**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*

**Answer:**

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)

AB^{2} = BC^{2}

[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:

AB^{2} + BC^{2} = AC^{2} (Pythagoras theorem)

2AB^{2} = AC^{2} (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

**Answer:**

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)

AB^{2} = BC^{2}

[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:

AB^{2} + BC^{2} = AC^{2} (Pythagoras theorem)

2AB^{2} = AC^{2} (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

**Answer:**

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)

AB^{2} = BC^{2}

[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:

AB^{2} + BC^{2} = AC^{2} (Pythagoras theorem)

2AB^{2} = AC^{2} (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

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