Question

Question 4
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Answer 1

O is the point of intersection of AC and BD. Coordinates of O can be calculated as follows:
$x=\frac{3-1}{2}=1$
$y=\frac{2+2}{2}=2$
Now, AC can be calculated as follows:
Diagonals of a square are equal and bisect each other.
$\Rightarrow AC=\sqrt{(3+1{)}^2+(2-2{)}^2}$
$=\sqrt{16}=4$
Hence, sides of square = 2√2 (Using Pythagoras theorem)
Let coordinates of point D be $(x_1,y_1)$
$AD=2\sqrt{2}=\sqrt{(x_1+1{)}^2+(y_1-2{)}^2}$
$\Rightarrow 8=(x_1+1{)}^2+(y_1-2{)}^2$
$CD=2\sqrt{2}=\sqrt{(x_1-3{)}^2+(y_1-2{)}^2}$
$\Rightarrow 8=(x_1-3{)}^2+(y_1-2{)}^2$
From these equations, it is clear that;
$(x_1+1{)}^2+(y_1-2{)}^2$
$=(x_1-3{)}^2+(y_1-2{)}^2$
$\Rightarrow (x_1+1{)}^2=(x_1-3{)}^2$
$\Rightarrow {x_1}^2+2x+1={x_1}^2-6x+9$
$\Rightarrow 2x+1=-6x+9$
$\Rightarrow 8x=8$
$\Rightarrow x=1$
Value of $y_1$ can be calculated as follows by using the value of x.
$CD=2\sqrt{2}=\sqrt{(x_1-3{)}^2+(y_1-2{)}^2}$
$\Rightarrow 8=(1-3{)}^2+(y_1-2{)}^2$
$\Rightarrow 4+(y_1-2{)}^2=8$
$\Rightarrow y_1-2=2$
$\Rightarrow y_1=4$
Hence, D = (1, 4)
Coordinates of B can be calculated using coordinates of O; as follows:
Earlier, we had calculated O = (1, 2)
For BD, using section formula,
$1=\frac{x+1}{2}$
$\Rightarrow x+1=2$
$\Rightarrow x=1$
$2=\frac{y+4}{2}$
$\Rightarrow y+4=4$
$\Rightarrow y=0$
Hence, B = (1, 0) and D = (1, 4)