Question

Find the point on x-axis which is equidistant from the point $A(3,2,2)$ and $B(5,5,4)$.

• A. $(16,0,0)$
• B. $\left(\frac{5}{4},0,0\right)$
• C. $(9,0,0)$
• D. $\left(\frac{49}{4},0,0\right)$

Answer 1

Answer: (D)

The point on the $x$-axis is of the form $P(x,0,0)$. Since the points $A$ and $B$ are equidistant from $P$. Therefore $P{A}^2=P{B}^2$
$\Rightarrow {\left(x-3\right)}^2+{\left(0-2\right)}^2+{\left(0-2\right)}^2$
$={\left(x-5\right)}^2+{\left(0-5\right)}^2+{\left(0-4\right)}^2$
$\Rightarrow 4x=25+25+16-17$
i.e., $x=\frac{49}{4}$
Thus, the point $P$ on the $x$-axis $\left(\frac{49}{4},0,0\right)$ is equidistant from $A$ and $B$.