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
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