(2a,−4b)⇒(x_1, y_1)=(6a,4b), (x_2, y_2)=(2a,-4b) \rArr (x1+x22,y1+y22)=(6a+2a2,4b+(−4b)2)=(8a2,02)=(4a,0)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{6a+2a}{2}, \dfrac{4b+(-4b)}{2}\Big) = \Big(\dfrac{8a}{2}, \dfrac{0}{2}\Big) = (4a, 0)
  8. (x1,y1)=(−4u,0),(x2,y2)=(3u,−2v)⇒(x_1, y_1) = (-4u, 0), (x_2, y_2) = (3u, -2v) \rArr (x1+x22,y1+y22)=(−4u+3u2,0+(−2v)2)=(−u2,−v)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{-4u+3u}{2}, \dfrac{0+(-2v)}{2}\Big) = \Big(\dfrac{-u}{2}, -v\Big)
  9. (x1,y1)=(a+b,2a−b),(x2,y2)=(3a−b,−b)⇒(x_1, y_1) = (a+b, 2a-b), (x_2, y_2) = (3a-b, -b) \rArr (x1+x22,y1+y22)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) =(a+b+3a−b2,2a−b+(−b)2)=(4a2,2a−2b2)=(2a,a−b)= \Big(\dfrac{a+b+3a-b}{2}, \dfrac{2a-b+(-b)}{2}\Big) = \Big(\dfrac{4a}{2}, \dfrac{2a-2b}{2}\Big) = (2a, a-b)
  10. (x1,y1)=(42,1),(x2,y2)=(22,7)⇒(x_1, y_1) = (4\sqrt{2}, 1), (x_2, y_2) = (2\sqrt{2}, 7) \rArr (x1+x22,y1+y22)=(42+222,1+72)=(622,82)=(32,4)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{4\sqrt{2}+2\sqrt{2}}{2}, \dfrac{1+7}{2}\Big) = \Big(\dfrac{6\sqrt{2}}{2}, \dfrac{8}{2}\Big) = (3\sqrt{2}, 4)
  11. (x1,y1)=(2−3,32+43),(x2,y2)=(32+3,−2+23)⇒(x_1, y_1) = (\sqrt{2}-\sqrt{3}, 3\sqrt{2}+4\sqrt{3}), (x_2, y_2) = (3\sqrt{2}+\sqrt{3}, -\sqrt{2}+2\sqrt{3}) \rArr (x1+x22,y1+y22)=(2−3+32+32,32+43+(−2+23)2)=(422,22+632)=(22,2+33)\Big(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Big) = \Big(\dfrac{\sqrt{2}-\sqrt{3}+3\sqrt{2}+\sqrt{3}}{2}, \dfrac{3\sqrt{2}+4\sqrt{3}+(-\sqrt{2}+2\sqrt{3})}{2}\Big) = \Big(\dfrac{4\sqrt{2}}{2}, \dfrac{2\sqrt{2}+6\sqrt{3}}{2}\Big) = (2\sqrt{2}, \sqrt{2}+3\sqrt{3})