** Q1=Two opposite vertices of a square are (-1,2)and (3,2).Find the coordinates of other two vertices. **

** Q2=If two opposite vertices of a square of a square are (5,4)and (1,-6) ,find the coordinates of its remaining two vertices. **

Dear Student!

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)

⇒ AB^{2} = BC^{2}

⇒ [*x* – (–1)] ^{2} + (*y* – 2)^{2 }= (*x* – 3)^{2} + (*y* – 2)^{2 }(Distance formula)

⇒ (*x* + 1)^{2} ^{ }= (*x* – 3)^{2}

⇒ *x*^{2} + 2*x* + 1 = *x*^{2 }– 6*x* + 9

⇒ 2*x *+ 6*x* = 9 – 1

⇒ 8*x *= 8

⇒ *x *= 1

In ΔABC, we have

AB^{2} + BC^{2} = AC^{2} (Pythagoras theorem)

⇒ 2AB^{2} = AC^{2} ( 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 vertices of the square ABCD are (1, 4) and (1, 0).

The other question is similar to the solved question. Try to solve the other question. If you have any problem in solving the question, then get back to us we will be happy to help you.

Cheers!

**
**