1.

Find the distances with the help of the number line given below.i. d(B, E) ii. d (J, J) iii. d(P, C)iv. d(J, H) v. d(K, O)vi. d(O, E) vii. d(P, J)viii. d(Q, B)

Answer»

i. Co-ordinate of the point B is 2.

 Co-ordinate of the point E is 5.

 Since, 5 > 2

∴ d(B, E) = 5 – 2 ∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.

Co-ordinate of the point A is 1. 

Since, 1 > -2 

∴ d(J, A) = 1 – (-2) = 1 + 2 

∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.

Co-ordinate of the point C is 3. 

Since, 3 > -4

∴ d(P,C) = 3 – (-4) = 3 + 4

∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.

 Co-ordinate of the point H is -1.

 Since, -1 > -2 

∴ d(J,H) = – 1 – (-2)

= -1 + 2 

∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.

Co-ordinate of the point O is 0. 

Since,0 > -3

∴ d(K, O) = 0 – (-3) = 0 + 3 

∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0. 

∴ Co-ordinate of the point E is 5.

 Since, 5 > 0 

∴ d(O, E) = 5 – 0

∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4. 

Co-ordinate of the point J is -2. 

Since -2 > -4 

∴ d(P, J) = -2 – (-4)

= – 2+ 4 

∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5. 

Co-ordinate of the point B is 2. 

Since,2 > -5 

∴ d(Q,B) = 2 – (-5) = 2 + 5 

∴ d(Q, B) = 7



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