Question

The two opposite vertices of a square are (-1,2) and(3,2).Find the coordinates of the other two vertices.

Answer 1

Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (x, y).

AB = BC  ( ABCD is a square)

⇒ AB2 = BC2

⇒ [x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2  (Distance formula)

⇒ (x + 1)2 = (x – 3)2

⇒ x2 + 2x + 1 = x2 – 6x + 9

⇒ 2x + 6x = 9 – 1

⇒ 8x = 8

⇒ x = 1

In ΔABC, we have

AB2 + BC2 = AC2  (Pythagoras theorem)

⇒ 2AB2 = AC2  ( AB = BC)

⇒ 2[(x – (–1))2 + (y – 2)2] = (3 – (–1))2 + (2 – 2)2

⇒ 2[(x + 1)2 + (y – 2)2] = (4)2 + (0)2

⇒ 2[(1 + 1)2 + (y – 2)2] = 16  ( x = 1)

⇒ 2[ 4 + (y – 2)2] = 16

⇒ 8 + 2 (y – 2)2 = 16

⇒ 2 (y – 2)2 = 16 – 8 = 8

⇒ (y – 2)2 = 4

⇒ y – 2 = ± 2

⇒ y – 2 = 2 or y – 2 = –2

⇒ y = 4 or y = 0

Thus, the other two verticles of the square ABCD are (1,4) and (1,0).

Answer 2

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