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Question

$$A(3,4)$$ and $$C(1, -1)$$ are the two opposite angular points of a square ABCD. Find the coordinates of the other two vertices.


A
(92,12)
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B
 (12,52)
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C
 (12,12)
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D
 (92,52)
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Solution

The correct options are
A $$\begin{pmatrix} \cfrac { 9 }{ 2 },\cfrac { 1 }{ 2 }\end{pmatrix}$$
B  $$\begin{pmatrix}- \cfrac { 1 }{ 2 },\cfrac { 5 }{ 2 }\end{pmatrix}$$
Given-
ABCD is a square with $$A(3,4)$$ & $$C(1,-1).$$ 
To find out-
The coordinates of B & C
Solution-
Let the coordinates of $$B=(x,y).$$ 
Since ABCD is a square,
$$ AB=BC\Longrightarrow { AB }^{ 2 }={ BC }^{ 2 }.$$ 
Using distance formula,
$$ { \left( x-3 \right)  }^{ 2 }+{ \left( y-4 \right)  }^{ 2 }={ \left( x-1 \right)  }^{ 2 }+{ \left( y+1 \right)  }^{ 2 }$$ 
$$\Longrightarrow 4x+10y-23=0$$ 
$$\Longrightarrow x=\left( \cfrac { 23-10y }{ 4 }  \right)$$ .........(i).
Now, in $$\Delta ABC $$,
$$ { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$ 
$$\Longrightarrow { \left( x-3 \right)  }^{ 2 }+{ \left( y-4 \right)  }^{ 2 }+{ \left( x-1 \right)  }^{ 2 }+{ \left( y+1 \right)  }^{ 2 }={ \left( 3-1 \right)  }^{ 2 }+{ \left( 4+1 \right)  }^{ 2 }$$
$$ \Longrightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-3y=0.$$ 
Substituting for x from (i),
$$ { \left( \cfrac { 23-10y }{ 4 }  \right)  }^{ 2 }+{ y }^{ 2 }-4\left( \cfrac { 23-10y }{ 4 }  \right) -3y=0$$ 
$$\Longrightarrow 4{ y }^{ 2 }-12y+5=0$$ 
$$\Longrightarrow \left( 2y-1 \right) \left( 2y-5 \right) =0$$ 
$$\Longrightarrow y=(\cfrac { 1 }{ 2 } ,\cfrac { 5 }{ 2 } ).$$ 
Substituting for y in (i),
$$ (x,y)=\left( \cfrac { 9 }{ 2 } ,\cfrac { 1 }{ 2 }  \right)$$ & $$ \left( -\cfrac { 1 }{ 2 } ,\cfrac { 5 }{ 2 }  \right)$$ 
Now abscissae of A & C are 3 & 1, respectively.
$$ \therefore$$  Abscissa of B will be in between 1 & 3.
So here,  $$\cfrac { 9 }{ 2 }$$ will be the abscissa of B.
$$ \therefore$$  Coordinates of B & D are $$\left( \cfrac { 9 }{ 2 } ,\cfrac { 1 }{ 2 }  \right)$$ & $$\left( -\cfrac { 1 }{ 2 } ,\cfrac { 5 }{ 2 }  \right) $$, respectively.

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Mathematics

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