Question

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

Answer 1

In the above Question , we have the following information -

The locus of the point (x y) is equidistant from the point (7, 1)and (3 , 5) .

To find -

Find a relation between x and y such that , this condition satisfies .

Solution -

Here , let the point P be ( h, k ) ,

Now ,

Let the point ( 7, 1 ) be labelled as A and (3, 5) as b respectively.

Now , by the condition ,

PA = PB .

Squaring both sides -

=> { PA }² = { PB }² .

=> ( h - 7 )² + ( k - 1 )² = ( h - 3 )² + ( k - 5 )².

=> h² - 14h + 49 + k² - 2k + 1 = h² - 6h + 9 + k² - 10k + 25.

Here , the h² and k² terms get cancelled.

=> - 14h - 2k + 50 = - 6h - 10k + 34

=> 14h - 6h - 10k + 2k = 50 - 34

=> 8h - 8k = 16

=> h - k = 2 .

Substituting h and k with x and y respectively .

=> x - y = 2 .

This is the required relation .

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