Question

if the point x,y is eqidistant from the points a+b,b-a and a-b ,a+b then prove that bx = ay​

Answer 1

Answer:

before solution of given question 1st trying to revise some formulas which will require to use on given question

1.distance formula :

if [math]A(x_1,y_1)[/math] and [math]B(x_2,y_2)[/math] are 2 points then distance AB is given by

[math]d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/math]

2) algebric property :

[math]A^2-B^2=(A+B)(A-B)[/math]

Step-by-step explanation:

see the pic

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

According to the question

The point (x,y) is eqidistant from the points (a+b,b-a )and (a-b ,a+b) ;

•therefore the distance between the points

(x,y) and (a+b,b-a )

$\implies \sqrt{(x - (a + b)) {}^{2} + (y - (b - a)) {}^{2} }$

•therefore the distance between the points

(x,y) and (a+b,b-a )

$\implies \sqrt{(x - (a - b) {}^{2} + (y - (a + b)) {}^{2} }$

As, the points are equidistant from the point (x,y)

then,,,,

The rest of calculation is in attachment.....

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \#\mathcal{answer with quality }\: \: \& \: \: \#BAL }$