Find the point on the x-axis which is equidistant from

Solution:

We know that any point on x-axis has its ordinate = 0

Let the required point be P(x, 0).

Let the given points be A(2, –5) and B(–2, 9)

$\therefore \quad \mathrm{AP}=\sqrt{(x-2)^{2}+5^{2}}=\sqrt{x^{2}-4 x+4+25}$

$=\sqrt{x^{2}-4 x+29}$

$\mathrm{BP}=\sqrt{[\mathbf{x}-(-2)]^{2}+(-\mathbf{9})^{2}}=\sqrt{(\mathbf{x}+2)^{2}+(-\mathbf{9})^{2}}$

$=\sqrt{x^{2}+4 x+4+81}=\sqrt{x^{2}+4 x+85}$

Since, $A$ and $B$ are equidistant from $P$,

$\therefore \quad \mathrm{AP}=\mathrm{BP}$

$\Rightarrow \sqrt{x^{2}-4 x+29}=\sqrt{x^{2}+4 x+85}$

$\Rightarrow x^{2}-4 x+29=x^{2}+4 x+85$

$\Rightarrow x^{2}-4 x-x^{2}-4 x=85-29$

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

$\therefore \quad$ The required point is $(-7,0)$