#### Answer

$y = 1.5x + 0.5$
$y = -\frac{2}{3}x + 2\frac{2}{3}$

#### Work Step by Step

first of all lets find the derivative of y:
$y = x-\sqrt x=x+x^{0.5}$
$y' = (x+x^{0.5})'= (x)'+(x^{0.5})' =1+0.5x^{-0.5}$
So the slope of the tangent line at $(1, 2)$ is (x = 1, and calculate the derivative)
$1+0.5\times 1^{-0.5}=1.5$
We use the point-slope form to write an equation of the tangent line at $(1,2)$
$y - 2 = 1.5 (x-1)$
$y = 1.5x - 1.5 + 2 = 1.5x + 0.5$
The slope of the normal line at $(1,2)$ is the negative reciprocal of $1.5$, namely $-\frac{2}{3}$, so an equation is
$y - 2 = -\frac{2}{3}(x-1)$
$y = -\frac{2}{3}x +\frac{2}{3} + 2 = -\frac{2}{3}x + 2\frac{2}{3}$