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Ex-4.3 ka 5 question number..

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Ex-4.3 ka 5 question number..
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Shehrebanu .... 4 years, 2 months ago

For Fig. 4. 6 (i) y = x (ii) x + y = 0 (iii) y = 2x (iv) 2 + 3y = 7x Solution: The points given in the figure 4.6 are (0,0), (-1,1), (1,-1) Substituting the values for x and y from these points in the equations, we get, (i) y = x (0,0) ⟹ 0 = 0 (-1, 1) ⟹ -1≠1 --------------------------- equation not satisfied (1, -1) ⟹ 1≠ -1 --------------------------- equation not satisfied (ii) x + y = 0 (0,0) ⟹ 0 + 0 = 0 (-1, 1) ⟹ -1+1= 0 (1, -1) ⟹ 1+ (-1)=0 (iii) y = 2x (0,0) ⟹ 0 = 2× 0 0 = 0 (-1, 1) ⟹ 1= 2×(-1) 1 ≠ -2 --------------------------- equation not satisfied (1, -1) ⟹ -1=2×1 -1≠2 --------------------------- equation not satisfied (iv) 2 + 3y = 7x (0,0) ⟹ 2+(3×0) = 7× 0 2 ≠ 0 --------------------------- equation not satisfied (-1, 1) ⟹ 2+(3×1) = 7× -1 5 ≠ -7 --------------------------- equation not satisfied (1, -1) ⟹ 2+(3×-1)= 7× 1 -1≠ 7 --------------------------- equation not satisfied Since, only equation x + y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is x + y = 0 For Fig. 4. 7 (i) y = x + 2 (ii) y = x – 2 (iii) y = –x + 2 (iv) x + 2y = 6 Solution: The points given in the figure 4.7 are (0,2), (2,0), (-1,3) Substituting the values for x and y from these points in the equations, we get, (i) y = x + 2 (0,2) ⟹ 2 = 0+2 2 = 2 (2, 0) ⟹ 0= 2+2 0≠4 --------------------------- equation not satisfied (-1, 3) ⟹ 3= -1+2 3≠1 --------------------------- equation not satisfied (ii) y = x – 2 (0,2) ⟹ 2 = 0 –2 2 ≠ – 2 --------------------------- equation not satisfied (2, 0) ⟹ 0= 2–2 0= 0 (-1, 3) ⟹ 3= –1–2 3≠ –3 --------------------------- equation not satisfied (iii) y = –x + 2 (0,2) ⟹ 2 = -0+2 2 = 2 (2, 0) ⟹ 0 = -2+2 0=0 (-1, 3) ⟹ 3= -(-1)+2 3=3 (iv) x + 2y = 6 (0,2) ⟹ 0+(2× 2) = 6 4 ≠ 6 --------------------------- equation not satisfied (2, 0) ⟹ 2+(2× 0) = 6 2 ≠ 64 --------------------------- equation not satisfied (-1, 3) ⟹ -1+(2× 3) = 6 5≠6 --------------------------- equation not satisfied Since, only equation y = –x + 2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is y = –x + 2.
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