# 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 $(8,-3)$.

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Updated on: 10-Oct-2022

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