在下图中，直线 $OP$ 与 $x$ 轴正方向成 $\alpha^\circ$ 的角，使得 $\tan \alpha = 0.2$，且 $Q$ 位于 $x$ 轴上。

1. Express the coordinates of $P$ in terms of a single variable using the given tangent ratio.

Let the coordinates of $P$ be $(x, y)$.
Since $P$ is in the first quadrant, $x > 0$ and $y > 0$.
$\tan \alpha = \frac{y}{x} = 0.2 = \frac{1}{5}$
$x = 5y$

2. Set up an equation using the length of $OP$.

$OP = \sqrt{x^2 + y^2} = \frac{\sqrt{26}}{2}$
$x^2 + y^2 = \left(\frac{\sqrt{26}}{2}\right)^2 = \frac{26}{4} = 6.5$

3. Solve for $x$ and $y$ to show the coordinates of $P$.

Substitute $x = 5y$ into the equation:
$(5y)^2 + y^2 = 6.5$
$25y^2 + y^2 = 6.5$
$26y^2 = 6.5$
$y^2 = \frac{6.5}{26} = 0.25$
Since $y > 0$, $y = \sqrt{0.25} = 0.5$.
$x = 5(0.5) = 2.5$
Therefore, $P = (2.5, 0.5)$.

1. Express the area of triangle OPQ in terms of the coordinates of Q

Let the coordinates of $Q$ be $(q, 0)$.
The base of $\Delta OPQ$ is $OQ = q$.
The perpendicular height of $\Delta OPQ$ from $P(2.5, 0.5)$ to the $x$-axis is $0.5$.
Area of $\Delta OPQ = \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times q \times 0.5 = 0.25q$

2. Equate to the given area and solve for q

$0.25q = 0.65$
$q = \frac{0.65}{0.25} = 2.6$
Therefore, the coordinates of $Q$ are $(2.6, 0)$.

1. Find the coordinates of $Q$ using the area of the triangle given in part (b)

The base of $\Delta OPQ$ is $OQ$ (which lies on the $x$-axis) and its perpendicular height is the $y$-coordinate of $P$, which is $0.5$.
$\text{Area} = \frac{1}{2} \times OQ \times 0.5 = 0.65$
$0.25 OQ = 0.65$
$OQ = 2.6$
Since $Q$ lies on the positive $x$-axis, $Q = (2.6, 0)$.

2. Calculate the gradients of $OP$ and $PQ$

Gradient of $OP$, $m_{OP} = \frac{0.5 - 0}{2.5 - 0} = 0.2$ (or given as $\tan \alpha = 0.2$).
Gradient of $PQ$, $m_{PQ} = \frac{0 - 0.5}{2.6 - 2.5} = \frac{-0.5}{0.1} = -5$.

3. Show that the lines are perpendicular by multiplying their gradients

$m_{OP} \times m_{PQ} = 0.2 \times (-5) = -1$
Since the product of their gradients is $-1$, $OP$ is perpendicular to $PQ$. Therefore, $\angle OPQ = 90^\circ$, making $\Delta OPQ$ a right-angled triangle.

1. Find the coordinates of $Q$ from the area of $\Delta OPQ$

Area of $\Delta OPQ = \frac{1}{2} \times OQ \times y_P = 0.65$
$\frac{1}{2} \times OQ \times 0.5 = 0.65$
$OQ = \frac{0.65}{0.25} = 2.6$
Since $Q$ lies on the positive $x$-axis, $Q = (2.6, 0)$.

2. Use the midpoint property of the diagonals of a rectangle to find $R$

In rectangle $OPQR$, the diagonals $OQ$ and $PR$ bisect each other.
Midpoint of $OQ = \left(\frac{0 + 2.6}{2}, \frac{0 + 0}{2}\right) = (1.3, 0)$
Let $R = (x, y)$.
Midpoint of $PR = \left(\frac{2.5 + x}{2}, \frac{0.5 + y}{2}\right)$
Equating the midpoints:
$\frac{2.5 + x}{2} = 1.3 \implies 2.5 + x = 2.6 \implies x = 0.1$
$\frac{0.5 + y}{2} = 0 \implies 0.5 + y = 0 \implies y = -0.5$
$R = (0.1, -0.5)$