Question

Example 4 : Find a relation between x and y such that the point (x, y) is equidistant
from the points (7.1) and (3,5).
Solution Tet Prahe equidistant from the points (71) and B(35)​

Answer 1

Step-by-step explanation:

Let the Points be

=> A ( x , y )

=> B ( 7 , 1 )

=> C ( 3, 5 )

According to the given question,

=> AB = AC

Applying Distance Formula ,

=> √ ( x2 - x1 )² + ( y2 - y1 )²

Distance AB Will be,

=> √ ( 7 - x )² + ( 1 - y )²

=> √ 49 + x² - 14x + 1 + y² - 2y

=> √ x² + y² + 50 - 14x - 2y

Distance AC Will be,

=> √ ( x - 3)² + ( y - 5 )²

=> √ x² + 9 - 6x + y² + 25 - 10y

=> √ x² + y² + 36 - 6x - 10 y

Square on both sides,

=> x² + y² + 36 - 6x - 10y = x² + y² + 50 - 14x -2y

=> -8x + 8y + 16 = 0

=> -8 ( x - y -2 ) = 0

=> x - y = 2

=> x = y + 2

Hence the Relation is x = y + 2

Hope it helps!