Question

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Answer 1

Point (x, y) is equidistant from (3, 6) and (−3, 4).
$\therefore$ $\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x-(-3))^2+(y-4)^2}$
$\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}$
${(x-3)^2+(y-6)^2}={(x+3)^2+(y-4)^2}$
$36-16=6x+6x+12y-8y$
$20=12x+4y$
$3x+y=5$
$3x+y-5=0$