Question

Equation to the locus of a point equidistant from the points $A(1,3)$ and $B(-2,1)$ is

• A. $4x+6y=5$
• B. $6x+4y=5$
• C. $-4x+6y=8$
• D. None of these

Answer 1

Answer: (B)

$6x+4y=5$

Let the point be $P(x,y)$

Distance between $P(x,y)$ and $A(1,3)=\sqrt{{(1-x)}^2+{(3-y)}^2}=\sqrt{1+x^2-2x+9+y^2-6y}=\sqrt{x^2+y^2-2x-6y+10}$
Distance between $(x,y)$ and $(-2,1)=\sqrt{{(-2-x)}^2+{(1-y)}^2}=\sqrt{4+x^2+4x+1+y^2-2y}=\sqrt{x^2+y^2+4x-2y+5}$
As the point $(x,y)$ is equidistant from the two points, both the distances calculated are equal.
$\Rightarrow \sqrt{x^2+y^2-2x-6y+10}=\sqrt{x^2+y^2+4x-2y+5}$
$\Rightarrow x^2+y^2-2x-6y+10=x^2+y^2+4x-2y+5$
$\Rightarrow 6x+4y=5$