Question

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?

• A. $\left(\frac{9}{2},\frac{1}{2}\right)$
• B. $\left(\frac{9}{2},\frac{5}{2}\right)$
• C. $\left(\frac{-1}{2},\frac{5}{2}\right)$
• D. None of these

Answer 1

Answer: (B)

Let $A(3,4)$ and $C(1,-1)$ be the given opposite angular points of a square $ABCD$ and let $B(x,y)$ be the unknown vertex. Then,
$AB=BC$
$\Rightarrow (AB{)}^2=(BC{)}^2$
$\Rightarrow (x-3{)}^2+(y-4{)}^2=(x-1{)}^2+(y+1{)}^2$
$\Rightarrow 4x+10y-23=0$
$\Rightarrow x=\left(\frac{23-10y}{4}\right).....$(i)

Also in right $△ABC$,
$(AB{)}^2+(BC{)}^2=(AC{)}^2$
$\Rightarrow (x-3{)}^2+(y-4{)}^2+(x-1{)}^2+(y+1{)}^2$
$=(3-1{)}^2+(4+1{)}^2$
$\Rightarrow x^2+y^2-4x-3y-1=\mathrm{0......}$(ii)
Substituting the value of $x$ from Eq. (i) into Eq. (ii), we get
${\left(\frac{23-10y}{4}\right)}^2+y^2-4\left(\frac{23-10y}{4}\right)-3y-1=0$
$\Rightarrow 4y^2-12y+5=0$
or $(2y-1)(2y-5)=0$
$\therefore y=\frac{1}{2}$ 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
$\left(-\frac{1}{2},\frac{5}{2}\right)$