The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Let ABCD is a square, where A(-1, 2) and B(3, 2). And Point O is the point of intersection of AC and BD
Coordinate of points B and D.
Step 1: Find distance between A and C and coordinates of point O.
AC = \(\sqrt{(3+1)^2 + (2-2)^2}\) = 4
Coordinates of O can be calculated as follows:
x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2
O(1,2)
Step 2: Find the side of the square using Pythagoras theorem
Let a be the side of square and AC = 4
From right triangle, ACD,
a = 2√2
Hence, each side of square = 2√2
Step 3: Find coordinates of point D
Equate length measure of AD and CD
Say, if coordinate of D are (x1, y1)
AD = \(\sqrt{(x_1 + 1)^2 + (y_1 - 2)^2}\)
Squaring both sides,
AD2 = (x1 + 1)2 + (y1 – 2)2
Similarly, CD2 = (x1 – 3)2 + (y1 – 2)2
Since all sides of a square are equal, which means AD = CD
(x1 + 1)2 + (y1 – 2)2 = (x1 – 3)2 + (y1 – 2)2
x12 + 1 + 2x1 = x12 + 9 – 6x1
8x1 = 8
x1 = 1
Value of y1 can be calculated as follows by using the value of x.
From step 2: each side of square = 2√2
CD2 = (x1 – 3)2 + (y1 – 2)2
8 = (1 – 3)2 + (y1 – 2)2
8 = 4 + (y1 – 2)2
y1 – 2 = 2
y1 = 4
Hence, D = (1, 4)
Step 4: Find coordinates of point B
From line segment, BOD
Earlier, we had calculated O = (1, 2)
Say B = (x2, y2)
For BD;
1 = (x2 + 1)/2
x2 = 1
And 2 = (y2 + 4)/2
=> y2 = 0
Therefore, the coordinates of required points are B = (1,0) and D = (1,4)