Question

The points A(3,2) and B(2,-3) are equidistant from a point P(x,y). Find the relation between x and y

Answer 1

SoluTion :-

A (3, 2) and B (2, -3) are equidistant from the point P (x, y).

∴ PA = PB

$\boxed {\sf {PA^{2} = PB^{2}}}\\\\\\\\\\\sf {\twoheadrightarrow \ (x - 3)^{2} + (y - 2)^{2} = (x - 2)^{2} + (y + 3)^{2}}\\\\\\\sf {\twoheadrightarrow \ x^{2} + 9 - 6x + y^{2} + 4 - 4y = x^{2} + 4 - 4x + y^{2} + 9 + 6y}\\\\\\\sf {\twoheadrightarrow \ -6x - 4y = -4x + 6y}\\\\\\\sf {\twoheadrightarrow \ -2x - 10y = 0}\\\\\\\sf {\twoheadrightarrow \ x + 5y = 0}$

Answer 2

Step-by-step explanation:

Question:-

The points A(3,2) and B(2,-3) are equidistant from a point P(x,y). Find the relation between x and y

SoluTion :-

Given us,

A (3, 2) and B(2, -3) are equidistant from the point P (x, y).

Find it :-

the relation between x and y

We know

∴ PA = PB

$\bf \red{squaring \: both \: sides}$

Then,

${\bf \red {PA^{2} = PB^{2}}}\\\ \: \bf{ = \ (x - 3)^{2} + (y - 2)^{2} = (x - 2)^{2} + (y + 3)^{2}}\\ \\ \: \bf { = \ x^{2} + 9 - 6x + y^{2} + 4 - 4y = x^{2} + 4 - 4x + y^{2} + 9 + 6y}\\ \\ \bf { = -6x - 4y = -4x + 6y}\\\\ \bf \: { = - 6x + 4x = 6y + 4y }$

Add subtraction

$\bf - 2x \: = 10y$

On moving 10 from one side to another.

$\bf \: - 2x - 10y = 0$

2 cut from the multiplication table

Then,

$\bf \red{ - x - 5y = 0}$

Then,

$\bf \red{ x </strong><strong>+</strong><strong> </strong><strong>5y = 0}$

HENCE,

Solved The Question,