Question

The two opposite vertices of a square as shown in the figure are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

• A. B= (0, 1) and D= (1, 4)
• B. B= (1, 4) and D= (1, 0)
• C. B= (4, 1) and D= (0, 1)
• D. B= (1, 0) and D= (1, 4)

Answer 1

The correct option is D

B= (1, 0) and D= (1, 4)

O is the point of intersection of AC and BD so O is the midpoint of AC. AC is parallel to X-axis since its y-coordinates are constant.

$\therefore$ $AC=|\text{difference of x coordinates of A and C}|$

=$|-1-3|$

= 4

$\therefore$ AO = $\frac{AC}{2}$

= 2

Let coordinates of 'O' be (x,2)

$\Rightarrow$ {x coordinate of O} - {x coordinate of A} = 2 ($\because$ AO = 2)

$\Rightarrow$ (x-(-1)) = 2

$\Rightarrow x=1$

$\therefore$ O = (1,2)

$\because$ Diagonals of square are perpendicular to each other so BD will be parallel to Y-axis.

Now AO = BO ($\because$ diagonals of a square are equal and bisect each other)
$\therefore$ BO = 2

Let B = $x_1,y_1$

$\Rightarrow$ {y coordinate of O} - {y coordinate of B} = 2

$\Rightarrow 2-y_1=2$

$\Rightarrow y_1=0$

and {x-coordinate of B} = {x-coordinate of O}

$\therefore x_1=1$

$\therefore$ B = (1, 0)
and $\Rightarrow$ {y coordinate of D} - {y coordinate of B} = 4 ($\because$ BD = 4)

$\Rightarrow$ y coordinate of D - 0 = 4

$\Rightarrow$ y coordinate of D = 4

$\therefore$ D = (1,4)
$\because x-coordinateofD=x-coordinateofB$)