I) Coordinates of point P a) (1,1) b) (1,2) c) (2,1) d) (1,0)

ii) Coordinates of point Q

a) (0,-3) b) (-3,0) c) (-3,-3) d) (0,3)

iii) Coordinates of point T

a) (-4,-2) b) (-2,-4) c) (4,-2) d) (-2,4)

iv) (Ordinate of T) – (Abscissa of R)

a) -4 b) 4 c) 0 d) -2

v) (Abscissa of Q) – (Ordinate of S)

a) -4 b) 4 c) 2 d) -2

Given:

Five points P, Q, R, S ,T on a coordinate plane.

To find:

The coordinates of (i) P, (ii) Q, (iii) T, (iv) (Ordinate of T) – (Abscissa of R) and (v) (Abscissa of Q) – (Ordinate of S).

Solution:

From the given graph it is clear that the coordinates of given points are

P = (1,1)

Q = (-3,0)

R = (-2,-3)

S = (2,1)

T = (4,-2)

(i)

Coordinates of point P are (1,1).

Therefore, the correct option is (a).

(ii)

Coordinates of point Q are (-3,0).

Therefore, the correct option is (b).

(iii)

Coordinates of point T are (4,-2).

Therefore, the correct option is (c).

(iv)

Ordinate is y-coordinate and abscissa is x-coordinate.

Ordinate of T = -2

Abscissa of R = -2

(Ordinate of T) – (Abscissa of R) = -2 - (-2)

                                                     = -2 + 2

                                                     = 0

Therefore, the correct option is (c).

(v)

Abscissa of Q = -3

Ordinate of S = 1

(Abscissa of Q) – (Ordinate of S) = -3 - 1

                                                      = -4

Therefore, the correct option is (a).

You are given the time and speed so from there you can get the points (2,50) and (6,54) once you got that you can see how the slope is 1 and 
b=48
x-y=-48 (standard form)
 
part B= 
1    49
2    50
3    51
4    52
5    53
6    54
7    55