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Q.
Let the opposite angular points of a square be $(3,4)$ and $(1,-1)$. Then, which of the following are not the coordinates of remaining angular points?
Straight Lines
Solution:
Let $A(3,4)$ and $C(1,-1)$ be the given opposite angular points of a square $A B C D$ and let $B(x, y)$ be the unknown vertex. Then,
$A B =B C$
$\Rightarrow (A B)^2 =(B C)^2 $
$\Rightarrow (x-3)^2+(y-4)^2 =(x-1)^2+(y+1)^2 $
$\Rightarrow 4 x+10 y-23 =0 $
$\Rightarrow x =\left(\frac{23-10 y}{4}\right) .....$(i)
Also in right $\triangle A B C$,
$(A B)^2+(B C)^2 =(A C)^2$
$\Rightarrow (x-3)^2+(y-4)^2+ (x-1)^2+(y+1)^2 $
$ =(3-1)^2+(4+1)^2$
$\Rightarrow x^2+y^2-4 x-3 y-1 =0 ......$(ii)
Substituting the value of $x$ from Eq. (i) into Eq. (ii), we get
$ \left(\frac{23-10 y}{4}\right)^2+y^2-4\left(\frac{23-10 y}{4}\right)-3 y-1=0$
$\Rightarrow 4 y^2-12 y+5=0$
or $ (2 y-1)(2 y-5)=0 $
$\therefore y=\frac{1}{2} $ or $=\frac{5}{2}$
Putting $y=\frac{1}{2}$ in Eq. (i), we get $x=\frac{9}{2}$
and putting $y=\frac{5}{2}$ in Eq. (i), we get $x=-\frac{1}{2}$
Hence, the required vertices of the square are $\left(\frac{9}{2}, \frac{1}{2}\right)$ and
$\left(-\frac{1}{2}, \frac{5}{2}\right)$
