The point on the curve $y = \sqrt {x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0$ is

$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {x - 1}} = m_{1}$ is the slope of tangent to $y=\sqrt{x-1}$

Slope of the line $2x + y - 5 = 0$ is $m_{2} = -2$

For lines are perpendicular

$m_{1} m_{2} = -1$

$\implies \left (\dfrac {1}{2\sqrt {x - 1}}\right )(-2) = -1$

$\implies \dfrac {2}{2\sqrt {x - 1}} = 1$

$\implies \sqrt {x - 1} = 1$

Squaring both sides,

$\implies x - 1 = 1$

$\implies x = 2$

$\therefore y = \sqrt {x - 1}$

$= \sqrt {2 - 1}$

$= \sqrt {1}$

$\therefore y = 1$

$\therefore (2, 1)$ is the point on the curve $y=\sqrt{x-1}$