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The two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$. Find the coordinates of the other two vertices.
Given:
Two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$.
To do:
We have to find the coordinates of other two vertices.
Solution:
Let ABCD be the given square and $A (-1, 2)$ and $C (3, 2)$ 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 $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Therefore,
$\mathrm{AB}=\mathrm{BC}$
Squaring on both sides, we get, $\mathrm{AB}^{2}=\mathrm{BC}^{2}$
$\Rightarrow (x+1)^{2}+(y-2)^{2}=(x-3)^{2}+(y-2)^{2}$
$\Rightarrow x^{2}+2 x+1+y^{2}-4 y+4 =x^{2}-6 x+9+y^{2}-4 y+4$
$\Rightarrow 2 x+6 x-4 y+4 y=13-5$
$\Rightarrow 8 x=8$
$\Rightarrow x=1$.........(i)
$\mathrm{ABC}$ is a right angled triangle.
$\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$\Rightarrow (3+1)^{2}+(2-2)^{2}=x^{2}+2 x+1+y^{2}-4 y +4+x^{2}-6 x+9+y^{2}-4 y+4$
$(4)^{2}=2 x^{2}+2 y^{2}-4 x-8 y+18$
$16=2 x^{2}+2 y^{2}-4 x-8 y+18$
$\Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+18-16=0$
$\Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+2=0$
$\Rightarrow 2 (x^{2}+y^{2}-2 x-4 y+1)=0$
$\Rightarrow x^{2}+y^{2}-2x-4y+1=0$
Substituting $x=1$, we get,
$(1)^{2}+y^{2}-2(1)-4 y+1=0$
$\Rightarrow 1+y^2-2-4y+1=0$
$\Rightarrow y^2-4y=0$
$\Rightarrow y(y-4)=0$
$\Rightarrow(y)(y-4)=0$
$y=0$ or $y-4=0$
$y=0$ or $y=4$
Therefore, the other points of the square are $(1,0)$ and $(1,4)$.
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