please answer my question if u know
the two opposite vertices of a square are (-1,2) and (3,2). FIND THE COORDINATES OF ANOTHER TWO VERTICES
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sahil573:
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Answered by
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Hy..
ANNU...
THANKYOU YOU ASKING QUESTION
[HERE YOUR ANSWER]
ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (x, y).
AB = BC (ABCD is a square)
⇒ AB^2 = BC^2
⇒ [x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2 (Distance formula)
⇒ (x + 1)2 = (x – 3)2
⇒ x2 + 2x + 1 = x2 – 6x + 9
⇒ 2x + 6x = 9 – 1
⇒ 8x = 8
⇒ x = 1
IN ΔABC
AB^2 + BC^2 = AC^2
USE
(PYTHAGORAS THEOREM)
⇒ 2AB^2 = AC^2 (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
BEST OF LUCK
CHEERS
ANNU...
THANKYOU YOU ASKING QUESTION
[HERE YOUR ANSWER]
ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (x, y).
AB = BC (ABCD is a square)
⇒ AB^2 = BC^2
⇒ [x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2 (Distance formula)
⇒ (x + 1)2 = (x – 3)2
⇒ x2 + 2x + 1 = x2 – 6x + 9
⇒ 2x + 6x = 9 – 1
⇒ 8x = 8
⇒ x = 1
IN ΔABC
AB^2 + BC^2 = AC^2
USE
(PYTHAGORAS THEOREM)
⇒ 2AB^2 = AC^2 (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
BEST OF LUCK
CHEERS
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