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# If two opposite vertices of a square are ( 5,4 ) and ( 1,6) find the coordinates of its remaining vertices

Asked by 1st March 2013, 12:07 PM

Answer : Given :If two opposite vertices of a square are ( 5,4 ) and ( 1,6)

To find : the coordinates of its remaining vertices

Let B (xy).

We have AB = BC

Squaring both sides

=> AB2 = BC2

=> (x-5)2 +(y-4)2 = (x-1)2 +(y-6)2

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

=> 4 - 8x +4y =0

=>2x - y =1 .........(1)

also AC2 = (5-1)2 + (4-6)2 = 16 + 4 = 20

In right triangle ABC, by Pythagoras theorem,

AC2 = AB2+BC2

=> AC2 = 2AB2

=> 20 = 2 ( (x-5)2 +(y-4)2 )

=> 20 = 2 ( x2 + 25 -10x+ y2 +16 -8y )

=> 10=  x2 + 25 -10x+ y2 +16 -8y

=>  x2  -10x+ y2  -8y + 31 =0

=>  x2  -10x+ (2x+1)2  - 8(2x+1) + 31 =0 {using eq 1}

=>  x2  -10x+ 4x2 + 1 + 4x  -16x-8 + 31 =0

=>5 x2 -22x +24 =0

D = b2 -4ac

=> D = (-22)2 - 4(5)(24)

=> D = 484 - 480

=> D =4

=> D1/2 = 2

x = (-b+D1/2) / 2a , (-b - D1/2) / 2a

=> x = (22 +2) / 10 , (22-2) / 10

=> x = 2.4 , 2

=> y = 2x+1 , => y = 5.8 , 5

therefore the points are ( 2.4 , 5.8) and ( 2,5) Answer

Answered by Expert 1st March 2013, 3:32 PM
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