1. The problem is to create a sample C1 (Coordinate Geometry) question for Sec 3 math midterm in Montreal. 2. A typical C1 problem involves finding the equation of a line, distance between points, or midpoint. 3. Let's create a problem: Find the equation of the line passing through points $A(2,3)$ and $B(5,11)$. 4. Formula for slope $m$ of line through points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$m=\frac{y_2 - y_1}{x_2 - x_1}$$ 5. Calculate slope: $$m=\frac{11 - 3}{5 - 2}=\frac{8}{3}$$ 6. Use point-slope form of line equation: $$y - y_1 = m(x - x_1)$$ 7. Substitute $m=\frac{8}{3}$ and point $A(2,3)$: $$y - 3 = \frac{8}{3}(x - 2)$$ 8. Simplify: $$y - 3 = \frac{8}{3}x - \frac{16}{3}$$ 9. Add 3 to both sides: $$y = \frac{8}{3}x - \frac{16}{3} + 3$$ 10. Convert 3 to fraction with denominator 3: $$3 = \frac{9}{3}$$ 11. Final equation: $$y = \frac{8}{3}x - \frac{16}{3} + \frac{9}{3} = \frac{8}{3}x - \frac{7}{3}$$ 12. So, the equation of the line is: $$y = \frac{8}{3}x - \frac{7}{3}$$ This is a typical C1 question for Sec 3 math midterm.