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

Given:

Two opposite vertices of a square are $(5, 4)$ and $(1, -6)$.

To do:

We have to find the coordinates of its remaining two vertices.

Solution:

Let ABCD be the given square and $A (5, 4)$ and $C (1, -6)$ 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,

\( \mathrm{AB}^{2}=\mathrm{BC}^{2} \)

\( \Rightarrow (x-5)^{2}+(y-4)^{2}=(x-1)^{2}+(y+6)^{2} \)

\( \Rightarrow x^{2}-10 x+25+y^{2}-8 y+16 =x^{2}-2 x+1+y^{2}+12 y+36 \)

\( \Rightarrow -10 x+2 x-8 y-12 y=37-41 \)

\( \Rightarrow -8 x-20 y=-4 \)

\( \Rightarrow -4(2 x+5 y)=-4 \)

\( \Rightarrow 2 x+5 y=1 \)

\( \Rightarrow 2 x=1-5 y \)
\( \Rightarrow x=\frac{1-5 y}{2} \)
\( \mathrm{ABC} \) is a right angled triangle.

\( \Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \)
\( \Rightarrow (5-1)^{2}+(4+6)^{2}=x^{2}-10 x+25+y^{2}-8 y +16+x^{2}-2 x+1+y^{2}+12 y+36 \)
\( (4)^{2}+(10)^{2}=2 x^{2}+2 y^{2}-12 x+4 y+78 \)

\( 16+100=2 x^{2}+2 y^{2}-12 x+4 y+78 \)

\( \Rightarrow  2 x^{2}+2 y^{2}-12 x+4 y+78-16-100=0 \)

\( \Rightarrow 2 x^{2}+2 y^{2}-12 x+4 y-38=0 \)

\( \Rightarrow x^{2}+y^{2}-6 x+2 y-19=0 \)

Substituting \( x=\frac{1-5 y}{2} \)

\( (\frac{1-5 y}{2})^{2}+y^{2}-6(\frac{1-5 y}{2})+2 y-19=0 \)

\( \Rightarrow \frac{1+25 y^{2}-10 y}{4}+y^{2}-3(1-5 y)+2 y-19=0 \)

\( \Rightarrow 1+25 y^{2}-10 y+4 y^{2}-12(1-15 y)+8 y-76=0 \)

\( \Rightarrow 1+25 y^{2}-10 y+4 y^{2}-12+60 y+8 y-76=0 \)

\( \Rightarrow 29 y^{2}+58 y-87=0 \)
\( \Rightarrow y^{2}+2 y-3=0 \)
\( \Rightarrow y^{2}+3 y-y-3=0 \)
\( \Rightarrow y(y+3)-1(y+3)=0 \)
\( \Rightarrow(y+3)(y-1)=0 \)

\( y+3=0 \) or \( y-1=0 \)

\( y=-3 \) or \( y=1 \)

If \( y=1, \) then,
\( x=\frac{1-5 y}{2} \)

\( =\frac{1-5 \times 1}{2} \)

\( =\frac{1-5}{2} \)

\( =\frac{-4}{2} \)

\( =-2 \)
If \( y=-3, \) then,

\( x=\frac{1-5(-3)}{2} \)

\( =\frac{1+15}{2} \)

\( =\frac{16}{2} \)

\( =8 \)
Therefore, the other points of the square are $(-2,1)$ and $(8,-3)$.