Question

Find the co-ordinates of the vertices of the square ABCD (side = a), taking AB and AD as axes.

Answer 1

Answer:

A = (0 , 0)

B = (a , 0)

C = (a , a)

D = (0 , a)

Step-by-step explanation:

Let

Square is in 1st quadrant

A be origin of plane ,

AB is along x-axis

and

AD is along y-axis

Then

We are given that

Length of side of square = a

So

Length of AB = a

Since

A is origin its co-ordinates are ( 0 , 0 )

and

B is on x-axis so its x co-ordinate "a"

And its y co-ordinate is "0"

Thus

Co-ordinates of B are ( a , 0 )

Since

D is on y-axis so its y co-ordinate is "a"

and its x co-ordinate is 0

NOW

Since ABCD is a square so Distance between C and D is equal to C and B

So

Co-ordinates of C are ( a , a )

Thus

A = (0 , 0)

B = (a , 0)

C = (a , a)

D = (0 , a)

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When square is in other quadrants

IN 2nd quadrant co-ordinates are following

A = (0 , 0)

B = (-a , 0)

C = (-a , a)

D = (0 , a)

IN 3rd quadrant co-ordinates are following

A = (0 , 0)

B = (-a , 0)

C = (-a , -a)

D = (0 , -a)

IN 4th quadrant co-ordinates are following

A = (0 , 0)

B = (a , 0)

C = (a , -a)

D = (0 , -a)

I hope it helpful.