

10th Maths SA2 Chapter wise Test Papers Links 

The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices
Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (
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
⇒ x^2 + 2x + 1 = x^2 – 6x + 9
⇒ 2x + 6x = 9 – 1
⇒ 8x = 8 ⇒ x = 1
In ΔABC, we have
AB ^2 + BC^2 = AC^2 (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
Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).
Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (
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
⇒ x^2 + 2x + 1 = x^2 – 6x + 9
⇒ 2x + 6x = 9 – 1
⇒ 8x = 8 ⇒ x = 1
In ΔABC, we have
AB ^2 + BC^2 = AC^2 (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
Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).
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