if 2 opposite vertices of a square are (5,4) and (1,-6) . find the co-ordinates of its remaining vertices ???can you explain in different method

Asked by Rutuja Mehta | 15th Oct, 2013, 09:49: AM

Expert Answer:

Let ABCD be a square and let A(5, 4) and C(1, -6) be the given two vertices.

Also, let (x, y) be the coordinates of point B. 
Since AB=BC

(AB)2 = (BC)2

(x - 5)+ (y - 4)= ­­(1 - x)+ (-6 - y)2

Or x-10x + 25 + y- 8y + 16 = 1 + x- 2x + 36 + 12y + y2

-10x + 25 - 8y + 16 = 1 - 2x + 36 + 12y

-8x - 20y = 37 - 41

-4(2x + 5y) = -4

2x + 5y = 1    ....(1)

In right angled triangle ABC,

AB2 + BC2 = AC2

(x - 5)2 + (y - 4)2 + (x - 1)2 + (y + 6)2 = (5 - 1)2 + (4 + 6)2

x2 - 10x + 25 + y2 - 8y + 16 + x2 - 2x + 1 + y2 + 12y + 36 = 16 + 100

2x2 + 2y2 - 12x + 4y + 78 - 116 = 0

x2 + y2 - 6x + 2y + 39 - 58 = 0

x2 + y2 - 6x + 2y - 19 = 0 (ii)

From (i) x =  , putting in (ii), we get,

 

25y2 - 10y + 1 + 4y2 + 68y - 88 = 0

29y2 + 58y - 87 = 0

 y2 + 2y - 3 = 0

(y + 3)(y - 1) = 0

    y = -3, 1

When y = -3, x = = 8 ;

when y = 1, x = 

Therefore, required points are (-2, 1), (8, -3)

Answered by Rashmi Khot | 15th Oct, 2013, 02:53: PM

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