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. $(-3 ,1 3 ,0 )$
• 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 $ $PA = PB$
$\Rightarrow PA^{2} = PB^{2}$
$\Rightarrow \left(x- 5\right)^{2} + \left(y - 0\right)^{2} + \left(0 - 6\right)^{2}$
$= \left(x- 0\right)^{2} + \left(y + 3\right)^{2} + \left(0 - 2\right)^{2}$
$\Rightarrow x^{2} + 25 - 10x + y^{2} + 36$
$= x^{2}+y^{2} + 9 + 6y + 4$
$\Rightarrow 5x + 3y = 24 \quad\ldots\left(i\right)$
Also, $PB = PC$
$\Rightarrow PB^{2} = PC^{2}$
$\Rightarrow \left(x - 0\right)^{2} + \left(y + 3\right)^{2} + \left(0 - 2\right)^{2}$
$= \left(x - 4\right)^{2} + \left(y - 5\right)^{2} + \left(0 - 0\right)^{2}$
$\Rightarrow x^{2}+y^{2} + 9 + 6y + 4$
$= x^{2}+ 1 6 -8 x + y^{2} + 25 - 10y$
$\Rightarrow 2x + 4y = 7\quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$x = \frac{75}{14}$, $y = \frac{-13}{14}$
$\therefore $ Coordinates of $P = \left(\frac{75}{14}, \frac{-13}{14}, 0\right)$.