Question

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

• A. $(2,-1)$
• B. $(10,3)$
• C. $(2,1)$
• D. $(5,-2)$

Answer 1

The correct option is C $(2,1)$

$\frac{dy}{dx}=\frac{1}{2\sqrt{x-1}}=m_1$ 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
$m_1{m}_2=-1$
$⟹\left(\frac{1}{2\sqrt{x-1}}\right)(-2)=-1$
$⟹\frac{2}{2\sqrt{x-1}}=1$
$⟹\sqrt{x-1}=1$
Squaring both sides,

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