Question

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

Answer 1

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,

$P{Q}^2=Q{R}^2$

$\Rightarrow$ $(x-3{)}^2+(y-4{)}^2=(x-1{)}^2+(y-1{)}^2$

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 )

$\Rightarrow$ $x=\frac{23-10y}{4}$ ----- ( 2 )

Length of the hypotenuse $=\sqrt{2}\times side$

$\Rightarrow$ $PR=\sqrt{2}PQ$

$\Rightarrow$ $P{R}^2=2P{Q}^2$

$\Rightarrow$ $(3-1{)}^2+(4+1{)}^2=2\left[\right(x-3{)}^2+(y-4{)}^2]$

$\Rightarrow$ $4+25=2[x^2-6x+9+y^2-8y+16]$

$\Rightarrow$ $29=2[x^2-6x+y^2-8y+25]$

$\Rightarrow$ $29=2x^2-12x+2y^2-16y+50$

$\Rightarrow$ $29=2{\left(\frac{23-10y}{4}\right)}^2-12\left(\frac{23-10y}{4}\right)+2y^2-16y+50$ [ From ( 2 ) ]

$\Rightarrow$ $29=\frac{2}{16}(529-460y+100y^2)-69+30y+2y^2-16y+50$

$\Rightarrow$ $29=66.125-57.5y+12.5y^2-69+30y+2y^2-16y+50$

$\Rightarrow$ $29=47.125-43.5y+14.5y^2$

$\Rightarrow$ $14.5y^2-43.5y+18.125=0$

By soving we get,

$y=\frac{5}{2}$ and $y=\frac{1}{2}$

Substituting value of $y$ in equation ( 1 ) we get,

$\Rightarrow$ $x=\frac{-1}{2}$ and $x=\frac{9}{2}$

Vertices of a square are $(\frac{9}{2},\frac{1}{2})$ and $(\frac{-1}{2},\frac{5}{2})$