Math, asked by choudharynoori26, 11 months ago

If $(7,9)$ and $(10,2)$ are the coordinates of two opposite vertices of a square, what is the sum of the $y$-coordinates of the other two vertices?

Answers

Answered by amitnrw
4

Answer:

11

Step-by-step explanation:

$(7,9)$ and $(10,2)$ are the coordinates of two opposite vertices of a square

Let say  Co-ordinates of Other Vertices (x , y)

=>(x - 7)² + (y-9)² = (x-10)² + (y - 2)²

=> x² + 49 -14x + y² +81 - 18y = x² + 100 - 20x + y² + 4 - 4y

=> 6x = 14y - 26

=> 3x = 7y - 13

=> x = (7y - 13)/3

Other two Vertices will Satisfy this

Length of Diagonal = √(10-7)² + (2-9)²  = √9 + 49 = √58 = √2*√29

Length of Square side = Diagonal /√2 = √29

Side = (x - 7)² + (y - 9)² = (√29)²

=>  ((7y - 13)/3  - 7)² + (y - 9)² = 29

=> (7y - 13 - 21)² + 9(y-9)² = 29*9

=> (7y - 34)² + 9(y² + 81 - 18y) = 261

=> 49y² + 1156 - 476y + 9y² + 729 - 162y = 261

=> 58y² - 638y + 1624 = 0

Sum of Y co-ordinates = Sum of roots = -(-638)/58

= 11

the sum of the Y-coordinates of the other two vertices = 11

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