Question

find a relation between x and y if the point p(x, y)is equidistant from A(4,5) and B(-2,3).please answer fast ​

Answer 1

Step-by-step explanation:

Given that a point P(x, y) is equidistant from two points A(4, 5) and B(-2, 3). Therefore, the distance between AP and AB will be same.

We are aware about the distance formula:

• $\boxed{\sf Distance=\sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2}}$

This is the formula to calculate distance between two points say (x1, y1) and (x2, y2).

By using this formula and equating AP = BP, we get:

$\sf \implies AP = BP$

$\sf \implies \sqrt{(x-4)^2 + (y-5)^2} = \sqrt{(x-(-2))^2 + (y-3)^2}$

$\sf \implies \sqrt{(x-4)^2 + (y-5)^2} = \sqrt{(x + 2)^2 + (y-3)^2}$

$\sf \implies (x-4)^2 + (y-5)^2= (x + 2)^2 + (y-3)^2$

Now, using the algebraic identities:-

• $\boxed{\sf(A+B)^2 = A^2 + B^2 + 2AB}$
• $\boxed{\sf(A-B)^2 = A^2 + B^2 - 2AB}$

$\sf \implies {x}^{2} + 16 - 8x+ {y}^{2} + 25 - 10y = {x}^{2} + 4 + 4x+ {y}^{2} + 9 - 6y$

$\sf \implies \not {x}^{2} + 16 - 8x+ \not{y}^{2} + 25 - 10y = \not {x}^{2} + 4 + 4x+ \not{y}^{2} + 9 - 6y$

$\sf \implies 16 - 8x + 25 - 10y = 4 + 4x + 9 - 6y$

$\sf \implies 0 = 4 + 4x + 9 - 6y - 16 + 8x - 25 + 10y$

$\sf \implies 0 = 12x + 4y - 12$

$\sf \implies 0 = 3x + y - 6$

$\boxed{\sf \implies 6 = 3x + y }$

This is the required relation between x and y