Finally, we evaluate the definite integral of the line equation from $x = 0$ to $x = 7$. This calculates the area under the line segment PQ, which is the area of the triangle POQ. This confirms the standard formula for the area of a right-angled triangle: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 6 = 21$. Area of $\triangle POQ = \frac{6}{7} \int_{0}^{7} (7-x) dx$ To integrate the function $\frac{6}{7}(7-x)$ with respect to $x$, we can first take the constant $\frac{6}{7}$ out of the integral sign. This is a property of integrals: $\int c \cdot f(x) dx = c \cdot \int f(x) dx$, where 'c' is a constant. This simplifies the integration process, allowing us to focus on integrating the polynomial $(7 - x)$. $\frac{6}{7} \left[ 7x - \frac{x^2}{2} \right]_{0}^{7}$ Now we integrate the expression $(7 - x)$ term by term. The integral of a constant, say 'a', with respect to $x$ is $ax$. So, the integral of 7 is $7x$. The integral of $x$ (which is $x^1$) with respect to $x$ is $\frac{x^{1+1}}{1+1} = \frac{x^2}{2}$. (Recall the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$). Since we are dealing with a definite integral, we don't add the constant of integration, $C$. After integrating, we denote the limits of integration from 0 to 7 using the square bracket notation. $\frac{6}{7} \left[ \left( 7(7) - \frac{7^2}{2} \right) - \left( 7(0) - \frac{0^2}{2} \right) \right] = \frac{6}{7} \left[ 49 - \frac{49}{2} - 0 \right] = \frac{6}{7} \left[ \frac{98 - 49}{2} \right] = \frac{6}{7} \left[ \frac{49}{2} \right] = 3 \cdot 7 = 21 \text{ units}^2$ This step involves evaluating the definite integral using the Fundamental Theorem of Calculus. First, we substitute the upper limit $(x = 7)$ into the integrated expression: $7(7) - \frac{7^2}{2} = 49 - \frac{49}{2}$. Next, we substitute the lower limit $(x = 0)$ and get $7(0) - \frac{0^2}{2} = 0$. We then subtract the result of the lower limit substitution from the result of the upper limit substitution: $(49 - \frac{49}{2}) - 0$. To simplify $49 - \frac{49}{2}$, we find a common denominator, which is 2. So, $49 = \frac{98}{2}$. Thus, $\frac{98}{2} - \frac{49}{2} = \frac{49}{2}$. Finally, multiply this result by the constant $\frac{6}{7}$: $\frac{6}{7} \times \frac{49}{2}$. We can simplify this by cancelling out common factors: 7 divides 49 to give 7, and 2 divides 6 to give 3. So, $3 \times 7 = 21$. This gives us the area of the triangle POQ as 21 square units. (definite integral, Fundamental Theorem of Calculus, area, right-angled triangle, power rule for integration)

Answer: The area of $\triangle POQ$ is 21 square units.

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