Find points on the x-axis which are at a distance of 5 units from the point (5, -4).

We know that,

y-coordinate of any point on x-axis = 0.

Let point on x-axis which is at a distance of 5 units from (5, -4) be P(x, 0).

By distance formula,

$\Rightarrow d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \Rightarrow 5 = \sqrt{(x - 5)^2 + [0 - (-4)]^2} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 25 - 10x + 16} \\[1em] \Rightarrow (5)^2 = x^2 - 10x + 41

On squaring both sides,

$\Rightarrow 25 = x^2 - 10x + 41 \\[1em] \Rightarrow x^2 - 10x + 41 - 25 = 0 \\[1em] \Rightarrow x^2 - 10x + 16 = 0 \\[1em] \Rightarrow x^2 - 8x - 2x + 16 = 0 \\[1em] \Rightarrow x(x - 8) - 2(x - 8) = 0 \\[1em] \Rightarrow (x - 2)(x - 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = 8.$

∴ P = (x, 0) = (2, 0) or (8, 0).

Hence, points on the x-axis which are at a distance of 5 units from the point (5, -4) are (2, 0) or (8, 0).