1. **State the problem:** We need to find the coordinates of the midpoint of points P and Q, where P and Q are the intersection points of the line $L: x - y = 3$ and the curve $C: 3x^2 - y^2 + xy = 9$. 2. **Express $y$ from the line equation:** From $x - y = 3$, we get $$y = x - 3.$$ 3. **Substitute $y$ into the curve equation:** Replace $y$ in $3x^2 - y^2 + xy = 9$ with $x - 3$: $$3x^2 - (x - 3)^2 + x(x - 3) = 9.$$ 4. **Expand and simplify:** $$(x - 3)^2 = x^2 - 6x + 9,$$ so the equation becomes $$3x^2 - (x^2 - 6x + 9) + x^2 - 3x = 9.$$ Simplify inside: $$3x^2 - x^2 + 6x - 9 + x^2 - 3x = 9,$$ which simplifies to $$(3x^2 - x^2 + x^2) + (6x - 3x) - 9 = 9,$$ $$3x^2 + 3x - 9 = 9.$$ 5. **Bring all terms to one side:** $$3x^2 + 3x - 9 - 9 = 0,$$ $$3x^2 + 3x - 18 = 0.$$ 6. **Divide entire equation by 3:** $$x^2 + x - 6 = 0.$$ 7. **Solve quadratic equation:** Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=1$, $c=-6$: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-6)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}.$$ So, - $x_1 = \frac{-1 + 5}{2} = 2$ - $x_2 = \frac{-1 - 5}{2} = -3$ 8. **Find corresponding $y$ values:** Using $y = x - 3$: - For $x=2$, $y = 2 - 3 = -1$ - For $x=-3$, $y = -3 - 3 = -6$ So, points are $P(2, -1)$ and $Q(-3, -6)$. 9. **Find midpoint coordinates:** Midpoint $M$ has coordinates $$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + (-3)}{2}, \frac{-1 + (-6)}{2} \right) = \left( \frac{-1}{2}, \frac{-7}{2} \right).$$ **Final answer:** The midpoint of $P$ and $Q$ is $$\boxed{\left(-\frac{1}{2}, -\frac{7}{2}\right)}.$$