d (R S) = 8. d (R,M) = 10. d (s,m) = 2
find which point
point is between the other
two
Answers
Question 1:
Find the distances with the help of the number line given below.
(i) d(B,E) (ii) d(J, A) (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:
It is known that, distance between the two points is obtained by subtracting the smaller co-ordinate from larger co-ordinate.
(i) The co-ordinates of points B and E are 2 and 5 respectively. We know that 5 > 2.
∴ d(B, E) = 5 − 2 = 3
(ii) The co-ordinates of points J and A are −2 and 1 respectively. We know that 1 > −2.
∴ d(J, A) = 1 − (−2) = 1 + 2 = 3
(iii) The co-ordinates of points P and C are −4 and 3 respectively. We know that 3 > −4.
∴ d(P, C) = 3 − (−4) = 3 + 4 = 7
(iv) The co-ordinates of points J and H are −2 and −1 respectively. We know that −1 > −2.
∴ d(J, H) = −1 − (−2) = −1 + 2 = 1
(v) The co-ordinates of points K and O are −3 and 0 respectively. We know that 0 > −3.
∴ d(K, O) = 0 − (−3) = 0 + 3 = 3
(vi) The co-ordinates of points O and E are 0 and 5 respectively. We know that 5 > 0.
∴ d(O, E) = 5 − 0 = 5
(vii) The co-ordinates of points P and J are −4 and −2 respectively. We know that −2 > −4.
∴ d(P, J) = −2 − (−4) = −2 + 4 = 2
(viii) The co-ordinates of points Q and B are −5 and 2 respectively. We know that 2 > −5.
∴ d(Q, B) = 2 − (−5) = 2 + 5 = 7
Question 2:
If the co-ordinate of A is x and that of B is y, find d(A, B) .
(i) x = 1, y = 7 (ii) x = 6, y = 2 (iii) x = 3, y = 7
(iv) x = 4, y = 5 (v) x = 3, y = 6 (vi) x = 4, y = 8
ANSWER:
It is known that, distance between the two points is obtained by subtracting the smaller co-ordinate from larger co-ordinate.
(i) The coordinates of A and B are x and y respectively. We have, x = 1 and y = 7. We know that 7 > 1.
∴ d(A, B) = y − x = 7 − 1 = 6
(ii) The coordinates of A and B are x and y respectively. We have, x = 6 and y = −2. We know that 6 > −2.
∴ d(A, B) = x − y = 6 − (−2) = 6 + 2 = 8
(iii) The coordinates of A and B are x and y respectively. We have, x = −3 and y = 7. We know that 7 > −3.
∴ d(A, B) = y − x = 7 − (−3) = 7 + 3 = 10
(iv) The coordinates of A and B are x and y respectively. We have, x = −4 and y = −5. We know that −4 > −5.
∴ d(A, B) = x − y = −4 − (−5) = −4 + 5 = 1
(v) The coordinates of A and B are x and y respectively. We have, x = −3 and y = −6. We know that −3 > −6.
∴ d(A, B) = x − y = −3 − (−6) = −3 + 6 = 3
(vi) The coordinates of A and B are x and y respectively. We have, x = 4 and y = −8. We know that 4 > −8.
∴ d(A, B) = x − y = 4 − (−8) = 4 + 8 = 12
Question 3:
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
(i) d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
(ii) d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
(iii) d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
(iv) d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
(v) d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
(vi) d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
ANSWER:
(i) We have, d(P, R) = 7; d(P, Q) = 10; d(Q, R) = 3
Now, d(P, R) + d(Q, R) = 7 + 3
Or, d(P, R) + d(R, Q) = 10
∴ d(P, Q) = d(P, R) + d(Q, R)
Hence, the points P, R and Q are collinear.
The point R is between P and Q i.e., P-R-Q.
(ii) We have, d(R, S) = 8; d(S, T) = 6; d(R, T) = 4
Now, 8 + 6 = 14, so 8 + 6 ≠ 4; 6 + 4 = 10, so 6 + 4 ≠ 8 and 8 + 4 = 12, so 8 + 4 ≠ 6
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points R, S and T are non-collinear.
(iii) We have, d(A, B) = 16; d(C, A) = 9; d(B, C) = 7
Now, d(C, A) + d(B, C) = 9 + 7
Or, d(A, C) + d(C, B) = 16
∴ d(A, B) = d(A, C) + d(C, B)
Hence, the points A, C and B are collinear.
The point C is between A and B i.e., A-C-B.
(iv) We have, d(L, M) = 11; d(M, N) = 12; d(N, L) = 8
Now, 11 + 12 = 23, so 11 + 12 ≠ 8; 12 + 8 = 20, so 12 + 8 ≠ 11 and 11 + 8 = 19, so 11 + 8 ≠ 12
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points L, M and N are non-collinear.
(v) We have, d(X, Y) = 15; d(Y, Z) = 7; d(X, Z) = 8
Now, d(X, Z) + d(Y, Z) = 7 + 8
Or, d(X, Z) + d(Z, Y) = 15
∴ d(X, Y) = d(X, Z) + d(Z, Y)
Hence, the points X, Z and Y are collinear.
The point Z is between X and Y i.e., X-Z-Y.
(vi) We have, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Now, 5 + 8 = 13, so 5 + 8 ≠ 6; 8 + 6 = 14, so 8 + 6 ≠ 5 and 5 + 6 = 11, so 5 + 6 ≠ 8
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points D, E and F are non-collinear.
Question 4:
On a number line, points A, B and C are such that d(A,C) = 10, d(C,B) = 8 . Find d(A, B) considering all possibilities.
ANSWER:
There are only two possibilities.
Case 1 : When point C is between the points A and B.
We have, d(A, C) = 10; d(C, B) = 8
Now, d(A, B) = d(A, C) + d(C, B) = 10 + 8
∴ d(A, B) = 18
Case 2 : When point B is between the points A and C.
We have, d(A, C) = 10; d(C, B) = 8
Now, d(A, C) = d(A, B) + d(B, C)
So, d(A, B) = d(A, C) − d(B, C) = 10 − 8
∴ d(A, B) = 2