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

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 & CSolution-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.Mathematics

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