Question

Find a relation between $x$ and $y$ such that the point $(x.y)$ is equidistant from the points $(7,1)and(3,5)$.

Answer 1

By using distance formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ we have,

$d(x.y,7.1)=\sqrt{(x-7{)}^2+(y-1{)}^2}$

$d(x.y,3.5)=\sqrt{(x-3{)}^2+(y-5{)}^2}$

Squaring above and equating we have,

$\therefore (x-7{)}^2+(y-1{)}^2=(x-3{)}^2+(y-5{)}^2$

$\therefore x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25$

$\therefore 50-14x-2y=34-6x-10y$

$\therefore 16=8x-8y$

$\therefore 2=x-y$

$\therefore y=x-2$