Question

Find the value of $a$ when the distance between the points $(3,a)$ and $(4,1)$ is $\sqrt{10}.$The points $(2,1)$ and $(1,-2)$ are equidistant from the point $\left(x,y\right),$Find locus of point

Answer 1

$A=(3,a),B=(4,1)$

$\therefore AB=\sqrt{{(4-3)}^2+{(1-a)}^2}=\sqrt{10}$

$\therefore 1^2+a^2+1-2a=10$

$a^2-2a-8=0$

$\Rightarrow a=4,-2$

$P=(2,1),Q=(1,-2),X=(x.y)$

$PX=QX$

$\Rightarrow {PX}^2={QX}^2$

$\Rightarrow {\left[\sqrt{{(x-2)}^2+{(y-1)}^2}\right]}^2={\left[\sqrt{{(x-1)}^2+{(y+2)}^2}\right]}^2$

$\Rightarrow {(x-2)}^2+{(y-1)}^2={(x-1)}^2+{(y+2)}^2$

$\Rightarrow x^2+4-4x+y^2+1-2y=x^2+1-2x-y^2+4+4y$

$\Rightarrow 2x+6y=0$

$\Rightarrow x+3y=0$ is required locus