The distance between two points on a number line is calculated by subtracting the smaller co-ordinate from the greater co-ordinate.
Co-ordinate of B: 2
Co-ordinate of E: 5
Now, 2 < 5
∴ d (B, E) = 5 − 2
∴ d (B, E) = 3
Co-ordinate of J: − 2
Co-ordinate of A: 1
Now, − 2 < 1
∴ d (J, A) = 1 − (− 2)
∴ d (J, A) = 1 + 2
∴ d (J, A) = 3
Co-ordinate of P: − 4
Co-ordinate of C: 3
Now, − 4 < 3
∴ d (P, C) = 3 − (− 4)
∴ d (P, C) = 3 + 4
∴ d (P, C) = 7
Co-ordinate of J: − 2
Co-ordinate of H: − 1
Now, − 2 < − 1
∴ d (J, H) = − 1 − (− 2)
∴ d (J, H) = − 1 + 2
∴ d (J, H) = 1
Co-ordinate of K: − 3
Co-ordinate of O: 0
Now, − 3 < 0
∴ d (K, O) = 0 − (− 3)
∴ d (K, O) = 0 + 3
∴ d (K, O) = 3
Co-ordinate of O: 0
Co-ordinate of E: 5
Now, 0 < 5
∴ d (O, E) = 5 − 0
∴ d (O, E) = 5
Co-ordinate of P: − 4
Co-ordinate of J: − 2
Now, − 4 < − 2
∴ d (P, J) = − 2 − (− 4)
∴ d (P, J) = − 2 + 4
∴ d (P, J) = 2
Co-ordinate of Q: − 5
Co-ordinate of B: 2
Now, − 5 < 2
∴ d (Q, B) = 2 − (− 5)
∴ d (Q, B) = 2 + 5
∴ d (Q, B) = 7
Co-ordinate of A: 1
Co-ordinate of B: 7
Now, 1 < 7
∴ d (A, B) = 7 − 1
∴ d (A, B) = 6
Co-ordinate of A: 6
Co-ordinate of B: − 2
Now, 6 > − 2
∴ d (A, B) = 6 − (− 2)
∴ d (A, B) = 6 + 2
∴ d (A, B) = 8
Co-ordinate of A: − 3
Co-ordinate of B: 7
Now, − 3 < 7
∴ d (A, B) = 7 − (− 3)
∴ d (A, B) = 7 + 3
∴ d (A, B) = 10
Co-ordinate of A: − 4
Co-ordinate of B: − 5
Now, − 4 > − 5
∴ d (A, B) = − 4 − (− 5)
∴ d (A, B) = − 4 + 5
∴ d (A, B) = 1
Co-ordinate of A: − 3
Co-ordinate of B: − 6
Now, − 3 > − 6
∴ d (A, B) = − 3 − (− 6)
∴ d (A, B) = − 3 + 6
∴ d (A, B) = 3
Co-ordinate of A: 4
Co-ordinate of B: − 8
Now, 4 > − 8
∴ d (A, B) = 4 − (− 8)
∴ d (A, B) = 4 + 8
∴ d (A, B) = 12
d (P, Q) = 10 ... (i)
and d (P, R) + d (Q, R) = = 7 + 3 = 10 ... (ii)
From (i) and (ii),
d (P, Q) = d (P, R) + d (Q, R)
∴ Point R is between Point P and Point Q.
∴ P – R – Q
d (R, S) = 8 ... (i)
and d (S, T) + d (R, T) = = 6 + 4 = 10 ... (ii)
From (i) and (ii),
d (R, S) ≠ d (S, T) + d (R, T)
∴ Points R, S and T are not collinear.
d (A, B) = 16 ... (i)
and d (C, A) + d (B, C) = = 9 + 7 = 16 ... (ii)
From (i) and (ii),
d (A, B) = d (C, A) + d (B, C)
∴ Point C is between Point A and Point B.
∴ A – C – B
d (M, N) = 12 ... (i)
and d (L, M) + d (N, L) = = 11 + 8 = 19 ... (ii)
From (i) and (ii),
d (M, N) ≠ d (L, M) + d (N, L)
∴ Points L, M and N are not collinear.
d (X, Y) = 15 ... (i)
and d (Y, Z) + d (X, Z) = = 7 + 8 = 15 ... (ii)
From (i) and (ii),
d (X, Y) = d (Y, Z) + d (X, Z)
∴ Point Z is between Point X and Point Y.
∴ X – Z – Y
d (E, F) = 8 ... (i)
and d (D, E) + d (D, F) = = 5 + 6 = 11 ... (ii)
From (i) and (ii),
d (E, F) ≠ d (D, E) + d (D, F)
∴ Points D, E and F are not collinear.
Here, we have three possibilities.
∴ d (A, C) = d (A, B) + d (B, C)
∴ 10 = d (A, B) + 8
∴ 10 − 8 = d (A, B)
∴ 2 = d (A, B)
i.e. d (A, B) = 2
∴ d (A, B) = d (A, C) + d (C, B)
∴ d (A, B) = 10 + 8 = 18
i.e. d (A, B) = 18
But, this is not possible as
d (A, C) > d (C, B)
∴ d (A, B) = 2 or 18
Here, we have two possibilities.
As seen in the figure,
d (X, Z) = d (X, Y) + d (Y, Z)
∴ d (X, Z) = 17 + 8
∴ d (X, Z) = 25 unit
As seen in the figure,
d (X, Z) + d (Z, Y) = d (X, Y)
∴ d (X, Z) + 8 = 17
∴ d (X, Z) = 17 − 8
∴ d (X, Z) = 9 unit
Therefore, d (X, Z) = 25 unit or 9 unit.
As seen in the figure,
l (AB) + l (BC) = l (AC)
∴ l (AB) + 6.5 = 11
∴ l (AB) = 11 − 6.5
∴ l (AB) = 4.5 unit
As seen in the figure,
l (RT) = l (RS) + l (ST)
∴ l (RT) = 2.5 + 3.7
∴ l (RT) = 6.2 unit
As seen in the figure,
l (XZ) = l (XY) + l (YZ)
∴ 3\(\sqrt 7\) = \(\sqrt 7\) + l (YZ)
∴ 3\(\sqrt 7\) − \(\sqrt 7\) = l (YZ)
∴ 2\(\sqrt 7\) = l (YZ)
i.e. l (YZ) = 2\(\sqrt 7\) unit
A triangle is formed by three non-collinear points.
This page was last modified on
16 April 2026 at 12:14