Question

If any two vertices of a square are (4, 4) and (4, –4), then the possible coordinates of the remaining vertices of square cannot be

(8, 0)

(–4, 0)

(12, –4)

(–4 ,4)

Answer 1

Answer:

Let ABCD is a square where two opposite vertices are A(−1,2)andC(3,2).

Let B(x,y) and D(x1.y1) ids the other two vertices.

In Square ABCD

AB=BC=CD=DA

Hence AB=BC

⇒(x+1)2+(y−2)2=(3−x)2+(2−y)2   [by distance formula]

Squaring both sides

⇒(x+1)2+(y−2)2=(3−x)2+(2−y)2

⇒x2+2x+1+y2+4−4y=9+x2−6x+4+y2−4y

⇒2x+5=13−6x

⇒2x+6x=13−5

⇒8x=8

⇒x=1

In △ABC,∠B=90∘  [all angles of square are 90