Question

What can you say about the position of the points (5,4), (8,4), (3,4), (0,4), (-4,4), (-2,4)?

Answer 1

Hint: We can locate the points on coordinate axes then find the points where it is located.

Complete step-by-step answer:

To locate the points create a coordinate axis.
Now there are 4 Quadrants in the coordinate axes named as 1st quadrant, 2nd quadrant, 3rd quadrant and 4rth quadrant.
Now as you see in the first quadrant x-axis is positive and y-axis is also positive therefore 1st quadrant has points $(x,y) = ( + , + )$
Similarly, 2nd quadrant $(x,y) = ( - , + )$
Similarly, 3rd quadrant $(x,y) = ( - , - )$
Similarly, 4rth quadrant $(x,y) = ( + , - )$
Now we can see the first point (5,4) as there we can see 5 is positive and 4 is positive, therefore it lies in the 1st Quadrant. Now start from 0 go to 5 units in x-axis then go 4 units perpendicular to x axis. Required point is $A(5,4)$.
For point (8, 4): 8 is the x-axis point which is positive and 4 is y-axis point which is positive, therefore it lies in the 1nd Quadrant. Now start from 0 go to 8 units in x-axis then go 4 units perpendicular to x axis. Required point is $B(8,4)$
For point (3, 4):3 is the x-axis point which is positive and 4 is y-axis point which is positive, therefore it lies in the 1st Quadrant. Now start from 0 go to 3 units in x-axis then go 4 units perpendicular to x axis. Required point is $C(3,4)$
For point (0, 4):0 is x-axis point which is neutral and 4 is y axis point which is positive, therefore it lies on the y-axis line i.e., x=0. We didn’t need to go for x-axis because the x point is neutral. Go 4 units perpendicular to x axis on line y-axis. Required point is $D(0,4)$
For point (-4, 4):-4 is the x-axis point which is negative and 4 is y-axis point which is positive, therefore it lies in the 2st Quadrant. Now start from 0 goes to -4 units in x-axis (move left) then go 4 units perpendicular to x axis. Required point is $E( - 4,4)$
For point (-2, 4):-2 is the x-axis point which is negative and 4 is y-axis point which is positive, therefore it lies in the 2st Quadrant. Now start from 0 goes to -2 units in x-axis (move left) then go 4 units perpendicular to x axis. Required point is $F( - 2,4)$

Note: While locating the point always remember the first element of point is x-axis point and second element point is always y-axis point.