Question

Find other two vertices of square if two opposite vertices are (-1,2) and (3,2) ?_________solve in two to three steps if not solve long?

Answer 1

`B(1,4) and D(1,0)`

Explanation:

`A(-1,2), B(1,4) , C(3,2) , D(1,0)` are the four vertices of the

square. Diagonal `AC= sqrt((3+1)^2+(2-2)^2) =4`

Sides of square are `4/sqrt(2)=2sqrt(2)` . Let `B=(x,y)`

then, `AB^2= (x+1)^2+(y-2)^2 =8`, similarly

`BC^2= (x-3)^2+(y-2)^2 =8`

`:.(x+1)^2+(y-2)^2 = (x-3)^2+(y-2)^2` or

`(x+1)^2 = (x-3)^2` or

`cancelx^2+2x+1= cancelx^2-6x+9` or

`8x=8:. x=1 :. (1-3)^2+(y-2)^2 =8` or

`(y-2)^2 =4 or y-2 = +-2 :. y=4 and y=0`

`:. B(1,4) and D(1,0)` [Ans}