Question

$A(3,4)$ and $C(1,-1)$ are the two opposite angular points of a square ABCD. Find the coordinates of the other two vertices.

• A. $\left(\begin{array}{c}\frac{9}{2},\frac{1}{2}\end{array}\right)$
• B. $\left(\begin{array}{c}-\frac{1}{2},\frac{5}{2}\end{array}\right)$
• C. $\left(\begin{array}{c}-\frac{1}{2},\frac{1}{2}\end{array}\right)$
• D. $\left(\begin{array}{c}-\frac{9}{2},\frac{5}{2}\end{array}\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $\left(\begin{array}{c}\frac{9}{2},\frac{1}{2}\end{array}\right)$
B $\left(\begin{array}{c}-\frac{1}{2},\frac{5}{2}\end{array}\right)$

Given-

ABCD is a square with $A(3,4)$ & $C(1,-1).$

To find out-

The coordinates of B & C

Solution-

Let the coordinates of $B=(x,y).$

Since ABCD is a square,

$AB=BC⟹{AB}^2={BC}^2.$

Using distance formula,

${(x-3)}^2+{(y-4)}^2={(x-1)}^2+{(y+1)}^2$

$⟹4x+10y-23=0$

$⟹x=\left(\frac{23-10y}{4}\right)$ .........(i).

Now, in $\Delta ABC$,

${AB}^2+{BC}^2={AC}^2$

$⟹{(x-3)}^2+{(y-4)}^2+{(x-1)}^2+{(y+1)}^2={(3-1)}^2+{(4+1)}^2$

$⟹x^2+y^2-4x-3y=0.$

Substituting for x from (i),

${\left(\frac{23-10y}{4}\right)}^2+y^2-4\left(\frac{23-10y}{4}\right)-3y=0$

$⟹4y^2-12y+5=0$

$⟹(2y-1)(2y-5)=0$

$⟹y=(\frac{1}{2},\frac{5}{2}).$

Substituting for y in (i),

$(x,y)=(\frac{9}{2},\frac{1}{2})$ & $(-\frac{1}{2},\frac{5}{2})$

Now abscissae of A & C are 3 & 1, respectively.

$\therefore$ Abscissa of B will be in between 1 & 3.

So here, $\frac{9}{2}$ will be the abscissa of B.

$\therefore$ Coordinates of B & D are $(\frac{9}{2},\frac{1}{2})$ & $(-\frac{1}{2},\frac{5}{2})$, respectively.