Question

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

• A. $3x+y-5=0$
• B. $3x-y-5=0$
• C. $3x+y+5=0$
• D. $3x-y+5=0$

Answer 1

The correct option is A $3x+y-5=0$

Sol. $A(3,6)$ and $B(-3,4)$ are the given points. Point $P(x,y)$ is equidistant from the points A and B.
$PA=PB$
$\Rightarrow \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x+3{)}^2+(y-4{)}^2}$
$\Rightarrow (x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$
$\Rightarrow x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$

$\Rightarrow x^2-6x+9+y^2-12y+36-x^2-6x-9-y^2+8y-16=0$
$\Rightarrow -12x-4y+20=0$
$\Rightarrow 12x+4y-20=0$
$\Rightarrow 3x+y-5=0$