Question

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

Answer:

Let B (x, y).

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

HOPE THIS HELPS YOU!!

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Answer 2

Step-by-step explanation:

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