Question

Two opposite vertices of a square are given as $(-1,1)$ and $(1,5)$, then one of other vertices is

• A. $(0,3)$
• B. $(4,-2)$
• C. $(2,2)$
• D. None of these

Answer 1

Answer: (C)

$(2,2)$

Let $A$ and $C$ be $(-1,1)$ and $(1,5)$ respectively.
Let $B$ be $(x,y)$
Equation of $AB$ is $\frac{x+1}{\cos \theta }=\frac{y-1}{\sin \theta }=r....\left(1\right)$
Where $\tan \theta$ is slope of line $AB$ and equation of $AC$ is
$\frac{2}{\cos (\frac{\pi }{4}+\theta )}=\frac{4}{\sin (\frac{\pi }{4}+\theta )}=\sqrt{2}r.....\left(2\right)$
$\Rightarrow 2(\frac{\sin \theta }{\sqrt{2}}+\frac{\cos \theta }{\sqrt{2}})=4(\frac{\cos \theta }{\sqrt{2}}-\frac{\sin \theta }{\sqrt{2}})$
$\Rightarrow \tan \theta =\frac{1}{3}$
$\Rightarrow \sin \theta =\frac{1}{\sqrt{10}}$ and $\cos \theta =\frac{3}{\sqrt{10}}$
Now $\frac{2\sqrt{2}}{\cos \theta -\sin \theta }=\sqrt{2}r.....using\left(2\right)$
$\Rightarrow \frac{2}{\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{10}}}=r\Rightarrow r=\sqrt{10}units$
Now, putting $r,\sin \theta ,\cos \theta$ is equation $\left(1\right)$
$\Rightarrow \frac{x+1}{\frac{3}{\sqrt{10}}}=\frac{y-1}{\frac{1}{\sqrt{10}}}=\sqrt{10}\Rightarrow x=2,y=2$
$\Rightarrow$ The other vertices $(2,2)$
Hence Choice (c) is correct.