Question

A point is at a distance of $\sqrt{10}$ unit from the point $(4,3)$. Find the coordinates of point $A$, if its ordinate is twice its abscissa.

Answer 1

The coordinates of point $A$ are such that its ordinate is twice its abscissa.
Then, let the coordinates of point $A=(x,2x)$
$\sqrt{{(x-4)}^2+{(2x-3)}^2}=\sqrt{10}$
${(x-4)}^2+{(2x-3)}^2=10\Rightarrow x^2+16-8x+4x^2+9-12x=10$
$\Rightarrow 5x^2-20x+15=0\Rightarrow x^2-4x+3=0\Rightarrow x^2-x-3x+3=0$
$\Rightarrow x(x-1)(-3(x-1)=0\Rightarrow (x-1\left)\right(x-3)=0$
$x=1,3$
So, the coordinates of the point $A$ are $(1,2)$ and $(3,6)$