if relation between x and y such that the point (x, y) is equidistant from the point(7,1) and (3,5)
Answers
Given : point (x, y) is equidistant from the point(7,1) and (3,5)
To Find : relation between x and y
Solution:
point (x, y) is equidistant from the point(7,1) and (3,5)
(x - 7)² + ( y - 1)² = (x - 3)² + ( y - 5)²
=> x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² -10y + 25
=> -14x - 2y + 50 = -6x - 10y + 34
=> 8x - 8y = 16
=> x - y = 2
Another way :
point (x, y) is equidistant from the point (7,1) and (3,5)
hence its perpendicular bisector
mid point = ( 7 +3)/2 , ( 1 + 5)/2
= 5 , 3
Slope between points = ( 3 - 1)/( 5 - 7) = - 1
Hence Slope of perpendicular line = 1
y - 3 = 1 (x - 5)
=> y - 3 = x - 5
=> x - y = 2
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