Wellcome!

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

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)2 = (x – 3)2 + (y – 2)2 (x + 1)2 = (x – 3)2x 2 + 2x + 1 = x 2 – 6x + 92x + 6x = 9 – 18x = 8x = 1In 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)22[(x + 1)2 + (y – 2)2] = (4)2 + (0)22[(1 + 1)2 + (y – 2)2] = 16  ( x = 1)2[ 4 + (y – 2)2] = 168 + 2 (y – 2)2 = 162 (y – 2)2 = 16 – 8 = 8(y – 2)2 = 4y – 2 = ± 2y – 2 = 2 or y – 2 = –2y = 4 or y = 0Thus, 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)

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)2 = (x – 3)2 + (y – 2)2 (x + 1)2 = (x – 3)2x 2 + 2x + 1 = x 2 – 6x + 92x + 6x = 9 – 18x = 8x = 1In 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)22[(x + 1)2 + (y – 2)2] = (4)2 + (0)22[(1 + 1)2 + (y – 2)2] = 16  ( x = 1)2[ 4 + (y – 2)2] = 168 + 2 (y – 2)2 = 162 (y – 2)2 = 16 – 8 = 8(y – 2)2 = 4y – 2 = ± 2y – 2 = 2 or y – 2 = –2y = 4 or y = 0Thus, 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)

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)2 = (x – 3)2 + (y – 2)2 (x + 1)2 = (x – 3)2x 2 + 2x + 1 = x 2 – 6x + 92x + 6x = 9 – 18x = 8x = 1In 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)22[(x + 1)2 + (y – 2)2] = (4)2 + (0)22[(1 + 1)2 + (y – 2)2] = 16  ( x = 1)2[ 4 + (y – 2)2] = 168 + 2 (y – 2)2 = 162 (y – 2)2 = 16 – 8 = 8(y – 2)2 = 4y – 2 = ± 2y – 2 = 2 or y – 2 = –2y = 4 or y = 0Thus, 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)

You can see here all the solutions of this question by various user for NCERT Solutions. We hope this try will help you in your study and performance.

This Solution may be usefull for your practice and CBSE Exams or All label exams of secondory examination. These solutions or answers are user based solution which may be or not may be by expert but you have to use this at your own understanding of your syllabus.

#### What do you have in your Mind....

* Now You can earn points on every asked question and Answer by you. This points make you a valuable user on this forum. This facility is only available for registered user and educators.

## Search your Question Or Keywords

#### Do you have a question to ask?

User Earned Point: Select

## All Tags by Subjects:

Science (2231)
History (278)
Geography (310)
Economics (257)
Political Science (96)
Mathematics (200)
General Knowledge (5686)
Biology (94)
Physical Education (20)
Chemistry (118)
Civics (114)
Home Science (12)
Sociology (9)
Hindi (45)
English (258)
Physics (1435)
Other (97)
Accountancy (378)