Question

if the two opposite vertices of a square are (3, 2) and(-1 ,-2) find the coordinates of other two vertics?

Answer 1

A = (-1, 2)

B = (3, 2)

C = (x1, y1)

D = (x, y)

Solution:

Since it's a square all the sides are equal.

So, AB = BD

=> AB2=BD2

Applying distance formula,

=>(x−(−1))2+(y−2)2=(x−3)2+(y−2)2

=>(x+1)2−(x−3)2=(y−2)2−(y−2)2

=>x2+1+2x−(x2+9−6x)=0

=>8x=8

=>x=1

Now, in ∆ABD,

AB2+BD2=AC2

=>(1+1)2+(y−2)2+(1−3)2+(y−2)2=(3+1)2+(2−2)2

=>4+y2+4−4y+4+y2−4y+4=16

=>2y2−8y=0

=>y2−4y=0

=>y(y−4)=0

=>y=0,y=4

We know that, in a square diagonals bisect each other.

Mid−pointofAD=[(−1+3)/2,(2+2)/2]

Mid−pointofAD=(1,2)

Mid−pointofBC=[(1+x1)/2,(y+y1)/2]=(1,2)

=>(1+x1)/2=1and(y+y1)/2=2

=>x1=1andy+y1=4

Now,if,y=0

=>y1=4

And,if,y=4

=>y1=0

Therefore, the coordinates of the square are A(-1, 2), B(1, 0), C(1, 4) and D(3, 2)