Question

The point in the $xy$-plane which is equidistant from the points $(5,0,6)$, $(0,-3,2)$ and $(4,5,0)$ is

• A. $(1,2,3)$
• B. $(-3,13,0)$
• C. $(13,-2,0)$
• D. None of these

Answer 1

Answer: (D)

Any point in $xy$-plane is of the form $(x,y,0)$. Let $P(x,y,0)$ is equidistant from $A(5,0,6)$, $B(0,-3,2)$ and $C(4,5,0)$.
$\therefore$ $PA=PB$
$\Rightarrow P{A}^2=P{B}^2$
$\Rightarrow {\left(x-5\right)}^2+{\left(y-0\right)}^2+{\left(0-6\right)}^2$
$={\left(x-0\right)}^2+{\left(y+3\right)}^2+{\left(0-2\right)}^2$
$\Rightarrow x^2+25-10x+y^2+36$
$=x^2+y^2+9+6y+4$
$\Rightarrow 5x+3y=24\dots \left(i\right)$
Also, $PB=PC$
$\Rightarrow P{B}^2=P{C}^2$
$\Rightarrow {\left(x-0\right)}^2+{\left(y+3\right)}^2+{\left(0-2\right)}^2$
$={\left(x-4\right)}^2+{\left(y-5\right)}^2+{\left(0-0\right)}^2$
$\Rightarrow x^2+y^2+9+6y+4$
$=x^2+16-8x+y^2+25-10y$
$\Rightarrow 2x+4y=7\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$x=\frac{75}{14}$, $y=\frac{-13}{14}$
$\therefore$ Coordinates of $P=\left(\frac{75}{14},\frac{-13}{14},0\right)$.