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Question

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


A
xy=2
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B
x+y=2
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C
xy=3
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D
x+y=3
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Solution

The correct option is A $$x-y = 2$$

Distance between two points $$
\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }
\right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{
x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Given, 
Distance between the points $$ (x,y) ; (7,1) = $$ Distance between $$ (x,y) ; (3,5) $$


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


$$\Rightarrow   \left( 7-x
\right) ^{ 2 }+\left( 1 - y \right) ^{ 2 }  = \left( 3-x
\right) ^{ 2 }+\left( 5 - y \right) ^{ 2 } $$

On expanding the squares and simplifying, we get

 $$ x - y = 2 $$


Mathematics

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