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

It is given that P (x,y) is equidistant from A(3, 6) and B(-3, 4).
Using Distance formula, we can write
PA = PB
$\sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} = \sqrt {{{[x - ( - 3)]}^2} + {{(y - 4)}^2}} $
$ \Rightarrow \sqrt {{x^2} + 9 - 6x + {y^2} + 36 - 12y} = \sqrt {{x^2} + 9 + 6x + {y^2} + 16 - 8y} $
Squaring both sides, we get
$ \Rightarrow $$ x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 8y$
$\Rightarrow −6x − 12y + 45 = 6x - 8y + 25$
$\Rightarrow 12x + 4y = 20$
$\Rightarrow 3x + y = 5$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free