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Answered by shweta8525
4

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+(30) = 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 ≠ 6 ————————— equation not satisfied

(-1, 3) ⟹ -1+(2×3) = 6

5 ≠ 6 ————————— equation not satisfied

Since, only equation y = –x+2satisfies all the points, the equation whose graphs are given in Fig. 4.7 is

y = –x+2

Step-by-step explanation:

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