Answer
a) $\color{blue}{\dfrac{y - 1/\sqrt{2}}{x-\sqrt{2}} = -\dfrac{1}{2}}\quad$ or $\quad\displaystyle \color{blue}{y = -\frac{x}{2} + \sqrt{2}}$.
b) The area under the tangent line in part a) over $[1,2]$ (area of trapezoid $A'BCD'$ in the attached diagram) is $\approx 0.664$, and is less than $\ln 2$, the area under $y=1/x,x\gt 0$ over the closed interval $[1,2]$. Thus $\ln 2 > 0.66$. See details below.
Work Step by Step
a)
The slope of secant line $\overleftrightarrow{AD}$ passing through the points $A(1,1)$ and $D(2,1/2)$ is
$\begin{align*}
m_{AD} &= \frac{\Delta y}{\Delta x} \\
&= \frac{1/2-1}{2-1} \\
m_{AD}&= -\frac{1}{2}
\end{align*}
$
Since a tangent line to $y=1/x, x\gt 0$, has slope $y' = -1/x^2$, then such a tangent line that is parallel to the secant line $\overleftrightarrow{AD}$ must satisfy
$\begin{align*}
-\frac{1}{x^2} &= -\frac{1}{2} \\
x^2 &= 2 \\
x &= \sqrt{2}\qquad [\text{disregard}\ -\sqrt{2}\ \text{since}\ x \gt 0 ]
\end{align*}
$
implying that such a tangent line must be tangent to $y = 1/x,x\gt 0$ at $x=\sqrt{2}$, so that the point of tangency must be $(x_T,y_T) = (\sqrt{2},1/\sqrt{2})$.
The point-slope form of the equation of such a tangent line $(\overleftrightarrow{A'D'})$ is thus
$\begin{align*}
\frac{y-y_T}{x-x_T} &= m_{AD} \\
\color{blue}{\frac{y - 1/\sqrt{2}}{x-\sqrt{2}}} &= \color{blue}{-\frac{1}{2}.}
\end{align*}$
An equivalent equation (slope-intercept form) is
$\displaystyle \color{blue}{y = -\frac{x}{2} + \sqrt{2}}$.
b)
Note that $\displaystyle \ln 2 = \int_1^2 1/x\ dx$ is the area under the curve $y=1/x,x\gt 0$ over the closed interval $[1,2]$. The area under the tangent line $(\overleftrightarrow{A'D'})$ that was obtained part a) over this same closed interval is the area of a trapezoid $A'BCD'$ (just a bit smaller than the trapezoid $ABCD$) and is totally contained in the area that represents $\ln 2$ so that the latter area is greater than the former area. That is,
$\begin{align*}
\text{Area under}\ y=1/x, x\gt 0,\ \text{over}\ [1,2] &\gt \text{Area under tangent line}\ \overleftrightarrow{A'D'} \\
& \qquad\quad(\text{area of trapezoid}\ A'BCD') \\
\int_1^2 1/x\ dx &\gt \int_1^2 \left(-\frac{x}{2} + \sqrt{2}\right)\ dx \\
\ln 2 &\gt \left( -\frac{x^2}{4} + \sqrt{2}x\ \right]_1^2 \\
&= \left( -\frac{2^2}{4} + \sqrt{2}(2)\right) - \left( -\frac{1}{4} + \sqrt{2}(1)\right) \\
&= (-1 + 2\sqrt{2}) - (-1/4 + \sqrt{2}) \\
&= (-1+1/4) +2\sqrt{2} -\sqrt{2} \\
&= -3/4 + \sqrt{2} \\
&\approx -.75 + 1.414 \\
&= 0.664 \\
\ln 2 &\gt 0.664\; \gt\; 0.66 \\
\color{blue}{\ln2}\ &\color{blue}{\gt 0.66.}
\end{align*}$
