Question

Each side of a square is of length 4. The centre of the square is (3, 7) and one of its diagonals is parallel to the line y = x. Find the coordinate of the vertices of square.

Answer 1

See Fig. ABCD is the square. M(3, 7) is the centre of the square. AC is parallel to the line y = x. Therefore, the equation of the diagonal AC is

y - 7 = (x - 3)

x - y + 4 = 0   ....(1)

Hence, the equation of the diagonal BD is

y − 7 = −1 (x − 3)

x + y - 10 = 0  ...(2)

Since the length of the side is 4, the lengths of the diagonals are 4√2. Let A = (h, h + 4) and MA = 2√2. This implies that

(h - 3)² + (h + 4 - 7)² = 8

(h - 3)² = 4

h = 3 ± 2

h = 5 or 1

Hence, A = (5,9) and C = (1,5). Let B = (k, 10 −k) and MD = 2√2. This implies that

(k - 3)² + (10 - k - 7)² = 8

(k - 3)² = ± 2

k = 5 or  1

Hence, B  = (55) and D = (19). Therefore, the vertices of the square are (5, 9), (5, 5), (1, 5), (1, 9).