Question

find the relation between x and y such yhat R(x,y) is equidistant from the point p(7,1) and Q(3,5)​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The relation between x and y such that R(x,y)

is equidistant from the point P (7,1) and Q(3,5)

CONCEPT TO BE IMPLEMENTED

The distance between two points

$\sf{(x_1,y_1) \: \: and \: \: (x_2,y_2) \: is \: }$

$\sf{ = \sqrt{ { (x_1 - x_2)}^{2} + { (y_1 - y_2)}^{2} \: \: } \: }$

EVALUATION

The given points are P (7,1) and Q(3,5) and

R (x, y)

The distance between P and R is

$\sf{ = \sqrt{ {(x - 7)}^{2} + {(y - 1)}^{2} } \: }$

The distance between Q and R is

$\sf{ = \sqrt{ {(x - 3)}^{2} + {(y - 5)}^{2} } \: }$

By the given condition

$\sf{ \sqrt{ {(x - 7)}^{2} + {(y - 1)}^{2} } \: } \sf{ = \sqrt{ {(x - 3)}^{2} + {(y - 5)}^{2} } \: }$

$\implies \sf{ {(x - 7)}^{2} + {(y - 1)}^{2} } \: \sf{ = {(x - 3)}^{2} + {(y - 5)}^{2} } \:$

$\implies \sf{ {x}^{2} - 14x + 49 + {y}^{2} - 2y + 1 = {x}^{2} - 6x + 9 + {y}^{2} - 10y + 25 }$

$\implies \sf{ 8x - 8y = 16 } \:$

$\implies \sf{ x - y = 2 } \:$

RESULT

Hence the required relation is

$\boxed{ \sf{ \: \: x - y = 2 } \: }$

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