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?

Question

2153 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T12:10:46

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

$A(-1,2), B(1,4) , C(3,2) , D(1,0)$ 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}