Question

Two opposite vertices of a square are ( - 1 , 2) and (3 , 2) find the coordinates of other two vertices

Answer 1

The coordinates of the other two vertices are (1,0) and (1,4).

Step-by-step explanation:

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

It is clear that AC line segment is parallel to the x-axis and it is a diagonal AC of the square ABCD.

The midpoint of AC is $(\frac{- 1 + 3}{2}, \frac{2 + 2}{2}) = (1,2)$

The other diagonal BD will have midpoint (1,2) also and it will be parallel to the y-axis.

{Since the diagonals of a square bisect each other perpendicularly}

So, we can have the coordinates of B(1,h) and D(1,k) and the midpoint of BD will be $(\frac{1 + 1}{2}, \frac{h + k}{2}) = (1,2)$

Hence, h + k = 4 ........... (1)

Now, Distance from O(1,2) to C(3,2) is equal to distance from O(1,2) to D(1,k) and also from O(1,2) and (B(1,h).

So, OC = OD = OB

⇒ 2² = (k - 2)²

⇒ (k - 2) = ± 2

⇒ k = 0, 4

So, the coordinates of B are (1,0) and D are (1,4). (Answer)