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Answer:
(1,4) (1,2) hope it's helpful for you
Answer:
other two vertices of the square ABCD are (1, 4) and (1, 0).
Step-by-step explanation:
Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinates of vertex B be (x, y).
AB = BC (As ABCD is a square)
AB2 = BC2
[x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2
(x + 1)2 = (x – 3)2
x 2 + 2x + 1 = x 2 – 6x + 9
2x + 6x = 9 – 1
8x = 8
x = 1
In ΔABC, we have:
AB2 + BC2 = AC2 (Pythagoras theorem)
2AB2 = AC2 (Since, AB = BC)
2[(x – (–1))2 + (y – 2)2] = (3 – (–1))2 + (2 – 2)2
2[(x + 1)2 + (y – 2)2] = (4)2 + (0)2
2[(1 + 1)2 + (y – 2)2] = 16 ( x = 1)
2[ 4 + (y – 2)2] = 16
8 + 2 (y – 2)2 = 16
2 (y – 2)2 = 16 – 8 = 8
(y – 2)2 = 4
y – 2 = ± 2
y – 2 = 2 or y – 2 = –2
y = 4 or y = 0
Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).