Question

Q.Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
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Answer 1

Answer:

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

Answer 2

Answer:

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