Question

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$\huge \underline{Question:-}$

1.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

tex] \huge \pink{Hello!! }[/tex]

$\huge \underline{Question:-}$

Let P(x,y) be equidistant from the points A(7,1) and B(3,5).

AP=BP

AP

2

=BP

2

(x−7)

2

+(y−1)

2

=(x−3)

2

+(y−5)

2

x

2

−14x+49+y

2

−2y+1=x

2

−6x+9+y

2

−10y+25

x−y=2

hope it's help you

Answer 2

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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