Question

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

Answer 1

Hope it helps you .....

This is the method.if any careless mistakes just correct it before you do and the distance formula is

= √ (x2- X1)^ 2 + (y2- y1)^2

Answer 2

Answer:

Let the point P(x, y), A(7, 1) and B(3, 5)

Step-by-step explanation:

According to statement

PA = PB

$\sqrt{ {(x - 7)}^{2} + {(y - 1)}^{2} } = \sqrt{ {(x - 3)}^{2} + {(y - 5)}^{2} } \\ squaring \: both \: sides \\ {x}^{2} + 49 - 14x + {y}^{2} + 1 - 2y = {x}^{2} + 9 - 6x + {y}^{2} + 25 - 10y \\ 50 - 14x - 2y = 34 - 6x - 10y \\ 50 - 34 = 14x - 6x + 2y - 10y \\ 16 = 8x - 8y \\ x - y = 2$