Math, asked by Kunj09, 1 year ago

the locus of a point equidistant from a given point (1,1) and (3,3)

Answers

Answered by rishu6845

Answer:

x+ y - 4 = 0

Step-by-step explanation:

Given---> Given point is equidistant from points

(1 , 1 ) and ( 3 , 3 )

To find ---> Locus of given point

Solution---> Let coordinates of P, A , B be ( x , y ) ,

( 1 , 1 ) and ( 3 , 3 )

Now , ATQ, point P is equidistance from A and B.

So, PA = PB

√{(x - 1)² + (y - 1)²} = √{(x-3)²+ (y-3)²}

Squaring both sides we get,

=> ( x-1 )² + ( y - 1 )² = ( x - 3 )² + ( y - 3 )²

We have a formula ,

( a - b )² = a² + b² - 2ab , applying it here , we get,

=> x²+1 - 2x + y² + 1 - 2y = x² + 9 - 6x + y² + 9 - 6y

=> -2x - 2y + 2 = -6x - 6y + 18

=> 6x - 2x + 6y - 2y + 2 - 18 = 0

=> 4x + 4y - 16 = 0

=> x + y - 4 = 0

Locus of given point is

x + y - 4 = 0

Answered by Aɾꜱɦ

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