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



Discussion

No Comment Found