Maths 4 Mark Question

i. Let the co-ordinate of point C be x.
Co-ordinate of point A is 1.
Since, point C is to the left of point A.
∴ 1 > x
∴ d(A, C) = 1 – x
∴ 7 = 1 -x
∴x = 1 – 7
∴ x = – 6

ii. Let the co-ordinate of point B be y.
Co-ordinate of point A is 1.
Since, point B is to the right of point A.
∴y > 1
∴ d(A, B) = 7 – 1
∴ 7 = y – 1
∴ 7 + 1 = 7
∴ 7 = 8
∴ The co-ordinates of the points at a distance of 7 units from A are -6 and 8.

i. Let the co-ordinate of point Q be x.
Co-ordinate of point P is -7.
Since, point Q is to the left of point P.
∴ -7 > x
∴ d(P, Q) = -7 -x
∴8 = -7 – x
∴ x = – 7 – 8
∴x = -15

ii. Let the co-ordinate of point R be y.
Co-ordinate of point P is -7.
Since, point R is to the right of point P.
∴ y > -7
∴ d(P, R) = 7- (-7)
∴ 8 = y + 7
∴ 8 – 7 = 7
∴ y = 1
∴ The co-ordinates of the points at a distance of 8 units from P are -15 and 1.

l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5

l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,

l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.

∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.

From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Case I: Points A, B, C are such that, A – B – C.

∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.

∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.

From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.