Let $(\mathrm{x}, 0)$ be the point on the $\mathrm{x}$ axis. Then as per the question, we have

$\sqrt{(x-2)^{2}+(0+5)^{2}}=\sqrt{(x+2)^{2}+(0-9)^{2}}$

$\Rightarrow \sqrt{(x-2)^{2}+(5)^{2}}=\sqrt{(x+2)^{2}+(-9)^{2}}$

$\Rightarrow(\mathrm{x}-2)^{2}+(5)^{2}=(\mathrm{x}+2)^{2}+(-9)^{2} \quad \text { (Squaring both sides) }$

$\Rightarrow x^{2}-4 x+4+25=x^{2}+4 x+4+81$

$\Rightarrow 8 \mathrm{x}=25-81$

$\Rightarrow \mathrm{x}=-\frac{56}{8}=-7$

Therefore, the point on the $\mathrm{x}$-axis is $(-7,0)$.