find a relation b/w x and y such that the point (x, y) is equidistant from points ( - 2,4) And (3,5)
Answers
Solution :-
Let P( x, y ) be the point equidistant from points A ( - 2, 4 ) and B ( 3, 5 )
Therefore, AP = BP
i ) A ( - 2, 4 ) P ( x, y )
d = √[ (x₂ - x₁)² + (y₂ - y₁)² ]
⇒ AP = √[ (x - ( - 2 ))² + (y - 4)² ]
⇒ AP = √[ (x + 2)² + y² - 2*y*4 + 4² ]
⇒ AP = √( x² + 2*x*2 + 2² + y² - 8y + 16 )
⇒ AP = √( x² + 4x + 4 + y² - 8y + 16 )
⇒ AP = √( x² + 4x + y² - 8y + 20 )
ii ) B ( 3, 5 ) P ( x, y )
d = √[ (x₂ - x₁)² + (y₂ - y₁)² ]
⇒ BP = √[ (x - 3)² + (y - 5)² ]
⇒ BP = √( x² - 2*x*3 + 3² + y² - 2*y*5 + 5² )
⇒ BP = √( x² - 6x + 9 + y² - 10y + 25 )
⇒ BP = √( x² - 6x + y² - 10y + 34 )
But, AP = BP
⇒ √( x² + 4x + y² - 8y + 20 ) = √( x² - 9x + y² - 10y + 34 )
Squaring on both sides
⇒ [ √( x² + 4x + y² - 8y + 20 ) ]² = [ √( x² - 9x + y² - 10y + 34 ) ]²
⇒ x² + 4x + y² - 8y + 20 = x² - 9x + y² - 10y + 34
⇒ 4x - 8y + 20 = - 9x - 10y + 34
⇒ 4x - 8y + 20 + 9x + 10y - 34 = 0
⇒ 13x + 2y - 14 = 0