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Two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$. Find the coordinates of other two vertices.
Given:
Two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$.
To do:
We have to find the coordinates of other two vertices.
Solution:
Let ABCD be the given square and $A (-1, 2)$ and $C (3, 2)$ are the opposite vertices.
Let the coordinates of $B$ be $(x, y)$. Join AC.
This implies,
$AB=BC=CD=DA$
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{AB}=\mathrm{BC} \)
Squaring on both sides, we get,
\( \mathrm{AB}^{2}=\mathrm{BC}^{2} \)
\( \Rightarrow (x+1)^{2}+(y-2)^{2}=(x-3)^{2}+(y-2)^{2} \)
\( \Rightarrow x^{2}+2 x+1+y^{2}-4 y+4 =x^{2}-6 x+9+y^{2}-4 y+4 \)
\( \Rightarrow 2 x+6 x-4 y+4 y=13-5 \)
\( \Rightarrow 8 x=8 \)
\( \Rightarrow x=1 \).........(i)
\( \mathrm{ABC} \) is a right angled triangle.
\( \Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \)
\( \Rightarrow (3+1)^{2}+(2-2)^{2}=x^{2}+2 x+1+y^{2}-4 y +4+x^{2}-6 x+9+y^{2}-4 y+4 \)
\( (4)^{2}=2 x^{2}+2 y^{2}-4 x-8 y+18 \)
\( 16=2 x^{2}+2 y^{2}-4 x-8 y+18 \)
\( \Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+18-16=0 \)
\( \Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+2=0 \)
\( \Rightarrow 2 (x^{2}+y^{2}-2 x-4 y+1)=0 \)
\( \Rightarrow x^{2}+y^{2}-2x-4y+1=0 \)
Substituting \( x=1 \), we get,
\( (1)^{2}+y^{2}-2(1)-4 y+1=0 \)
\( \Rightarrow 1+y^2-2-4y+1=0 \)
\( \Rightarrow y^2-4y=0 \)
\( \Rightarrow y(y-4)=0 \)
\( \Rightarrow(y)(y-4)=0 \)
\( y=0 \) or \( y-4=0 \)
\( y=0 \) or \( y=4 \)
Therefore, the other points of the square are $(1,0)$ and $(1,4)$.