If two opposite vertices of a square are (5, 4) and (1, -6) then the coordinates of its remaining two vertices are

$Solution-\\ ABCD\quad is\quad a\quad square.\quad \\ So\quad AB=BC=CD=DA\quad \& \quad the\quad diagonals\quad AC=BD..........(i).\\ We\quad shall\quad apply\quad distance\quad formula\quad to\quad get\quad the\quad above\quad lengths.\\ d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } .\quad \\ Now\quad \Delta ABC\quad is\quad a\quad right\quad one\quad with\quad AC\quad as\quad hypotenuse(\angle ABC={ 90 }^{ o }).\\ \therefore \quad AC=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 4+6 \right) }^{ 2 } } units=\sqrt { 116 } units=BD\quad (by\quad i)........(ii).\\ We\quad know\quad that\quad side\quad of\quad a\quad square=\frac { diagonal }{ \sqrt { 2 } } .\\ \therefore \quad AB=BC=CD=DA\quad =\frac { \sqrt { 116 } }{ \sqrt { 2 } } units=\sqrt { 58 } units.........(iii).\quad \\ Now,\quad using\quad the\quad distance\quad formula\quad d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } ,\\ { AB }^{ 2 }={ BC }^{ 2 }\\ \Longrightarrow { \left( { x }_{ 1 }-5 \right) }^{ 2 }+{ \left( { y }_{ 1 }-4 \right) }^{ 2 }={ \left( { x }_{ 1 }-1 \right) }^{ 2 }+{ \left( { y }_{ 1 }+6 \right) }^{ 2 }\\ \Longrightarrow 8{ x }_{ 1 }+20y_{ 1 }-4=0\\ \Longrightarrow 2{ x }_{ 1 }+5{ y }_{ 1 }-1=0\\ \Longrightarrow y=\frac { 1-2{ x }_{ 1 } }{ 5 } ........(iv).\\ \therefore \quad AB^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }\\ \Longrightarrow { \left( { x }_{ 1 }-5 \right) }^{ 2 }+{ \left( y_{ 1 }-4 \right) }^{ 2 }+58=116(from\quad ii\quad \& \quad iii)\\ \Longrightarrow { \left( { x }_{ 1 }-5 \right) }^{ 2 }+{ \left( y_{ 1 }-4 \right) }^{ 2 }=58\\ \Longrightarrow { \left( { x }_{ 1 }-5 \right) }^{ 2 }+{ \left( \frac { 1-2{ x }_{ 1 } }{ 5 } -4 \right) }^{ 2 }=58\\ \Longrightarrow 29{ { x }_{ 1 } }^{ 2 }-174{ x }_{ 1 }-464=0\\ \Longrightarrow { { x }_{ 1 } }^{ 2 }-6{ x }_{ 1 }-16=0\\ \Longrightarrow \left( { x }_{ 1 }-8 \right) \left( { x }_{ 1 }+2 \right) =0\\ \Longrightarrow { x }_{ 1 }=(8,-2).\\ So,\quad from\quad (iv),\\ { y }_{ 1 }=\left( \frac { 1-2\times 8 }{ 5 } ,\frac { 1-2(-2) }{ 5 } \right) =\left( -3,1 \right) .\\ \therefore \quad B\left( { x }_{ 1 },y_{ 1 } \right) =\left( 8,-3 \right) \quad and\quad \left( -2,1\quad \right) .\\ Similarly\quad { AD }^{ 2 }={ DC }^{ 2 }\\ \Longrightarrow { \left( { x }_{ 2 }-5 \right) }^{ 2 }+{ \left( { y }_{ 2 }-4 \right) }^{ 2 }={ \left( { x }_{ 2 }-1 \right) }^{ 2 }+{ \left( { y }_{ 2 }+6 \right) }^{ 2 }\\ \Longrightarrow 8{ x }_{ 2 }+20y_{ 2 }-4=0\\ \Longrightarrow 2{ x }_{ 2 }+5{ y }_{ 2 }-1=0\\ \Longrightarrow { y }_{ 2 }=\frac { 1-2{ x }_{ 2 } }{ 5 } ........(iv).\\ \therefore \quad AD^{ 2 }+{ DC }^{ 2 }={ AC }^{ 2 }\\ \Longrightarrow { \left( { x }_{ 2 }-5 \right) }^{ 2 }+{ \left( y_{ 2 }-4 \right) }^{ 2 }+58=116(from\quad ii\quad \& \quad iii)\\ \Longrightarrow { \left( { x }_{ 2 }-5 \right) }^{ 2 }+{ \left( y_{ 2 }-4 \right) }^{ 2 }=58\\ \Longrightarrow { \left( { x }_{ 2 }-5 \right) }^{ 2 }+{ \left( \frac { 1-2{ x }_{ 2 } }{ 5 } -4 \right) }^{ 2 }=58\\ \Longrightarrow 29{ { x }_{ 2 } }^{ 2 }-174{ x }_{ 2 }-464=0\\ \Longrightarrow { { x }_{ 2 } }^{ 2 }-6{ x }_{ 2 }-16=0\\ \Longrightarrow \left( { x }_{ 2 }-8 \right) \left( { x }_{ 2 }+2 \right) =0\\ \Longrightarrow { x }_{ 2 }=(8,-2).\\ So,\quad from\quad (iv),\\ { y }_{ 2 }=\left( \frac { 1-2\times 8 }{ 5 } ,\frac { 1-2(-2) }{ 5 } \right) =\left( -3,1 \right) .\\ \therefore \quad D\left( { x }_{ 2 },y_{ 2 } \right) =\left( 8,-3 \right) \quad and\quad \left( -2,1\quad \right) .\\ So\quad the\quad coordinates\quad of\quad other\quad two\quad vertices\quad are\quad \\ \left( 8,-3 \right) \quad and\quad \left( -2,1\quad \right) .\\ ans-\quad Option\quad B.\\ \\ \quad \\$