Question

# Two opposite vertices of a square are $$\left( {3,4} \right)$$ and $$\left( {1, - 1} \right)$$ Find the coordinates of the other vertices.

Solution

## Let $$PQRS$$ be a square.$$P(3,4)$$ and $$R(1,-1)$$Let co-ordinates of a point $$Q(x,y)$$$$PQ=QR$$Squaring on both sides,$$PQ^2=QR^2$$$$\Rightarrow$$  $$(x-3)^2+(y-4)^2=(x-1)^2+(y-1)^2$$Solving this we get,$$\Rightarrow$$  $$x^2-3x+9+y^2-8y+16=x^2-2x+1+y^2+2y+1$$$$\Rightarrow$$  $$4x+10y=23$$        ---- ( 1 )$$\Rightarrow$$  $$x=\dfrac{23-10y}{4}$$        ----- ( 2 )Length of the hypotenuse $$=\sqrt{2}\times side$$$$\Rightarrow$$  $$PR=\sqrt{2}PQ$$$$\Rightarrow$$  $$PR^2=2PQ^2$$$$\Rightarrow$$  $$(3-1)^2+(4+1)^2=2[(x-3)^2+(y-4)^2]$$$$\Rightarrow$$  $$4+25=2[x^2-6x+9+y^2-8y+16]$$$$\Rightarrow$$  $$29=2[x^2-6x+y^2-8y+25]$$$$\Rightarrow$$  $$29=2x^2-12x+2y^2-16y+50$$$$\Rightarrow$$  $$29=2\left(\dfrac{23-10y}{4}\right)^2-12\left(\dfrac{23-10y}{4}\right)+2y^2-16y+50$$            [ From ( 2 ) ]$$\Rightarrow$$  $$29=\dfrac{2}{16}(529-460y+100y^2)-69+30y+2y^2-16y+50$$$$\Rightarrow$$  $$29=66.125-57.5y+12.5y^2-69+30y+2y^2-16y+50$$$$\Rightarrow$$  $$29=47.125-43.5y+14.5y^2$$$$\Rightarrow$$  $$14.5y^2-43.5y+18.125=0$$By soving we get,$$y=\dfrac{5}{2}$$ and $$y=\dfrac{1}{2}$$Substituting value of $$y$$ in equation ( 1 ) we get,$$\Rightarrow$$  $$x=\dfrac{-1}{2}$$ and $$x=\dfrac{9}{2}$$Vertices of a square are $$\left(\dfrac{9}{2},\dfrac{1}{2}\right)$$ and $$\left(\dfrac{-1}{2},\dfrac{5}{2}\right)$$Mathematics

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