Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

Let the co-ordinates of the point on the x-axis be (x, 0).

Since, distance = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Let (x, 0) = (x₁, y₁) and (11, -8) = (x₂, y₂)

⇒ Distance between the given points =

$\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8 - 0)^2}\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8)^2}\\[1em] ⇒ 17^2 = (11 - x)^2 + (-8)^2\\[1em] ⇒ 289 = 121 + x^2 - 22x + 64\\[1em] ⇒ 289 = 185 + x^2 - 22x\\[1em] ⇒ 185 + x^2 - 22x - 289 = 0\\[1em] ⇒ x^2 - (26x - 4x) - 104 = 0\\[1em] ⇒ x^2 - 26x + 4x - 104 = 0\\[1em] ⇒ (x^2 - 26x) + (4x - 104) = 0\\[1em] ⇒ x(x - 26) + 4(x - 26) = 0\\[1em] ⇒ (x - 26)(x + 4) = 0\\[1em] ⇒ x = 26 \text{ and} -4$

Hence, the co-ordinates of points are (26, 0) and (-4, 0).