Question

If P(x,y) is equidistant from Q(-2,5) and R(6,-1) then determine a relationship between x and y

Answer 1

Answer:

$\text{The relation between x and y is }\frac{12y+8}{16}$

Step-by-step explanation:

Given P(x,y) is equidistant from Q(-2,5) and R(6,-1).

we have to determine the relation between x and y.

As given PQ=PR

By distance formula,

Distance between the coordinates $(x_1,y_1)\text{ and }(x_2,y_2)$ is

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

As,  PQ=PR

$\sqrt{(-2-x)^2+(5-y)^2}=\sqrt{(6-x)^2+(-1-y)^2}\\\\\text{Squaring on both sides}\\\\4+x^2+4x+25+y^2-10y=36+x^2-12x+1+y^2+2y\\\\29+4x-10y=37-12x+2y\\\\16x=12y+8\\\\x=\frac{12y+8}{16}$

which is the required relation.