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Question 1. Asked on :23 September 2019:09:47:39 AM

The two opposite vertices of a square are (-1,2)and (3,2).find the coordinate of the two other vertices.

-Added by Rohit Rajput Mathematics » Cooridinate Triangle

Answer:

Akki chauhan

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)= (x – 3)2 + (y – 2)

(x + 1)2 = (x – 3)2

2 + 2x + 1 = x – 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).

-Answered by Akki chauhan On 23 September 2019:08:46:23 PM(10898Average Rating Based on rating)

Answer:

priyanshu kumar


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)= (x – 3)2 + (y – 2)

(x + 1)2 = (x – 3)2

2 + 2x + 1 = x – 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).

-Answered by priyanshu kumar On 02 October 2019:12:30:02 PM(11529Average Rating Based on rating)

Answer:

Suraj Kumar

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)= (x – 3)2 + (y – 2)

(x + 1)2 = (x – 3)2

2 + 2x + 1 = x – 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).

-Answered by Suraj Kumar On 07 November 2019:08:37:47 AM(16535Average Rating Based on rating)

 

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