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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

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

Question 2. Asked on :23 September 2019:09:44:14 AM

#### find the center of circle passing though the points (6,-6),(3,-7)and (3,3).

-Added by Rohit Rajput Mathematics » Cooridinate Triangle

priyanshu kumar

let O (x, y) is the point of circle

if three given points A (3,-7) B (3,3) and C (6,-6)

we know distance between circumference and center is always same. i.e radius .
now,
OA^2=OB^2=OC^2

OA^2=OB^2
=>(x-3)^2+(y+7)^2=(x-3)^2+(y-3)^2

=>(x-3)^2-(x-3)^2=(y-3)^2-(y+7)^2

=> 0=(2y+4)(3)

=> y= -2
now again ,
OB^2=OC^2
(x-3)^2+(y-3)^2=(x-6)^2+(y+6)^2
put y=-2

=>(x-3)^2+(-2-3)^2=(x-6)^2+(-2+6)^2

=>(x-3)^2-(x-6)^2=16-25

=>(2x-9)(3)=-9

=> 2x= -3+9=6

=> x=3

hence center co-ordinate is (3,-2)

-Answered by priyanshu kumar On 02 October 2019:12:31:07 PM

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