If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that ax = by.

Hint:We will use the distance formula to find the length of AP and PB and then equate them, since they are given to be equidistant. Then we will get the desired result by simplifying them.

Complete step-by-step answer:
Let us first write the distance formula for two given points.
We know that the distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by:
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ units ……..(1)
Now, if we have to find the length of AP, we have \[{x_1} = a + b,{y_1} = b - a,{x_2} = x\] and ${y_2} = y$.
Putting these values in (1), we will have:-
$AP = \sqrt {{{\{ x - (a + b)\} }^2} + {{\{ y - (b - a)\} }^2}} $ units …………(2)
Now, if we have to find the length of BP, we have \[{x_1} = a - b,{y_1} = a + b,{x_2} = x\] and ${y_2} = y$.
Putting these values in (1), we will have:-
$BP = \sqrt {{{\{ x - (a - b)\} }^2} + {{\{ y - (b + a)\} }^2}} $ units …………(3)
Now, since the points A and B are equidistant from P, hence AP = BP ……….(4)
Therefore, by (2), (3) and (4), we will get:-
$ \Rightarrow \sqrt {{{\{ x - (a + b)\} }^2} + {{\{ y - (b - a)\} }^2}} = \sqrt {{{\{ x - (a - b)\} }^2} + {{\{ y - (b + a)\} }^2}} $
Squaring both sides, we will get:-
$ \Rightarrow {\{ x - (a + b)\} ^2} + {\{ y - (b - a)\} ^2} = {\{ x - (a - b)\} ^2} + {\{ y - (b + a)\} ^2}$
Now, we will use the formula: ${(a - b)^2} = {a^2} + {b^2} - 2ab$, we will now have:-
\[ \Rightarrow {x^2} + {(a + b)^2} - 2x(a + b) + {y^2} + {(b - a)^2} - 2y(b - a) = {x^2} + {(a - b)^2} - 2x(a - b) + {y^2} + {(a + b)^2} - 2y(a + b)\]On simplifying this, we will have:-
\[ \Rightarrow - 2x(a + b) - 2y(b - a) = - 2x(a - b) - 2y(a + b)\]
On simplifying it further, we will get:-
\[ \Rightarrow - 2ax - 2bx + 2ay - 2by = - 2ax + 2bx - 2ay - 2by\]
Cutting off the common terms from both the sides, we will get:-
\[ \Rightarrow - 2bx + 2ay = 2bx - 2ay\]
Taking 2 common from both the sides to cut it off, we will get:-
\[ \Rightarrow - bx + ay = bx - ay\]
Rearranging the terms, we will get:-
\[ \Rightarrow 2ay = 2bx\]
Taking 2 common from both the sides to cut it off again, we will get:-
\[ \Rightarrow ay = bx\]
Hence, proved.

Note:A point is said to be equidistant if the distances between from that point to given points are equal.For example,In circle the center of a circle is equidistant from every point on the circle.Similarly , In Rectangle the centre of rectangle is equidistant from all four vertices, and it is equidistant from two opposite sides and also equidistant from the other two opposite sides.Students should remember the formula for calculating the distance between two points i.e $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ for solving these types of problems.