Question

Find a relation between x and y such that the point $P(x,y)$ is equidistant from the points $A(7,0)$ and $B(0,7)$ The relation is $x=py$ then $p=?$

Answer 1

$P$ is equidistant from $A(7,0)$ and $B(0,7)$. So, using distance formula:

$(x-7{)}^2+(y{)}^2=(x{)}^2+(y-7{)}^2$..............................[$PA=PB$]

$⟹x^2+49-14x+y^2=x^2+y^2+49-14y$

$⟹-14x=-14y$

$⟹x=1\times y$

Hence, $p=1$