Question

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

• A. $(1,4)and(-1,1)$
• B. $(1,4)and(1,0)$
• C. $(1,-4)and(1,1)$
• D. $(1,-4)and(-1,0)$

Answer 1

The correct option is B $(1,4)and(1,0)$

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

$AB=BC$

$A{B}^2=B{C}^2$

$[x-(-1){]}^2+(y-2{)}^2=(x-3{)}^2+(y-2{)}^2$

$(x+1{)}^2=(x-3{)}^2\Rightarrow (x{)}^2+1+2x=(x{)}^2+9-6x$

$8x=8\Rightarrow x=1$

In $△ABC$ using Pythagoras theorem,

$A{B}^2+B{C}^2=A{C}^2$

$2A{B}^2=A{C}^2$ $[\because AB=BC]$

$2[x-(-1{)}^2+(y-2{)}^2]=(3-(-1){)}^2+(2-2{)}^2$

$2\left[\right(x+1{)}^2+(y-2{)}^2]=16$

$(x+1{)}^2+(y-2{)}^2=8$

$4+(y-2{)}^2=8\Rightarrow (y-2{)}^2=4$

$y-2=\pm 2\Rightarrow y=4,0$

Thus other two vertices of square $ABCD$ are $(1,4)and(1,0)$