Question

The two opposite vertices of a square are (–1, 2) and (3, 2). Find the co-ordinates of the other two vertices.

Answer 1

Let ABCD be a square and two opposite vertices of it are A(-1, 2) and C(3, 2).

Let $(x_1,y_1),(x_2,y_2)$ be the coordinate of vertex B and D respectively. We know that in a square all sides are equal to each other.

$\therefore AB=BC\sqrt{(x_1+1{)}^2+(y_1–2{)}^2}=\sqrt{(x_1-3{)}^2+(y_1–2{)}^2}$

Squaring on both sides we get

$(x_1+1{)}^2+(y_1-2{)}^2=(x_1-3{)}^2+(y_1-2{)}^2\Rightarrow x_1^2+2x_1+1=x_1^2-6x_1+9\Rightarrow 8x_1=8\Rightarrow x_1=1$

We know that the square, in a ΔABC,

$A{B}^2+B{C}^2=A{C}^2{\left[\sqrt{(1+1{)}^2+(y1–2{)}^2}\right]}^2+{\left[\sqrt{(1-3{)}^2+(y_1–2{)}^2}\right]}^2={\left[\sqrt{(3+1{)}^2+(2–2{)}^2}\right]}^2$
$\Rightarrow 4+y_1^2+4-4y_1+4+y_1^2-4y_1+4=16$
$\Rightarrow 2y_1^2+16-8y_1=16\Rightarrow 2y_1^2-8y_1=0\Rightarrow y_1(y_1-4)=0\Rightarrow y_1=0or4$

We know that in a square, the diagonals are of equal length and let O be the mid-point of AC. It will also be the mid-point of BD.

O = Mid point of AC $=(\frac{-1+3}{2},\frac{2+2}{2})=(1,2)$

Mid point of AC = Mid point of BD $(\frac{1+x_2}{2},\frac{y_1+y_2}{2})=(1,2)$

$\Rightarrow \frac{1+x_2}{2}=1$

$x_2=2$ And $\frac{y_1+y_2}{2}=2\Rightarrow y_1+y_2=4\text{If}{y}_1=0or4\text{Then}{y}_2=4\text{or}0$

Therefore, the required coordinates are B (1, 0) or B (1,4) and D (2, 4) or D (2,0)